Explanation:
Okay, so I need to find p(8) where p(x) is a degree 5 polynomial that satisfies p(n) = n/(n² - 1) for n = 2, 3, 4, 5, 6, 7. Hmm, let's think about how to approach this.

First, I remember that if a polynomial of degree d passes through d+1 points, then it's uniquely determined. But here, we have a degree 5 polynomial passing through 6 points (from n=2 to n=7), so that makes sense. But how do we find p(8)?

Maybe I can use some interpolation method? Or perhaps there's a clever trick here. Let me recall that sometimes when dealing with polynomials that take specific values at certain points, we can consider constructing a difference between the polynomial and the given function.

Wait, the given function is n/(n² -1). Let me simplify that. n/(n² -1) can be written as n/[(n-1)(n+1)]. Also, note that this is equivalent to [1/(n-1)] - [1/(n+1)] divided by something? Wait, let me check:

Let me try partial fractions. Let me write n/(n² - 1) as A/(n - 1) + B/(n + 1). Let's solve for A and B.

Multiply both sides by (n - 1)(n + 1):

n = A(n + 1) + B(n - 1)

Expanding:

n = A n + A + B n - B

Grouping terms:

n = (A + B)n + (A - B)

Therefore, equating coefficients:

For n: 1 = A + B

For constant term: 0 = A - B

Solving these equations: From the second equation, A = B. Then substituting into the first equation: 1 = A + A => 2A = 1 => A = 1/2, so B = 1/2.

Therefore, n/(n² -1) = 1/(2(n - 1)) + 1/(2(n + 1)). Wait, but that would be:

Wait, if A and B are both 1/2, then:

1/(2(n - 1)) + 1/(2(n + 1)) = [ (n + 1) + (n - 1) ] / [2(n² - 1)] = [2n]/[2(n² -1)] = n/(n² - 1). Yes, correct. So that's the partial fraction decomposition.

So, p(n) = [1/(2(n - 1)) + 1/(2(n + 1))] for n = 2,3,4,5,6,7.

But p(x) is a polynomial of degree 5, while the right-hand side is not a polynomial—it has denominators. So, how does that help?

Maybe consider the difference between p(x) and that expression. Let me define a function q(x) = p(x) - [1/(2(x - 1)) + 1/(2(x + 1))]. Then, for x = 2, 3, 4, 5, 6, 7, we have q(x) = 0. But q(x) is not a polynomial because of the denominators. So that might not be helpful directly.

Alternatively, maybe I need to consider multiplying both sides by something to make it a polynomial. Let me think.

Let me think about the original equation p(n) = n/(n² - 1) for n = 2,...,7. Let me rearrange this equation:

p(n) * (n² - 1) = n.

So for each n in 2 to 7, p(n) * (n² -1) - n = 0. Let me define another polynomial, say, Q(x) = p(x)(x² -1) - x. Then Q(n) = 0 for n = 2,3,4,5,6,7. So Q(x) has roots at x = 2,3,4,5,6,7. Therefore, Q(x) is a polynomial of degree 5 + 2 = 7? Wait, p(x) is degree 5, multiplied by x² - 1 (degree 2), so degree 7, then subtract x (degree 1), so overall degree 7. Therefore, Q(x) is a degree 7 polynomial with roots at 2 through 7. So Q(x) must be equal to k(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - something). Wait, but there are 6 roots, so maybe it's k(x - 2)...(x - 7)? But that would be 6 roots, so degree 6. However, Q(x) is degree 7. Therefore, there's one more root. Wait, but maybe there's a factor that I'm missing?

Wait, let's check. Let me check the degree again. p(x) is degree 5. Multiply by (x² - 1), so degree 5 + 2 = 7. Then subtract x, which is degree 1, so Q(x) is degree 7. Therefore, Q(x) has degree 7 and has roots at x = 2,3,4,5,6,7. Therefore, it must have another root, say, α. So Q(x) = k(x - 2)(x - 3)...(x - 7)(x - α). But we need to find α? Wait, but how?

Alternatively, maybe Q(x) is divisible by (x - 2)(x - 3)...(x - 7), but since Q(x) is degree 7 and the product is degree 6, then Q(x) = c(x - 2)...(x - 7)(x - α). But how do we find c and α?

Alternatively, maybe there's another approach. Since Q(x) is degree 7, and we know its roots at 2 through 7, but we need one more root. However, maybe there's a way to determine Q(x).

But perhaps there's more information we can use. Let me think. Let me note that Q(x) = p(x)(x² - 1) - x. Since p(x) is a degree 5 polynomial, and Q(x) is a degree 7 polynomial, but we can also note that Q(x) = 0 when x = 2,...,7. So maybe we can write Q(x) as (x - 2)(x - 3)...(x - 7) * something. However, since Q(x) is degree 7, and (x - 2)...(x - 7) is degree 6, then we need to multiply by a linear factor. Let me define Q(x) = (x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(Ax + B). Then, we can find A and B such that Q(x) = p(x)(x² - 1) - x. But how?

Alternatively, maybe we can consider that when x approaches infinity, Q(x) ~ p(x)x² - x ~ x² * x^5 - x = x^7 - x. Therefore, the leading term of Q(x) is x^7. On the other hand, if Q(x) = (x - 2)...(x - 7)(Ax + B), then the leading term is (x^6)(Ax) = A x^7. Therefore, A must be equal to 1. So A = 1. Then Q(x) = (x - 2)...(x - 7)(x + B). Now, we need to find B. How?

To find B, perhaps plug in a specific value of x where we know Q(x). Wait, but we know Q(x) for x = 2,...,7, which are already roots, so they are zero. But maybe plug in x = -1 or something? Let me try x = -1. Let's compute Q(-1). Let's compute Q(-1) = p(-1)( (-1)^2 - 1 ) - (-1) = p(-1)(1 - 1) + 1 = 0 + 1 = 1. On the other hand, Q(-1) = (-1 - 2)(-1 - 3)...(-1 - 7)(-1 + B). Let me compute that.

First, calculate the product: (-1 - 2) = -3, (-1 -3) = -4, (-1 -4) = -5, (-1 -5) = -6, (-1 -6) = -7, (-1 -7) = -8. So the product is (-3)(-4)(-5)(-6)(-7)(-8). Let me compute the sign: there are six negative terms, so (-1)^6 = 1. The absolute value is 3*4*5*6*7*8. Then multiplied by (-1 + B). Therefore, Q(-1) = 3*4*5*6*7*8*(B - 1). And we know Q(-1) = 1. Therefore:

3*4*5*6*7*8*(B - 1) = 1.

Let me compute 3*4*5*6*7*8. Let's compute step by step:

3*4 = 12

12*5 = 60

60*6 = 360

360*7 = 2520

2520*8 = 20160

Therefore, 20160*(B - 1) = 1 => B - 1 = 1/20160 => B = 1 + 1/20160 = 20161/20160.

So, Q(x) = (x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x + 20161/20160).

But maybe we can keep it as B for now.

But perhaps there's another way to find B? Or maybe we can use another value of x?

Alternatively, maybe we can use x = 1? Let's try that. Q(1) = p(1)(1² - 1) - 1 = p(1)(0) - 1 = -1. On the other hand, Q(1) = (1 - 2)(1 - 3)...(1 - 7)(1 + B). Let's compute that.

The product is (1 - 2)(1 - 3)(1 - 4)(1 - 5)(1 - 6)(1 - 7) = (-1)(-2)(-3)(-4)(-5)(-6). Let me compute the sign: there are six negative terms, so (-1)^6 = 1. The absolute value is 1*2*3*4*5*6 = 720. Therefore, Q(1) = 720*(1 + B). And we know Q(1) = -1, so:

720*(1 + B) = -1 => 1 + B = -1/720 => B = -1/720 - 1 = -721/720. Wait, but earlier I had B = 20161/20160? That's a contradiction. Which one is correct?

Wait, this suggests that there is inconsistency, which probably means that my assumption about Q(x) is wrong?

Wait, but how? Because earlier, when I plugged x = -1, I got B = 20161/20160, and when I plug x = 1, I get B = -721/720. These two results for B must be the same? Which is not the case. Therefore, my mistake must be somewhere.

Wait, let me check the value of Q(1). Q(x) = p(x)(x² - 1) - x. When x = 1, p(1)(1 - 1) - 1 = 0 - 1 = -1. That seems right.

But according to my expression Q(x) = (x - 2)...(x - 7)(x + B). So Q(1) = (1 - 2)...(1 - 7)(1 + B). Which is (-1)(-2)(-3)(-4)(-5)(-6). Let me compute that again:

Each term is 1 - n for n from 2 to 7:

1 - 2 = -1

1 - 3 = -2

1 - 4 = -3

1 - 5 = -4

1 - 6 = -5

1 - 7 = -6

Therefore, the product is (-1)(-2)(-3)(-4)(-5)(-6). There are six negative terms, so the product is positive. Let me compute the absolute value:

1*2*3*4*5*6 = 720. Therefore, the product is 720. Then Q(1) = 720*(1 + B) = -1, so 1 + B = -1/720, so B = -1/720 -1 = -(1 + 1/720) = -721/720. So that is correct.

But earlier, when I used x = -1, I found B = 20161/20160. These two values for B are different. That must mean that my assumption that Q(x) = (x - 2)...(x - 7)(x + B) is invalid. So what's wrong?

Wait, perhaps Q(x) isn't just that? Wait, Q(x) is a degree 7 polynomial with roots at x = 2, ..., 7. Therefore, it must be equal to (x - 2)...(x - 7)(Ax + B). However, when I tried x = -1 and x = 1, I get different values of B, which is a contradiction. Therefore, my mistake must be in assuming that Q(x) can be written as (x - 2)...(x - 7)(Ax + B). But why?

Wait, maybe Q(x) is not just that? Wait, perhaps there is another root? Wait, no. If Q(x) is degree 7, and has roots at x = 2, ..., 7, which is 6 roots, so it must have one more root. Therefore, it can be written as (x - 2)...(x - 7)(Ax + B). However, the problem arises when trying to find A and B via different x values. So maybe my mistake is in the calculation of Q(-1)?

Let me check again. Q(-1) = p(-1)( (-1)^2 - 1 ) - (-1) = p(-1)(1 - 1) + 1 = 0 + 1 = 1. Correct.

On the other hand, Q(-1) is equal to (-1 - 2)(-1 - 3)...(-1 - 7)(A*(-1) + B). Let me compute the product again:

(-1 - 2) = -3

(-1 - 3) = -4

(-1 - 4) = -5

(-1 - 5) = -6

(-1 - 6) = -7

(-1 - 7) = -8

So the product is (-3)(-4)(-5)(-6)(-7)(-8). Let me compute that. There are six negative numbers, so the product is positive. The absolute value is 3*4*5*6*7*8 = 20160 as before. Therefore, Q(-1) = 20160*(-A + B) = 1. Wait, hold on: the linear factor is A*(-1) + B = -A + B. Therefore, 20160*(-A + B) = 1. Wait, earlier I thought the linear factor was (x + B), but actually, if Q(x) = (x - 2)...(x - 7)(Ax + B), then at x = -1, it is (product)*(A*(-1) + B). Therefore, my previous mistake was in the earlier step when I thought it was (B - 1). Actually, it's (-A + B). So, in my previous calculation, I had B - 1, but actually it's (-A + B). Therefore, my previous calculation was wrong. Therefore, I need to correct that.

So, Q(-1) = 20160*(-A + B) = 1.

Also, when we plug x = 1, we had Q(1) = 720*(A*1 + B) = 720*(A + B) = -1.

Therefore, now we have two equations:

20160*(-A + B) = 1

720*(A + B) = -1

Let me write them as:

-20160A + 20160B = 1

720A + 720B = -1

Let me simplify the second equation by dividing by 720:

A + B = -1/720

So the second equation is A + B = -1/720

The first equation can be written as -20160A + 20160B = 1. Let me divide the first equation by 20160:

- A + B = 1/20160

So we have:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

So we can solve this system. Let me add the two equations:

(-A + B) + (A + B) = 1/20160 - 1/720

Simplify:

(-A + A) + (B + B) = 2B = 1/20160 - 1/720

Compute the right-hand side:

Convert 1/720 to denominator 20160. Since 20160 / 720 = 28. Therefore, 1/720 = 28/20160.

Therefore:

2B = 1/20160 - 28/20160 = (-27)/20160 = -9/6720 = -3/2240

Wait, let me check:

Wait, 1 - 28 = -27, over 20160. So -27/20160. Simplify: divide numerator and denominator by 3: -9/6720. Then divide by 3 again: -3/2240. So 2B = -3/2240, so B = -3/(2 * 2240) = -3/4480.

Wait, but let me check:

Wait, 2B = -27/20160. Let me simplify -27/20160:

Divide numerator and denominator by 3: -9/6720

Divide numerator and denominator by 3 again: -3/2240

Divide numerator and denominator by... well, 3 and 2240 are coprime? 2240 divided by 3? No. So B = -3/(2*2240) is not quite. Wait, 2B = -27/20160 => B = -27/(2*20160) = -9/(2*6720) = -3/(2*2240). Yes, so B = -3/(4480).

But let me compute -27 divided by 20160:

-27/20160 = -9/6720 = -3/2240 ≈ -0.001339...

But let me also check equation 2: A + B = -1/720 ≈ -0.001388...

So if B = -3/2240 ≈ -0.001339, then A = -1/720 - B ≈ -0.001388 - (-0.001339) ≈ -0.000049. Which is small. Let me compute A:

From equation 2: A = -1/720 - B

We have B = -3/2240, so:

A = -1/720 - (-3/2240) = -1/720 + 3/2240

Find common denominator. Let's compute 720 and 2240.

Prime factors:

720 = 2^4 * 3^2 * 5

2240 = 2^6 * 5 * 7

Common denominator would be LCM of 720 and 2240.

Factors needed: 2^6, 3^2, 5, 7. Therefore:

2^6 * 3^2 * 5 * 7 = 64 * 9 * 5 * 7 = 64 * 9 * 35 = 64 * 315 = 20160

Therefore, convert -1/720 to -28/20160 (since 20160 / 720 = 28)

Convert 3/2240 to 3 * (20160 / 2240) / 20160. Compute 20160 / 2240 = 9. Therefore, 3/2240 = 27/20160.

Therefore, A = -28/20160 + 27/20160 = (-28 + 27)/20160 = -1/20160.

So A = -1/20160 and B = -3/2240? Wait, no. Wait, earlier we had B = -3/2240? Wait, no. Wait, from solving 2B = -27/20160, so B = -27/(2*20160) = -9/(2*6720) = -3/(2*2240) = -3/4480. Wait, but let me check that again:

From earlier steps:

We had 2B = -27/20160 => B = -27/(2*20160) = -27/40320 = -9/13440 = -3/4480. Yes, so B = -3/4480.

Wait, but then A is -1/720 - B = (-1/720) - (-3/4480). Let me calculate:

Convert to common denominator 20160 again:

-1/720 = -28/20160

-3/4480 = Let me compute 20160 / 4480 = 4.5? Wait 4480 * 4 = 17920, 4480*4.5= 4480*4 + 4480*0.5 = 17920 + 2240 = 20160. Yes! Therefore, -3/4480 = - (3 * 4.5)/20160 = -13.5/20160. Wait, but fractions should be integers. Wait, perhaps I need to do it properly:

Let me write -3/4480 as x/20160. So x = -3 * (20160 / 4480) = -3 * 4.5 = -13.5. Hmm, which is not an integer. So this suggests that maybe I made a mistake in calculation earlier.

Wait, but how? Let me re-express:

From equation 2: A + B = -1/720

From equation 1: -A + B = 1/20160

Let me solve these equations again:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

Let me add equations 1 and 2:

(-A + B) + (A + B) = 1/20160 - 1/720

As before, left side is 2B.

Right side: 1/20160 - 1/720.

Convert 1/720 to denominator 20160:

Multiply numerator and denominator by 28: 1/720 = 28/20160.

So right side is 1/20160 - 28/20160 = (-27)/20160 = -9/6720 = -3/2240. Therefore:

2B = -3/2240 => B = -3/(2*2240) = -3/4480. So that's correct. Then equation 2:

A = -1/720 - B = -1/720 - (-3/4480) = -1/720 + 3/4480.

Convert to decimals to check:

-1/720 ≈ -0.00138888...

3/4480 ≈ 0.00066964...

Adding them: approx -0.00138888 + 0.00066964 ≈ -0.00071924. Which is approx -0.00072. Let me check if this equals to -1/20160? Wait, no, that was A?

Wait, no. Wait, A was found from equation 1 and 2? Wait, we had:

Equation 1: -A + B = 1/20160. So once we have B, we can find A:

From equation 1: -A + B = 1/20160 => -A = 1/20160 - B => A = B - 1/20160.

Since B = -3/4480:

A = -3/4480 - 1/20160.

Convert to common denominator 20160:

-3/4480 = - (3 * 4.5)/20160 = -13.5/20160.

Wait, but fractions need to be integers. So let me convert -3/4480:

Multiply numerator and denominator by 4.5 to get denominator 20160? Wait, 4480 * 4.5 = 20160. So yes:

-3/4480 = - (3 * 4.5)/20160 = -13.5 / 20160. But that's not a fraction with integer numerator. So maybe I made a mistake in the calculation?

Alternatively, let me compute:

Let me find A:

A = B - 1/20160 = (-3/4480) - (1/20160). Let me convert -3/4480 to denominator 20160:

Since 20160 / 4480 = 4.5, so multiply numerator and denominator by 4.5:

-3 * 4.5 = -13.5 over 20160. Therefore:

A = (-13.5 - 1)/20160 = -14.5 / 20160 = -29/40320. Wait, is that correct?

Yes, because -14.5/20160 is equal to -29/40320.

Therefore, A = -29/40320 and B = -3/4480.

But this seems messy, but maybe manageable. So now, Q(x) = (x - 2)(x - 3)...(x - 7)(Ax + B) where A = -29/40320 and B = -3/4480.

But maybe there's a better way to write this. Alternatively, maybe we can keep these fractions and use them in our calculations. However, this seems complicated. But perhaps we can proceed.

Now, the original goal is to find p(8). Recall that Q(x) = p(x)(x² - 1) - x. Therefore, solving for p(x):

p(x) = [Q(x) + x]/(x² - 1)

Therefore, to find p(8), we can compute:

p(8) = [Q(8) + 8]/(8² - 1) = [Q(8) + 8]/(64 - 1) = [Q(8) + 8]/63.

Therefore, we need to compute Q(8). Since Q(x) = (x - 2)...(x - 7)(Ax + B), so:

Q(8) = (8 - 2)(8 - 3)(8 - 4)(8 - 5)(8 - 6)(8 - 7)(A*8 + B)

Compute each term:

(8 - 2) = 6

(8 - 3) = 5

(8 - 4) = 4

(8 - 5) = 3

(8 - 6) = 2

(8 - 7) = 1

Therefore, the product is 6*5*4*3*2*1 = 6! = 720.

Then, Q(8) = 720*(A*8 + B). Therefore:

p(8) = [720*(8A + B) + 8]/63.

So we need to compute 8A + B.

From earlier, A = -29/40320 and B = -3/4480. Let me compute 8A + B:

First, compute 8A:

8 * (-29/40320) = -232/40320 = simplify. Divide numerator and denominator by 8: -29/5040.

Then, B = -3/4480.

Therefore, 8A + B = -29/5040 - 3/4480.

Find common denominator for 5040 and 4480. Let me compute LCM of 5040 and 4480.

Factorize 5040: 5040 = 7! = 2^4 * 3^2 * 5 * 7

Factorize 4480: 4480 = 448 * 10 = 2^6 * 5 * 7

Therefore, LCM is max exponents: 2^6, 3^2, 5, 7. So LCM = 64 * 9 * 5 * 7 = 64 * 315 = 20160.

Convert -29/5040 to denominator 20160: multiply numerator and denominator by 4 (since 5040 * 4 = 20160). So:

-29/5040 = -29*4 / 20160 = -116/20160.

Convert -3/4480 to denominator 20160: 4480 * 4.5 = 20160. Therefore, multiply numerator and denominator by 4.5:

-3/4480 = -3 * 4.5 / 20160 = -13.5 / 20160. But fractions with decimal points are not ideal. Let me do it properly:

Let me write -3/4480 as x/20160:

x = -3 * (20160 / 4480) = -3 * 4.5 = -13.5. Which is same as above.

But to write this as a fraction:

-13.5 = -27/2. Therefore, -13.5 / 20160 = -27/(2*20160) = -27/40320.

Therefore, 8A + B = -116/20160 -27/40320. Let me convert both terms to denominator 40320:

-116/20160 = -232/40320

-27/40320 remains as is.

Therefore, total is (-232 - 27)/40320 = -259/40320.

Therefore, Q(8) = 720 * (-259/40320) = Let me compute that:

720 divided by 40320 = 720 / 40320 = 1 / 56.

Therefore, 720 * (-259)/40320 = (-259)/56.

Therefore, p(8) = [ (-259)/56 + 8 ] / 63.

Compute (-259)/56 + 8:

Convert 8 to 8 * 56/56 = 448/56.

Therefore, (-259 + 448)/56 = 189/56.

Therefore, p(8) = (189/56)/63 = (189 / 56) * (1 / 63) = (189 / (56 * 63)).

Simplify numerator and denominator:

189 ÷ 63 = 3. So 3 / 56.

Therefore, p(8) = 3/56.

Wait, is that correct? Let me check my calculations again.

Let me confirm step-by-step:

First, Q(8) = 720*(8A + B) = 720*(-259/40320) = (720 / 40320)*(-259) = (1/56)*(-259) = -259/56.

Then, p(8) = [ -259/56 + 8 ] / 63. 8 is 448/56.

So -259/56 + 448/56 = (448 - 259)/56 = 189/56.

Then divide by 63: (189/56)/63 = (189)/(56*63). Simplify:

Divide numerator and denominator by 63: numerator becomes 189/63 = 3; denominator becomes 56. So 3/56. Yes.

Therefore, p(8) = 3/56. So is that the answer? Wait, that seems possible. But let me verify if this answer makes sense.

Alternatively, maybe there's a simpler way to reach this result, or maybe I made an error in calculation. Let me check with another method.

Alternatively, consider that since p(x) is a degree 5 polynomial, and we have Q(x) = p(x)(x² - 1) - x = (x - 2)...(x - 7)(Ax + B). Therefore, when x = 8, we can compute Q(8) = p(8)(8² -1) - 8 = p(8)*63 - 8. Therefore, p(8) = [Q(8) + 8]/63, which is what I did. And Q(8) = 720*(8A + B) = -259/56. Therefore, p(8) = (-259/56 + 8)/63 = 3/56. So that's correct.

But let me check with another approach. Maybe using finite differences or something else?

Alternatively, maybe using the concept of constructing the polynomial p(x) such that p(x) - [n/(n² - 1)] has roots at 2,...,7. But that seems similar to what I did before.

Alternatively, since we have a polynomial p(x) of degree 5 passing through those points, and we need to find p(8). Maybe we can use Lagrange interpolation? Although Lagrange interpolation for 6 points would be tedious, but maybe there's a pattern.

Alternatively, note that the function n/(n² - 1) can be written as [1/(2(n - 1))] + [1/(2(n + 1))], as we did before. Therefore, maybe consider that p(x) - [1/(2(x - 1)) + 1/(2(x + 1))] has roots at x = 2,...,7. However, since p(x) is a polynomial and the other terms are not, we need to multiply by (x - 1)(x + 1) to eliminate denominators. Let me try this.

Let me define R(x) = (x - 1)(x + 1)p(x) - [ (x + 1)/2 + (x - 1)/2 ].

Wait, let's see:

Original equation: p(x) = [1/(2(x - 1)) + 1/(2(x + 1))] for x = 2,...,7.

Multiply both sides by (x - 1)(x + 1):

(x² - 1)p(x) = [ (x + 1)/2 + (x - 1)/2 ] = [ (x + 1 + x -1)/2 ] = (2x)/2 = x.

Therefore, (x² - 1)p(x) - x = 0 for x = 2,...,7, which is exactly the Q(x) we previously defined. Therefore, that gives us the same equation. So we are back to the same place.

Therefore, our previous calculation seems correct. Therefore, the answer is 3/56. But just to confirm, maybe plug in a value.

Alternatively, maybe consider that p(x) is a degree 5 polynomial. Suppose we define another polynomial, say, p(x) = (x² - 1) * something + [n/(n² -1)] ?

Wait, not sure. Alternatively, think about considering the difference between p(x) and the function f(x) = x/(x² - 1). Since p(x) agrees with f(x) at x=2,...,7, then the difference p(x) - f(x) has zeros at those points. However, since p(x) is a polynomial and f(x) is not, this difference is not a polynomial. However, if we consider multiplying by (x² -1), as before, we get a polynomial.

Therefore, given that our previous steps are correct, then the answer is 3/56. Wait, but 3/56 seems a bit strange. Let me check with x=1. Wait, if x=1, then p(1) is undefined, but maybe we can compute Q(1) as before?

Earlier, we had Q(1) = -1, and also Q(1) = (1 - 2)...(1 - 7)(A*1 + B) = 720*(A + B) = -1. Which gave us the equation A + B = -1/720. Which we used. So that seems correct.

Another check: Suppose we take x = 0. Then Q(0) = p(0)*(0 -1) - 0 = -p(0). On the other hand, Q(0) = (0 -2)(0 -3)...(0 -7)(A*0 + B) = (-2)(-3)(-4)(-5)(-6)(-7)*B. Which is [(-2)(-3)(-4)(-5)(-6)(-7)]*B. The product is (-2)(-3)=6; (-4)(-5)=20; (-6)(-7)=42. Then 6*20=120; 120*42=5040. Therefore, Q(0) = 5040*B. Therefore, -p(0) = 5040*B => p(0) = -5040*B. Since B = -3/4480, then p(0) = -5040*(-3/4480) = (5040 / 4480)*3 = (5040 / 4480) *3. Simplify 5040 / 4480 = 504 / 448 = 63 / 56 = 9/8. Therefore, p(0) = (9/8)*3 = 27/8. Not sure if that helps, but seems consistent.

Alternatively, maybe plug in x = 8 into the original equation? Wait, not applicable. But since we have p(8) = 3/56, let's check if this is correct.

Alternatively, maybe consider that the answer is 3/56, and that seems plausible. But to be thorough, let me check with x=8 in the Q(x) equation.

Wait, we had Q(x) = p(x)(x² - 1) - x. So if p(8) = 3/56, then Q(8) = (3/56)(64 - 1) - 8 = (3/56)(63) - 8 = (189/56) - 8 = (189/56) - (448/56) = (-259)/56. Which matches our previous result of Q(8) = -259/56. So that checks out. Therefore, seems correct.

Therefore, I think the answer is 3/56.

But let me check for possible miscalculations when solving for A and B. Earlier, when solving for A and B, we had:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

Solving, we found B = -3/4480 and A = -29/40320. Let me verify these values:

From Equation 2: A + B = -1/720

Let me compute A + B = (-29/40320) + (-3/4480). Convert -3/4480 to denominator 40320:

Multiply numerator and denominator by 9: -3*9 / 4480*9 = -27/40320. Therefore, A + B = -29/40320 -27/40320 = -56/40320 = -1/720. Correct.

From Equation 1: -A + B = 1/20160. Compute -A + B = -(-29/40320) + (-3/4480) = 29/40320 - 3/4480 = convert -3/4480 to -27/40320. So total is 29/40320 -27/40320 = 2/40320 = 1/20160. Correct. Therefore, these values for A and B are correct.

Therefore, the calculation of Q(8) is correct, and hence p(8) is indeed 3/56.

Therefore, the answer is 3/56, so boxed answer is \boxed{\dfrac{3}{56}}.

**Final Answer**
\boxed{\dfrac{3}{56}}

To solve the problem, we are given a polynomial $ p(x) $ of degree 5 such that:

$$
p(n) = \frac{n}{n^2 - 1} \quad \text{for } n = 2, 3, 4, 5, 6, 7
$$

Our goal is to find $ p(8) $. Here’s how we proceed:

---

### Step 1: Understand the Structure of the Given Function

Let’s note that the function

$$
f(n) = \frac{n}{n^2 - 1}
$$

can be simplified using **partial fractions**:

$$
\frac{n}{n^2 - 1} = \frac{n}{(n - 1)(n + 1)} = \frac{1}{2(n - 1)} + \frac{1}{2(n + 1)}
$$

This decomposition helps simplify the structure of the function, though it’s not a polynomial.

---

### Step 2: Define a Related Polynomial

We define a polynomial function $ Q(x) $ such that:

$$
Q(x) = p(x)(x^2 - 1) - x
$$

This is a degree 7 polynomial since $ p(x) $ is degree 5 and $ x^2 - 1 $ is degree 2, so their product is degree 7. Subtracting $ x $ (degree 1) doesn’t change the degree.

Now, consider that for $ x = 2, 3, 4, 5, 6, 7 $, we have:

$$
Q(n) = p(n)(n^2 - 1) - n = 0
$$

So $ Q(x) $ has roots at $ x = 2, 3, 4, 5, 6, 7 $. Thus, we can write:

$$
Q(x) = k(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(Ax + B)
$$

Where $ A $ and $ B $ are constants to be determined.

---

### Step 3: Determine the Constants $ A $ and $ B $

We use the condition $ Q(x) = p(x)(x^2 - 1) - x $, and then plug in values of $ x $ to find $ A $ and $ B $.

Using values at $ x = -1 $ and $ x = 1 $, where $ Q(x) $ is known:

- At $ x = 1 $: $ Q(1) = -1 $, since $ p(1)(1^2 - 1) - 1 = -1 $
- At $ x = -1 $: $ Q(-1) = p(-1)(-1^2 - 1) - (-1) = p(-1)(0) + 1 = 1 $

Compute:

At $ x = 1 $:

$$
Q(1) = (1 - 2)(1 - 3)(1 - 4)(1 - 5)(1 - 6)(1 - 7)(A(1) + B) = (-1)(-2)(-3)(-4)(-5)(-6)(A + B)
$$

$$
= 720(A + B) = -1 \Rightarrow A + B = -\frac{1}{720}
$$

At $ x = -1 $:

$$
Q(-1) = (-1 - 2)(-1 - 3)(-1 - 4)(-1 - 5)(-1 - 6)(-1 - 7)(A(-1) + B) = (-3)(-4)(-5)(-6)(-7)(-8)(-A + B)
$$

$$
= 20160(-A + B) = 1 \Rightarrow -A + B = \frac{1}{20160}
$$

Now solve the system:

$$
\begin{cases}
A + B = -\frac{1}{720} \\
-A + B = \frac{1}{20160}
\end{cases}
$$

Add the two equations:

$$
2B = -\frac{1}{720} + \frac{1}{20160} = \frac{-28 + 1}{20160} = -\frac{27}{20160}
\Rightarrow B = -\frac{27}{40320}
$$

Substitute back to get:

$$
A = -\frac{1}{720} - B = -\frac{1}{720} + \frac{27}{40320} = -\frac{56}{40320} + \frac{27}{40320} = -\frac{29}{40320}
$$

---

### Step 4: Use This to Find $ p(8) $

Recall:

$$
Q(x) = p(x)(x^2 - 1) - x \Rightarrow p(x) = \frac{Q(x) + x}{x^2 - 1}
$$

Thus:

$$
p(8) = \frac{Q(8) + 8}{8^2 - 1} = \frac{Q(8) + 8}{63}
$$

Compute $ Q(8) $:

$$
Q(8) = (8 - 2)(8 - 3)(8 - 4)(8 - 5)(8 - 6)(8 - 7)(A \cdot 8 + B)
= 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot (A \cdot 8 + B)
= 720 \cdot (8A + B)
$$

Plug in values:

$$
8A + B = 8 \cdot \left(-\frac{29}{40320}\right) + \left(-\frac{27}{40320}\right)
= \frac{-232 - 27}{40320} = \frac{-259}{40320}
$$

So:

$$
Q(8) = 720 \cdot \left(\frac{-259}{40320}\right) = \frac{-259}{56}
$$

Now compute:

$$
p(8) = \frac{-259/56 + 8}{63} = \frac{-259 + 448}{56 \cdot 63} = \frac{189}{56 \cdot 63}
$$

Simplify:

$$
p(8) = \frac{189}{3528} = \frac{3}{56}
$$

---

### ✅ Final Answer

$$
\boxed{\dfrac{3}{56}}
$$

[Assistant]
Explanation:
Okay, so I need to find p(8) where p(x) is a degree 5 polynomial that satisfies p(n) = n/(n² - 1) for n = 2, 3, 4, 5, 6, 7. Hmm, let's think about how to approach this.

First, I remember that if a polynomial of degree d passes through d+1 points, then it's uniquely determined. But here, we have a degree 5 polynomial passing through 6 points (from n=2 to n=7), so that makes sense. But how do we find p(8)?

Maybe I can use some interpolation method? Or perhaps there's a clever trick here. Let me recall that sometimes when dealing with polynomials that take specific values at certain points, we can consider constructing a difference between the polynomial and the given function.

Wait, the given function is n/(n² -1). Let me simplify that. n/(n² -1) can be written as n/[(n-1)(n+1)]. Also, note that this is equivalent to [1/(n-1)] - [1/(n+1)] divided by something? Wait, let me check:

Let me try partial fractions. Let me write n/(n² - 1) as A/(n - 1) + B/(n + 1). Let's solve for A and B.

Multiply both sides by (n - 1)(n + 1):

n = A(n + 1) + B(n - 1)

Expanding:

n = A n + A + B n - B

Grouping terms:

n = (A + B)n + (A - B)

Therefore, equating coefficients:

For n: 1 = A + B

For constant term: 0 = A - B

Solving these equations: From the second equation, A = B. Then substituting into the first equation: 1 = A + A => 2A = 1 => A = 1/2, so B = 1/2.

Therefore, n/(n² -1) = 1/(2(n - 1)) + 1/(2(n + 1)). Wait, but that would be:

Wait, if A and B are both 1/2, then:

1/(2(n - 1)) + 1/(2(n + 1)) = [ (n + 1) + (n - 1) ] / [2(n² - 1)] = [2n]/[2(n² -1)] = n/(n² - 1). Yes, correct. So that's the partial fraction decomposition.

So, p(n) = [1/(2(n - 1)) + 1/(2(n + 1))] for n = 2,3,4,5,6,7.

But p(x) is a polynomial of degree 5, while the right-hand side is not a polynomial—it has denominators. So, how does that help?

Maybe consider the difference between p(x) and that expression. Let me define a function q(x) = p(x) - [1/(2(x - 1)) + 1/(2(x + 1))]. Then, for x = 2, 3, 4, 5, 6, 7, we have q(x) = 0. But q(x) is not a polynomial because of the denominators. So that might not be helpful directly.

Alternatively, maybe I need to consider multiplying both sides by something to make it a polynomial. Let me think.

Let me think about the original equation p(n) = n/(n² - 1) for n = 2,...,7. Let me rearrange this equation:

p(n) * (n² - 1) = n.

So for each n in 2 to 7, p(n) * (n² -1) - n = 0. Let me define another polynomial, say, Q(x) = p(x)(x² -1) - x. Then Q(n) = 0 for n = 2,3,4,5,6,7. So Q(x) has roots at x = 2,3,4,5,6,7. Therefore, Q(x) is a polynomial of degree 5 + 2 = 7? Wait, p(x) is degree 5, multiplied by x² - 1 (degree 2), so degree 7, then subtract x (degree 1), so overall degree 7. Therefore, Q(x) is a degree 7 polynomial with roots at 2 through 7. So Q(x) must be equal to k(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - something). Wait, but there are 6 roots, so maybe it's k(x - 2)...(x - 7)? But that would be 6 roots, so degree 6. However, Q(x) is degree 7. Therefore, there's one more root. Wait, but maybe there's a factor that I'm missing?

Wait, let's check. Let me check the degree again. p(x) is degree 5. Multiply by (x² - 1), so degree 5 + 2 = 7. Then subtract x, which is degree 1, so Q(x) is degree 7. Therefore, Q(x) has degree 7 and has roots at x = 2,3,4,5,6,7. Therefore, it must have another root, say, α. So Q(x) = k(x - 2)(x - 3)...(x - 7)(x - α). But we need to find α? Wait, but how?

Alternatively, maybe Q(x) is divisible by (x - 2)(x - 3)...(x - 7), but since Q(x) is degree 7 and the product is degree 6, then Q(x) = c(x - 2)...(x - 7)(x - α). But how do we find c and α?

Alternatively, maybe there's another approach. Since Q(x) is degree 7, and we know its roots at 2 through 7, but we need one more root. However, maybe there's a way to determine Q(x).

But perhaps there's more information we can use. Let me think. Let me note that Q(x) = p(x)(x² - 1) - x. Since p(x) is a degree 5 polynomial, and Q(x) is a degree 7 polynomial, but we can also note that Q(x) = 0 when x = 2,...,7. So maybe we can write Q(x) as (x - 2)(x - 3)...(x - 7) * something. However, since Q(x) is degree 7, and (x - 2)...(x - 7) is degree 6, then we need to multiply by a linear factor. Let me define Q(x) = (x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(Ax + B). Then, we can find A and B such that Q(x) = p(x)(x² - 1) - x. But how?

Alternatively, maybe we can consider that when x approaches infinity, Q(x) ~ p(x)x² - x ~ x² * x^5 - x = x^7 - x. Therefore, the leading term of Q(x) is x^7. On the other hand, if Q(x) = (x - 2)...(x - 7)(Ax + B), then the leading term is (x^6)(Ax) = A x^7. Therefore, A must be equal to 1. So A = 1. Then Q(x) = (x - 2)...(x - 7)(x + B). Now, we need to find B. How?

To find B, perhaps plug in a specific value of x where we know Q(x). Wait, but we know Q(x) for x = 2,...,7, which are already roots, so they are zero. But maybe plug in x = -1 or something? Let me try x = -1. Let's compute Q(-1). Let's compute Q(-1) = p(-1)( (-1)^2 - 1 ) - (-1) = p(-1)(1 - 1) + 1 = 0 + 1 = 1. On the other hand, Q(-1) = (-1 - 2)(-1 - 3)...(-1 - 7)(-1 + B). Let me compute that.

First, calculate the product: (-1 - 2) = -3, (-1 -3) = -4, (-1 -4) = -5, (-1 -5) = -6, (-1 -6) = -7, (-1 -7) = -8. So the product is (-3)(-4)(-5)(-6)(-7)(-8). Let me compute the sign: there are six negative terms, so (-1)^6 = 1. The absolute value is 3*4*5*6*7*8. Then multiplied by (-1 + B). Therefore, Q(-1) = 3*4*5*6*7*8*(B - 1). And we know Q(-1) = 1. Therefore:

3*4*5*6*7*8*(B - 1) = 1.

Let me compute 3*4*5*6*7*8. Let's compute step by step:

3*4 = 12

12*5 = 60

60*6 = 360

360*7 = 2520

2520*8 = 20160

Therefore, 20160*(B - 1) = 1 => B - 1 = 1/20160 => B = 1 + 1/20160 = 20161/20160.

So, Q(x) = (x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x + 20161/20160).

But maybe we can keep it as B for now.

But perhaps there's another way to find B? Or maybe we can use another value of x?

Alternatively, maybe we can use x = 1? Let's try that. Q(1) = p(1)(1² - 1) - 1 = p(1)(0) - 1 = -1. On the other hand, Q(1) = (1 - 2)(1 - 3)...(1 - 7)(1 + B). Let's compute that.

The product is (1 - 2)(1 - 3)(1 - 4)(1 - 5)(1 - 6)(1 - 7) = (-1)(-2)(-3)(-4)(-5)(-6). Let me compute the sign: there are six negative terms, so (-1)^6 = 1. The absolute value is 1*2*3*4*5*6 = 720. Therefore, Q(1) = 720*(1 + B). And we know Q(1) = -1, so:

720*(1 + B) = -1 => 1 + B = -1/720 => B = -1/720 - 1 = -721/720. Wait, but earlier I had B = 20161/20160? That's a contradiction. Which one is correct?

Wait, this suggests that there is inconsistency, which probably means that my assumption about Q(x) is wrong?

Wait, but how? Because earlier, when I plugged x = -1, I got B = 20161/20160, and when I plug x = 1, I get B = -721/720. These two results for B must be the same? Which is not the case. Therefore, my mistake must be somewhere.

Wait, let me check the value of Q(1). Q(x) = p(x)(x² - 1) - x. When x = 1, p(1)(1 - 1) - 1 = 0 - 1 = -1. That seems right.

But according to my expression Q(x) = (x - 2)...(x - 7)(x + B). So Q(1) = (1 - 2)...(1 - 7)(1 + B). Which is (-1)(-2)(-3)(-4)(-5)(-6). Let me compute that again:

Each term is 1 - n for n from 2 to 7:

1 - 2 = -1

1 - 3 = -2

1 - 4 = -3

1 - 5 = -4

1 - 6 = -5

1 - 7 = -6

Therefore, the product is (-1)(-2)(-3)(-4)(-5)(-6). There are six negative terms, so the product is positive. Let me compute the absolute value:

1*2*3*4*5*6 = 720. Therefore, the product is 720. Then Q(1) = 720*(1 + B) = -1, so 1 + B = -1/720, so B = -1/720 -1 = -(1 + 1/720) = -721/720. So that is correct.

But earlier, when I used x = -1, I found B = 20161/20160. These two values for B are different. That must mean that my assumption that Q(x) = (x - 2)...(x - 7)(x + B) is invalid. So what's wrong?

Wait, perhaps Q(x) isn't just that? Wait, Q(x) is a degree 7 polynomial with roots at x = 2, ..., 7. Therefore, it must be equal to (x - 2)...(x - 7)(Ax + B). However, when I tried x = -1 and x = 1, I get different values of B, which is a contradiction. Therefore, my mistake must be in assuming that Q(x) can be written as (x - 2)...(x - 7)(Ax + B). But why?

Wait, maybe Q(x) is not just that? Wait, perhaps there is another root? Wait, no. If Q(x) is degree 7, and has roots at x = 2, ..., 7, which is 6 roots, so it must have one more root. Therefore, it can be written as (x - 2)...(x - 7)(Ax + B). However, the problem arises when trying to find A and B via different x values. So maybe my mistake is in the calculation of Q(-1)?

Let me check again. Q(-1) = p(-1)( (-1)^2 - 1 ) - (-1) = p(-1)(1 - 1) + 1 = 0 + 1 = 1. Correct.

On the other hand, Q(-1) is equal to (-1 - 2)(-1 - 3)...(-1 - 7)(A*(-1) + B). Let me compute the product again:

(-1 - 2) = -3

(-1 - 3) = -4

(-1 - 4) = -5

(-1 - 5) = -6

(-1 - 6) = -7

(-1 - 7) = -8

So the product is (-3)(-4)(-5)(-6)(-7)(-8). Let me compute that. There are six negative numbers, so the product is positive. The absolute value is 3*4*5*6*7*8 = 20160 as before. Therefore, Q(-1) = 20160*(-A + B) = 1. Wait, hold on: the linear factor is A*(-1) + B = -A + B. Therefore, 20160*(-A + B) = 1. Wait, earlier I thought the linear factor was (x + B), but actually, if Q(x) = (x - 2)...(x - 7)(Ax + B), then at x = -1, it is (product)*(A*(-1) + B). Therefore, my previous mistake was in the earlier step when I thought it was (B - 1). Actually, it's (-A + B). So, in my previous calculation, I had B - 1, but actually it's (-A + B). Therefore, my previous calculation was wrong. Therefore, I need to correct that.

So, Q(-1) = 20160*(-A + B) = 1.

Also, when we plug x = 1, we had Q(1) = 720*(A*1 + B) = 720*(A + B) = -1.

Therefore, now we have two equations:

20160*(-A + B) = 1

720*(A + B) = -1

Let me write them as:

-20160A + 20160B = 1

720A + 720B = -1

Let me simplify the second equation by dividing by 720:

A + B = -1/720

So the second equation is A + B = -1/720

The first equation can be written as -20160A + 20160B = 1. Let me divide the first equation by 20160:

- A + B = 1/20160

So we have:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

So we can solve this system. Let me add the two equations:

(-A + B) + (A + B) = 1/20160 - 1/720

Simplify:

(-A + A) + (B + B) = 2B = 1/20160 - 1/720

Compute the right-hand side:

Convert 1/720 to denominator 20160. Since 20160 / 720 = 28. Therefore, 1/720 = 28/20160.

Therefore:

2B = 1/20160 - 28/20160 = (-27)/20160 = -9/6720 = -3/2240

Wait, let me check:

Wait, 1 - 28 = -27, over 20160. So -27/20160. Simplify: divide numerator and denominator by 3: -9/6720. Then divide by 3 again: -3/2240. So 2B = -3/2240, so B = -3/(2 * 2240) = -3/4480.

Wait, but let me check:

Wait, 2B = -27/20160. Let me simplify -27/20160:

Divide numerator and denominator by 3: -9/6720

Divide numerator and denominator by 3 again: -3/2240

Divide numerator and denominator by... well, 3 and 2240 are coprime? 2240 divided by 3? No. So B = -3/(2*2240) is not quite. Wait, 2B = -27/20160 => B = -27/(2*20160) = -9/(2*6720) = -3/(2*2240). Yes, so B = -3/(4480).

But let me compute -27 divided by 20160:

-27/20160 = -9/6720 = -3/2240 ≈ -0.001339...

But let me also check equation 2: A + B = -1/720 ≈ -0.001388...

So if B = -3/2240 ≈ -0.001339, then A = -1/720 - B ≈ -0.001388 - (-0.001339) ≈ -0.000049. Which is small. Let me compute A:

From equation 2: A = -1/720 - B

We have B = -3/2240, so:

A = -1/720 - (-3/2240) = -1/720 + 3/2240

Find common denominator. Let's compute 720 and 2240.

Prime factors:

720 = 2^4 * 3^2 * 5

2240 = 2^6 * 5 * 7

Common denominator would be LCM of 720 and 2240.

Factors needed: 2^6, 3^2, 5, 7. Therefore:

2^6 * 3^2 * 5 * 7 = 64 * 9 * 5 * 7 = 64 * 9 * 35 = 64 * 315 = 20160

Therefore, convert -1/720 to -28/20160 (since 20160 / 720 = 28)

Convert 3/2240 to 3 * (20160 / 2240) / 20160. Compute 20160 / 2240 = 9. Therefore, 3/2240 = 27/20160.

Therefore, A = -28/20160 + 27/20160 = (-28 + 27)/20160 = -1/20160.

So A = -1/20160 and B = -3/2240? Wait, no. Wait, earlier we had B = -3/2240? Wait, no. Wait, from solving 2B = -27/20160, so B = -27/(2*20160) = -9/(2*6720) = -3/(2*2240) = -3/4480. Wait, but let me check that again:

From earlier steps:

We had 2B = -27/20160 => B = -27/(2*20160) = -27/40320 = -9/13440 = -3/4480. Yes, so B = -3/4480.

Wait, but then A is -1/720 - B = (-1/720) - (-3/4480). Let me calculate:

Convert to common denominator 20160 again:

-1/720 = -28/20160

-3/4480 = Let me compute 20160 / 4480 = 4.5? Wait 4480 * 4 = 17920, 4480*4.5= 4480*4 + 4480*0.5 = 17920 + 2240 = 20160. Yes! Therefore, -3/4480 = - (3 * 4.5)/20160 = -13.5/20160. Wait, but fractions should be integers. Wait, perhaps I need to do it properly:

Let me write -3/4480 as x/20160. So x = -3 * (20160 / 4480) = -3 * 4.5 = -13.5. Hmm, which is not an integer. So this suggests that maybe I made a mistake in calculation earlier.

Wait, but how? Let me re-express:

From equation 2: A + B = -1/720

From equation 1: -A + B = 1/20160

Let me solve these equations again:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

Let me add equations 1 and 2:

(-A + B) + (A + B) = 1/20160 - 1/720

As before, left side is 2B.

Right side: 1/20160 - 1/720.

Convert 1/720 to denominator 20160:

Multiply numerator and denominator by 28: 1/720 = 28/20160.

So right side is 1/20160 - 28/20160 = (-27)/20160 = -9/6720 = -3/2240. Therefore:

2B = -3/2240 => B = -3/(2*2240) = -3/4480. So that's correct. Then equation 2:

A = -1/720 - B = -1/720 - (-3/4480) = -1/720 + 3/4480.

Convert to decimals to check:

-1/720 ≈ -0.00138888...

3/4480 ≈ 0.00066964...

Adding them: approx -0.00138888 + 0.00066964 ≈ -0.00071924. Which is approx -0.00072. Let me check if this equals to -1/20160? Wait, no, that was A?

Wait, no. Wait, A was found from equation 1 and 2? Wait, we had:

Equation 1: -A + B = 1/20160. So once we have B, we can find A:

From equation 1: -A + B = 1/20160 => -A = 1/20160 - B => A = B - 1/20160.

Since B = -3/4480:

A = -3/4480 - 1/20160.

Convert to common denominator 20160:

-3/4480 = - (3 * 4.5)/20160 = -13.5/20160.

Wait, but fractions need to be integers. So let me convert -3/4480:

Multiply numerator and denominator by 4.5 to get denominator 20160? Wait, 4480 * 4.5 = 20160. So yes:

-3/4480 = - (3 * 4.5)/20160 = -13.5 / 20160. But that's not a fraction with integer numerator. So maybe I made a mistake in the calculation?

Alternatively, let me compute:

Let me find A:

A = B - 1/20160 = (-3/4480) - (1/20160). Let me convert -3/4480 to denominator 20160:

Since 20160 / 4480 = 4.5, so multiply numerator and denominator by 4.5:

-3 * 4.5 = -13.5 over 20160. Therefore:

A = (-13.5 - 1)/20160 = -14.5 / 20160 = -29/40320. Wait, is that correct?

Yes, because -14.5/20160 is equal to -29/40320.

Therefore, A = -29/40320 and B = -3/4480.

But this seems messy, but maybe manageable. So now, Q(x) = (x - 2)(x - 3)...(x - 7)(Ax + B) where A = -29/40320 and B = -3/4480.

But maybe there's a better way to write this. Alternatively, maybe we can keep these fractions and use them in our calculations. However, this seems complicated. But perhaps we can proceed.

Now, the original goal is to find p(8). Recall that Q(x) = p(x)(x² - 1) - x. Therefore, solving for p(x):

p(x) = [Q(x) + x]/(x² - 1)

Therefore, to find p(8), we can compute:

p(8) = [Q(8) + 8]/(8² - 1) = [Q(8) + 8]/(64 - 1) = [Q(8) + 8]/63.

Therefore, we need to compute Q(8). Since Q(x) = (x - 2)...(x - 7)(Ax + B), so:

Q(8) = (8 - 2)(8 - 3)(8 - 4)(8 - 5)(8 - 6)(8 - 7)(A*8 + B)

Compute each term:

(8 - 2) = 6

(8 - 3) = 5

(8 - 4) = 4

(8 - 5) = 3

(8 - 6) = 2

(8 - 7) = 1

Therefore, the product is 6*5*4*3*2*1 = 6! = 720.

Then, Q(8) = 720*(A*8 + B). Therefore:

p(8) = [720*(8A + B) + 8]/63.

So we need to compute 8A + B.

From earlier, A = -29/40320 and B = -3/4480. Let me compute 8A + B:

First, compute 8A:

8 * (-29/40320) = -232/40320 = simplify. Divide numerator and denominator by 8: -29/5040.

Then, B = -3/4480.

Therefore, 8A + B = -29/5040 - 3/4480.

Find common denominator for 5040 and 4480. Let me compute LCM of 5040 and 4480.

Factorize 5040: 5040 = 7! = 2^4 * 3^2 * 5 * 7

Factorize 4480: 4480 = 448 * 10 = 2^6 * 5 * 7

Therefore, LCM is max exponents: 2^6, 3^2, 5, 7. So LCM = 64 * 9 * 5 * 7 = 64 * 315 = 20160.

Convert -29/5040 to denominator 20160: multiply numerator and denominator by 4 (since 5040 * 4 = 20160). So:

-29/5040 = -29*4 / 20160 = -116/20160.

Convert -3/4480 to denominator 20160: 4480 * 4.5 = 20160. Therefore, multiply numerator and denominator by 4.5:

-3/4480 = -3 * 4.5 / 20160 = -13.5 / 20160. But fractions with decimal points are not ideal. Let me do it properly:

Let me write -3/4480 as x/20160:

x = -3 * (20160 / 4480) = -3 * 4.5 = -13.5. Which is same as above.

But to write this as a fraction:

-13.5 = -27/2. Therefore, -13.5 / 20160 = -27/(2*20160) = -27/40320.

Therefore, 8A + B = -116/20160 -27/40320. Let me convert both terms to denominator 40320:

-116/20160 = -232/40320

-27/40320 remains as is.

Therefore, total is (-232 - 27)/40320 = -259/40320.

Therefore, Q(8) = 720 * (-259/40320) = Let me compute that:

720 divided by 40320 = 720 / 40320 = 1 / 56.

Therefore, 720 * (-259)/40320 = (-259)/56.

Therefore, p(8) = [ (-259)/56 + 8 ] / 63.

Compute (-259)/56 + 8:

Convert 8 to 8 * 56/56 = 448/56.

Therefore, (-259 + 448)/56 = 189/56.

Therefore, p(8) = (189/56)/63 = (189 / 56) * (1 / 63) = (189 / (56 * 63)).

Simplify numerator and denominator:

189 ÷ 63 = 3. So 3 / 56.

Therefore, p(8) = 3/56.

Wait, is that correct? Let me check my calculations again.

Let me confirm step-by-step:

First, Q(8) = 720*(8A + B) = 720*(-259/40320) = (720 / 40320)*(-259) = (1/56)*(-259) = -259/56.

Then, p(8) = [ -259/56 + 8 ] / 63. 8 is 448/56.

So -259/56 + 448/56 = (448 - 259)/56 = 189/56.

Then divide by 63: (189/56)/63 = (189)/(56*63). Simplify:

Divide numerator and denominator by 63: numerator becomes 189/63 = 3; denominator becomes 56. So 3/56. Yes.

Therefore, p(8) = 3/56. So is that the answer? Wait, that seems possible. But let me verify if this answer makes sense.

Alternatively, maybe there's a simpler way to reach this result, or maybe I made an error in calculation. Let me check with another method.

Alternatively, consider that since p(x) is a degree 5 polynomial, and we have Q(x) = p(x)(x² - 1) - x = (x - 2)...(x - 7)(Ax + B). Therefore, when x = 8, we can compute Q(8) = p(8)(8² -1) - 8 = p(8)*63 - 8. Therefore, p(8) = [Q(8) + 8]/63, which is what I did. And Q(8) = 720*(8A + B) = -259/56. Therefore, p(8) = (-259/56 + 8)/63 = 3/56. So that's correct.

But let me check with another approach. Maybe using finite differences or something else?

Alternatively, maybe using the concept of constructing the polynomial p(x) such that p(x) - [n/(n² - 1)] has roots at 2,...,7. But that seems similar to what I did before.

Alternatively, since we have a polynomial p(x) of degree 5 passing through those points, and we need to find p(8). Maybe we can use Lagrange interpolation? Although Lagrange interpolation for 6 points would be tedious, but maybe there's a pattern.

Alternatively, note that the function n/(n² - 1) can be written as [1/(2(n - 1))] + [1/(2(n + 1))], as we did before. Therefore, maybe consider that p(x) - [1/(2(x - 1)) + 1/(2(x + 1))] has roots at x = 2,...,7. However, since p(x) is a polynomial and the other terms are not, we need to multiply by (x - 1)(x + 1) to eliminate denominators. Let me try this.

Let me define R(x) = (x - 1)(x + 1)p(x) - [ (x + 1)/2 + (x - 1)/2 ].

Wait, let's see:

Original equation: p(x) = [1/(2(x - 1)) + 1/(2(x + 1))] for x = 2,...,7.

Multiply both sides by (x - 1)(x + 1):

(x² - 1)p(x) = [ (x + 1)/2 + (x - 1)/2 ] = [ (x + 1 + x -1)/2 ] = (2x)/2 = x.

Therefore, (x² - 1)p(x) - x = 0 for x = 2,...,7, which is exactly the Q(x) we previously defined. Therefore, that gives us the same equation. So we are back to the same place.

Therefore, our previous calculation seems correct. Therefore, the answer is 3/56. But just to confirm, maybe plug in a value.

Alternatively, maybe consider that p(x) is a degree 5 polynomial. Suppose we define another polynomial, say, p(x) = (x² - 1) * something + [n/(n² -1)] ?

Wait, not sure. Alternatively, think about considering the difference between p(x) and the function f(x) = x/(x² - 1). Since p(x) agrees with f(x) at x=2,...,7, then the difference p(x) - f(x) has zeros at those points. However, since p(x) is a polynomial and f(x) is not, this difference is not a polynomial. However, if we consider multiplying by (x² -1), as before, we get a polynomial.

Therefore, given that our previous steps are correct, then the answer is 3/56. Wait, but 3/56 seems a bit strange. Let me check with x=1. Wait, if x=1, then p(1) is undefined, but maybe we can compute Q(1) as before?

Earlier, we had Q(1) = -1, and also Q(1) = (1 - 2)...(1 - 7)(A*1 + B) = 720*(A + B) = -1. Which gave us the equation A + B = -1/720. Which we used. So that seems correct.

Another check: Suppose we take x = 0. Then Q(0) = p(0)*(0 -1) - 0 = -p(0). On the other hand, Q(0) = (0 -2)(0 -3)...(0 -7)(A*0 + B) = (-2)(-3)(-4)(-5)(-6)(-7)*B. Which is [(-2)(-3)(-4)(-5)(-6)(-7)]*B. The product is (-2)(-3)=6; (-4)(-5)=20; (-6)(-7)=42. Then 6*20=120; 120*42=5040. Therefore, Q(0) = 5040*B. Therefore, -p(0) = 5040*B => p(0) = -5040*B. Since B = -3/4480, then p(0) = -5040*(-3/4480) = (5040 / 4480)*3 = (5040 / 4480) *3. Simplify 5040 / 4480 = 504 / 448 = 63 / 56 = 9/8. Therefore, p(0) = (9/8)*3 = 27/8. Not sure if that helps, but seems consistent.

Alternatively, maybe plug in x = 8 into the original equation? Wait, not applicable. But since we have p(8) = 3/56, let's check if this is correct.

Alternatively, maybe consider that the answer is 3/56, and that seems plausible. But to be thorough, let me check with x=8 in the Q(x) equation.

Wait, we had Q(x) = p(x)(x² - 1) - x. So if p(8) = 3/56, then Q(8) = (3/56)(64 - 1) - 8 = (3/56)(63) - 8 = (189/56) - 8 = (189/56) - (448/56) = (-259)/56. Which matches our previous result of Q(8) = -259/56. So that checks out. Therefore, seems correct.

Therefore, I think the answer is 3/56.

But let me check for possible miscalculations when solving for A and B. Earlier, when solving for A and B, we had:

Equation 1: -A + B = 1/20160

Equation 2: A + B = -1/720

Solving, we found B = -3/4480 and A = -29/40320. Let me verify these values:

From Equation 2: A + B = -1/720

Let me compute A + B = (-29/40320) + (-3/4480). Convert -3/4480 to denominator 40320:

Multiply numerator and denominator by 9: -3*9 / 4480*9 = -27/40320. Therefore, A + B = -29/40320 -27/40320 = -56/40320 = -1/720. Correct.

From Equation 1: -A + B = 1/20160. Compute -A + B = -(-29/40320) + (-3/4480) = 29/40320 - 3/4480 = convert -3/4480 to -27/40320. So total is 29/40320 -27/40320 = 2/40320 = 1/20160. Correct. Therefore, these values for A and B are correct.

Therefore, the calculation of Q(8) is correct, and hence p(8) is indeed 3/56.

Therefore, the answer is 3/56, so boxed answer is \boxed{\dfrac{3}{56}}.

**Final Answer**
\boxed{\dfrac{3}{56}}

To solve the problem, we are given a polynomial $ p(x) $ of degree 5 such that:

$$
p(n) = \frac{n}{n^2 - 1} \quad \text{for } n = 2, 3, 4, 5, 6, 7
$$

Our goal is to find $ p(8) $. Here’s how we proceed:

---

### Step 1: Understand the Structure of the Given Function

Let’s note that the function

$$
f(n) = \frac{n}{n^2 - 1}
$$

can be simplified using **partial fractions**:

$$
\frac{n}{n^2 - 1} = \frac{n}{(n - 1)(n + 1)} = \frac{1}{2(n - 1)} + \frac{1}{2(n + 1)}
$$

This decomposition helps simplify the structure of the function, though it’s not a polynomial.

---

### Step 2: Define a Related Polynomial

We define a polynomial function $ Q(x) $ such that:

$$
Q(x) = p(x)(x^2 - 1) - x
$$

This is a degree 7 polynomial since $ p(x) $ is degree 5 and $ x^2 - 1 $ is degree 2, so their product is degree 7. Subtracting $ x $ (degree 1) doesn’t change the degree.

Now, consider that for $ x = 2, 3, 4, 5, 6, 7 $, we have:

$$
Q(n) = p(n)(n^2 - 1) - n = 0
$$

So $ Q(x) $ has roots at $ x = 2, 3, 4, 5, 6, 7 $. Thus, we can write:

$$
Q(x) = k(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(Ax + B)
$$

Where $ A $ and $ B $ are constants to be determined.

---

### Step 3: Determine the Constants $ A $ and $ B $

We use the condition $ Q(x) = p(x)(x^2 - 1) - x $, and then plug in values of $ x $ to find $ A $ and $ B $.

Using values at $ x = -1 $ and $ x = 1 $, where $ Q(x) $ is known:

- At $ x = 1 $: $ Q(1) = -1 $, since $ p(1)(1^2 - 1) - 1 = -1 $
- At $ x = -1 $: $ Q(-1) = p(-1)(-1^2 - 1) - (-1) = p(-1)(0) + 1 = 1 $

Compute:

At $ x = 1 $:

$$
Q(1) = (1 - 2)(1 - 3)(1 - 4)(1 - 5)(1 - 6)(1 - 7)(A(1) + B) = (-1)(-2)(-3)(-4)(-5)(-6)(A + B)
$$

$$
= 720(A + B) = -1 \Rightarrow A + B = -\frac{1}{720}
$$

At $ x = -1 $:

$$
Q(-1) = (-1 - 2)(-1 - 3)(-1 - 4)(-1 - 5)(-1 - 6)(-1 - 7)(A(-1) + B) = (-3)(-4)(-5)(-6)(-7)(-8)(-A + B)
$$

$$
= 20160(-A + B) = 1 \Rightarrow -A + B = \frac{1}{20160}
$$

Now solve the system:

$$
\begin{cases}
A + B = -\frac{1}{720} \\
-A + B = \frac{1}{20160}
\end{cases}
$$

Add the two equations:

$$
2B = -\frac{1}{720} + \frac{1}{20160} = \frac{-28 + 1}{20160} = -\frac{27}{20160}
\Rightarrow B = -\frac{27}{40320}
$$

Substitute back to get:

$$
A = -\frac{1}{720} - B = -\frac{1}{720} + \frac{27}{40320} = -\frac{56}{40320} + \frac{27}{40320} = -\frac{29}{40320}
$$

---

### Step 4: Use This to Find $ p(8) $

Recall:

$$
Q(x) = p(x)(x^2 - 1) - x \Rightarrow p(x) = \frac{Q(x) + x}{x^2 - 1}
$$

Thus:

$$
p(8) = \frac{Q(8) + 8}{8^2 - 1} = \frac{Q(8) + 8}{63}
$$

Compute $ Q(8) $:

$$
Q(8) = (8 - 2)(8 - 3)(8 - 4)(8 - 5)(8 - 6)(8 - 7)(A \cdot 8 + B)
= 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot (A \cdot 8 + B)
= 720 \cdot (8A + B)
$$

Plug in values:

$$
8A + B = 8 \cdot \left(-\frac{29}{40320}\right) + \left(-\frac{27}{40320}\right)
= \frac{-232 - 27}{40320} = \frac{-259}{40320}
$$

So:

$$
Q(8) = 720 \cdot \left(\frac{-259}{40320}\right) = \frac{-259}{56}
$$

Now compute:

$$
p(8) = \frac{-259/56 + 8}{63} = \frac{-259 + 448}{56 \cdot 63} = \frac{189}{56 \cdot 63}
$$

Simplify:

$$
p(8) = \frac{189}{3528} = \frac{3}{56}
$$

---

### ✅ Final Answer

$$
\boxed{\dfrac{3}{56}}
$$

Solution:
Let $q(x) = (x^2 - 1) p(x) - x.$  Then $q(x)$ has degree 7, and $q(n) = 0$ for $n = 2$, 3, 4, $\dots,$ 7, so
\[q(x) = (ax + b)(x - 2)(x - 3) \dotsm (x - 7)\]for some constants $a$ and $b.$

We know that $q(1) = (1^2 - 1)p(1) - 1 = -1.$  Setting $x = 1$ in the equation above, we get
\[q(1) = 720(a + b),\]so $a + b = -\frac{1}{720}.$

We also know that $q(-1) = ((-1)^2 - 1)p(-1) + 1 = 1.$  Setting $x = -1$ in the equation above, we get
\[q(-1) = 20160(-a + b),\]so $-a + b = \frac{1}{20160}.$  Solving for $a$ and $b,$ we find $a = -\frac{29}{40320}$ and $b = -\frac{3}{4480}.$  Hence,
\begin{align*}
q(x) &= \left( -\frac{29}{40320} x - \frac{3}{4480} \right) (x - 2)(x - 3) \dotsm (x - 7) \\
&= -\frac{(29x + 27)(x - 2)(x - 3) \dotsm (x - 7)}{40320}.
\end{align*}In particular,
\[q(8) = -\frac{(29 \cdot 8 + 27)(6)(5) \dotsm (1)}{40320} = -\frac{37}{8},\]so
\[p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\]


Explanation:
Okay, so I need to find the sum of the proper divisors of the sum of the proper divisors of 284. Let me break this down step by step. First, I should figure out what the proper divisors of 284 are, then sum them up, and then find the proper divisors of that sum and add those together. Let me start with the first part: finding the proper divisors of 284.

Alright, proper divisors of a number are all the divisors except the number itself. So for 284, I need to find all numbers that divide 284 without leaving a remainder, but exclude 284 itself. Let me think about how to find them. Maybe I can start by factoring 284 into its prime factors. That might make it easier to list all the divisors.

Let me try dividing 284 by small primes. First, check if it's even: 284 divided by 2 is 142. So 2 is a factor. Then divide 142 by 2 again: 142 ÷ 2 = 71. Now, 71... is that a prime number? Let me check. The square root of 71 is about 8.4, so I need to check divisibility by primes up to 7. 71 divided by 2? No, it's odd. Divided by 3? 3*23=69, 71-69=2, so no. Divided by 5? Ends with 1, so no. Divided by 7? 7*10=70, 71-70=1, so no. So 71 is prime. Therefore, the prime factorization of 284 is 2² * 71¹.

Now, to find all the divisors, I can use the exponents of the prime factors. For each prime factor, take exponents from 0 up to its exponent in the factorization. For 2, exponents can be 0, 1, or 2. For 71, exponents can be 0 or 1. Then multiply them together.

So the divisors would be:

For 2^0 * 71^0 = 1

2^0 * 71^1 = 71

2^1 * 71^0 = 2

2^1 * 71^1 = 2*71=142

2^2 * 71^0 = 4

2^2 * 71^1 = 4*71=284

So the divisors are 1, 2, 4, 71, 142, and 284. But since we want proper divisors, we exclude 284. Therefore, the proper divisors are 1, 2, 4, 71, and 142.

Let me double-check that. Let me list all numbers up to 284/2, which is 142, and see which divide 284. Starting from 1:

1 divides 284.

2 divides 284 (284/2=142)

3? 2+8+4=14, 14 isn't divisible by 3, so no.

4? 284 divided by 4 is 71, which is an integer. So yes.

5? Ends with 4, so no.

6? 284 divided by 6 is about 47.33, so no.

7? Let me check. 7*40=280, 284-280=4, so remainder 4. Not divisible.

8? 284 divided by 8 is 35.5. Not an integer.

9? 2+8+4=14 again, not divisible by 9.

10? Ends with 4, nope.

11? Let me do 284 divided by 11. 11*25=275, 284-275=9, not divisible.

12? 284/12 ≈ 23.666... nope.

13? 13*21=273, 284-273=11, not divisible.

14? 14*20=280, 284-280=4, nope.

15? 15*18=270, 284-270=14, nope.

16? 16*17=272, 284-272=12, nope.

17? 17*16=272, same as above.

18? 18*15=270, same.

19? 19*14=266, 284-266=18, not divisible.

20? 20*14=280, 284-280=4, nope.

21? 21*13=273, 284-273=11, nope.

22? 22*12=264, 284-264=20, nope.

23? 23*12=276, 284-276=8, nope.

24? 24*11=264, same as above.

25? 25*11=275, 284-275=9, nope.

26? 26*10=260, 284-260=24, nope.

27? 27*10=270, 284-270=14, nope.

28? 28*10=280, 284-280=4, nope.

29? 29*9=261, 284-261=23, nope.

30? 30*9=270, same as before.

Continuing up to 71, but since 71 is a prime factor, we already have 71 as a divisor. Let me check 71: 71*4=284, so yes. Then 71*2=142, which is also a divisor. Then 142*2=284, which is excluded. So the divisors are 1, 2, 4, 71, 142, 284. So the proper divisors are indeed 1, 2, 4, 71, 142. So that seems correct.

Now, the next step is to sum these proper divisors. Let me calculate that. So adding 1 + 2 + 4 + 71 + 142. Let me do this step by step.

1 + 2 = 3

3 + 4 = 7

7 + 71 = 78

78 + 142 = 220

So the sum of the proper divisors of 284 is 220. Wait, is that right? Let me check again.

1 + 2 = 3. Then +4 is 7. Then +71 is 78. Then +142. 78 + 142. Let me compute 78 + 142. 70 + 140 = 210, 8 + 2 = 10, so total 220. Yes, that seems correct.

Okay, so the sum of the proper divisors of 284 is 220. Now, the problem asks for the sum of the proper divisors of this sum, which is 220. So now I need to find the proper divisors of 220, then sum them up.

Alright, let me find the proper divisors of 220. Again, first, maybe find its prime factorization. Let me do that.

220 is even, so divide by 2: 220 ÷ 2 = 110.

110 ÷ 2 = 55.

55 is 5*11. So the prime factors are 2² * 5¹ * 11¹.

So, the prime factorization is 2² × 5 × 11.

Now, to find all divisors, we can take exponents for each prime:

For 2: exponents 0, 1, 2;

For 5: exponents 0, 1;

For 11: exponents 0, 1.

So the number of divisors is (2+1)*(1+1)*(1+1) = 3*2*2=12. So 12 divisors in total. Let me list them:

Start with 2^0 * 5^0 * 11^0 = 1

2^0 * 5^0 * 11^1 = 11

2^0 * 5^1 * 11^0 = 5

2^0 * 5^1 * 11^1 = 5*11=55

2^1 * 5^0 * 11^0 = 2

2^1 * 5^0 * 11^1 = 2*11=22

2^1 * 5^1 * 11^0 = 2*5=10

2^1 * 5^1 * 11^1 = 2*5*11=110

2^2 * 5^0 * 11^0 = 4

2^2 * 5^0 * 11^1 = 4*11=44

2^2 * 5^1 * 11^0 = 4*5=20

2^2 * 5^1 * 11^1 = 4*5*11=220

So the divisors are: 1, 11, 5, 55, 2, 22, 10, 110, 4, 44, 20, 220. Now, proper divisors exclude 220, so the proper divisors are all of these except 220. Let me write them down:

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110.

Let me check if there are 11 of them, which makes sense since total divisors were 12.

Now, let me confirm these by checking if they divide 220.

1 divides 220, yes.

2 divides 220, yes.

4 divides 220 (220 ÷ 4 = 55).

5 divides 220, yes.

10 divides 220, yes.

11 divides 220 (220 ÷ 11 = 20), yes.

20 divides 220 (220 ÷ 20 = 11), yes.

22 divides 220 (220 ÷ 22 = 10), yes.

44 divides 220 (220 ÷ 44 = 5), yes.

55 divides 220 (220 ÷ 55 = 4), yes.

110 divides 220 (220 ÷ 110 = 2), yes.

So all of these are indeed proper divisors. Now, I need to sum them up. Let me do that step by step.

Adding them one by one:

Start with 1 + 2 = 3.

3 + 4 = 7.

7 + 5 = 12.

12 + 10 = 22.

22 + 11 = 33.

33 + 20 = 53.

53 + 22 = 75.

75 + 44 = 119.

119 + 55 = 174.

174 + 110 = 284.

Wait, so the sum of the proper divisors of 220 is 284?

But hold on, that seems interesting because 284 is the original number we started with. Is this a coincidence? Wait, 284 and 220 are known as amicable numbers. Amicable numbers are pairs of numbers where each is the sum of the proper divisors of the other. So 220 and 284 form an amicable pair. That's why the sum of the proper divisors of 284 is 220, and then the sum of the proper divisors of 220 is back to 284. That's pretty cool.

But regardless, the problem is just asking for the sum of the proper divisors of the sum of the proper divisors of 284, which we found to be 284. Therefore, the answer should be 284. Let me double-check my addition steps to make sure I didn't make an arithmetic error.

Let me write the list of proper divisors again: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110.

Let me add them in pairs or groups to check.

Group 1: 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110.

Alternatively, perhaps group them as follows:

First, group the smaller ones:

1 + 2 = 3

4 + 5 = 9

10 + 11 = 21

20 + 22 = 42

44 + 55 = 99

And then the remaining is 110.

So total sum would be 3 + 9 + 21 + 42 + 99 + 110.

Calculating step by step:

3 + 9 = 12

12 + 21 = 33

33 + 42 = 75

75 + 99 = 174

174 + 110 = 284. Yep, same result. So the sum is indeed 284.

Therefore, the answer is 284. But let me just make sure I didn't skip any steps or make a mistake in the initial sum.

Earlier, when I summed the proper divisors of 284, I had 1 + 2 + 4 + 71 + 142 = 220. Let me check that again.

1 + 2 = 3

3 + 4 = 7

7 + 71 = 78

78 + 142 = 220. Correct. Then the proper divisors of 220 sum to 284. So the answer is 284. That seems correct.

Alternatively, maybe there's another way to approach this problem? Let me think. For instance, knowing that 284 and 220 are amicable numbers. But since the problem is presented in a way that requires computation, maybe I should have done the same steps. However, given that the problem is from a math competition or something similar, knowing about amicable numbers might be helpful. But since I didn't know that initially, I just did the calculations step by step.

But just to make sure, let me confirm once more that the divisors of 220 are indeed as I listed. Let me list them again:

Divisors of 220:

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220. So removing 220, the proper divisors are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. So that's 11 numbers. Their sum is 284. So that's correct. Therefore, the answer is 284.

But let me check once again the sum of these numbers. Maybe another way. For example, using the formula for the sum of divisors. The sum of all divisors of a number is given by multiplying (p^(k+1) - 1)/(p - 1) for each prime factor p with exponent k. Then subtract the number itself to get the sum of proper divisors.

So for 220, prime factorization is 2² * 5¹ * 11¹.

Sum of all divisors is (2³ - 1)/(2 - 1) * (5² - 1)/(5 - 1) * (11² - 1)/(11 - 1).

Calculating each part:

For 2²: (2³ - 1)/1 = (8 - 1)/1 = 7.

For 5¹: (5² - 1)/4 = (25 - 1)/4 = 24/4 = 6.

For 11¹: (11² - 1)/10 = (121 - 1)/10 = 120/10 = 12.

Multiply them all together: 7 * 6 * 12. Let's compute that:

7 * 6 = 42. 42 * 12 = 504. Therefore, the sum of all divisors of 220 is 504. Therefore, the sum of proper divisors is 504 - 220 = 284. Which matches what I got earlier. Therefore, that confirms that the sum is indeed 284.

So the answer is 284. Therefore, the problem's answer is 284, which I can put in a box.

**Final Answer**
\boxed{284}

To solve the problem, we need to determine the **sum of the proper divisors of the sum of the proper divisors of 284**. Let's approach it step by step.

---

### Step 1: Find the Proper Divisors of 284

The **proper divisors** of a number $ N $ are all the positive integers that divide $ N $ without a remainder, **excluding $ N $** itself.

We begin by factoring 284 into prime factors:

$$
284 = 2^2 \times 71
$$

Using the prime factorization, we generate all divisors:

- $2^0 = 1$
- $2^1 = 2$
- $2^2 = 4$
- $71^1 = 71$
- $2 \times 71 = 142$
- $4 \times 71 = 284$

So all the divisors of 284 are:
$$
1, 2, 4, 71, 142, 284
$$

Excluding 284 (as it's not a proper divisor), the **proper divisors** of 284 are:
$$
1, 2, 4, 71, 142
$$

**Sum of proper divisors of 284:**

$$
1 + 2 + 4 + 71 + 142 = 220
$$

---

### Step 2: Find the Proper Divisors of 220

Using the same method, we find the prime factorization of 220:

$$
220 = 2^2 \times 5 \times 11
$$

Now, list all divisors of 220 (excluding the number itself for proper divisors):

$$
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110
$$

**Sum of these divisors:**

$$
1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284
$$

---

### Conclusion

We've found that:

- The sum of the proper divisors of 284 is $ \boxed{220} $
- The sum of the proper divisors of 220 is $ \boxed{284} $

Thus, the **final answer** is:

$$
\boxed{284}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the proper divisors of the sum of the proper divisors of 284. Let me break this down step by step. First, I should figure out what the proper divisors of 284 are, then sum them up, and then find the proper divisors of that sum and add those together. Let me start with the first part: finding the proper divisors of 284.

Alright, proper divisors of a number are all the divisors except the number itself. So for 284, I need to find all numbers that divide 284 without leaving a remainder, but exclude 284 itself. Let me think about how to find them. Maybe I can start by factoring 284 into its prime factors. That might make it easier to list all the divisors.

Let me try dividing 284 by small primes. First, check if it's even: 284 divided by 2 is 142. So 2 is a factor. Then divide 142 by 2 again: 142 ÷ 2 = 71. Now, 71... is that a prime number? Let me check. The square root of 71 is about 8.4, so I need to check divisibility by primes up to 7. 71 divided by 2? No, it's odd. Divided by 3? 3*23=69, 71-69=2, so no. Divided by 5? Ends with 1, so no. Divided by 7? 7*10=70, 71-70=1, so no. So 71 is prime. Therefore, the prime factorization of 284 is 2² * 71¹.

Now, to find all the divisors, I can use the exponents of the prime factors. For each prime factor, take exponents from 0 up to its exponent in the factorization. For 2, exponents can be 0, 1, or 2. For 71, exponents can be 0 or 1. Then multiply them together.

So the divisors would be:

For 2^0 * 71^0 = 1

2^0 * 71^1 = 71

2^1 * 71^0 = 2

2^1 * 71^1 = 2*71=142

2^2 * 71^0 = 4

2^2 * 71^1 = 4*71=284

So the divisors are 1, 2, 4, 71, 142, and 284. But since we want proper divisors, we exclude 284. Therefore, the proper divisors are 1, 2, 4, 71, and 142.

Let me double-check that. Let me list all numbers up to 284/2, which is 142, and see which divide 284. Starting from 1:

1 divides 284.

2 divides 284 (284/2=142)

3? 2+8+4=14, 14 isn't divisible by 3, so no.

4? 284 divided by 4 is 71, which is an integer. So yes.

5? Ends with 4, so no.

6? 284 divided by 6 is about 47.33, so no.

7? Let me check. 7*40=280, 284-280=4, so remainder 4. Not divisible.

8? 284 divided by 8 is 35.5. Not an integer.

9? 2+8+4=14 again, not divisible by 9.

10? Ends with 4, nope.

11? Let me do 284 divided by 11. 11*25=275, 284-275=9, not divisible.

12? 284/12 ≈ 23.666... nope.

13? 13*21=273, 284-273=11, not divisible.

14? 14*20=280, 284-280=4, nope.

15? 15*18=270, 284-270=14, nope.

16? 16*17=272, 284-272=12, nope.

17? 17*16=272, same as above.

18? 18*15=270, same.

19? 19*14=266, 284-266=18, not divisible.

20? 20*14=280, 284-280=4, nope.

21? 21*13=273, 284-273=11, nope.

22? 22*12=264, 284-264=20, nope.

23? 23*12=276, 284-276=8, nope.

24? 24*11=264, same as above.

25? 25*11=275, 284-275=9, nope.

26? 26*10=260, 284-260=24, nope.

27? 27*10=270, 284-270=14, nope.

28? 28*10=280, 284-280=4, nope.

29? 29*9=261, 284-261=23, nope.

30? 30*9=270, same as before.

Continuing up to 71, but since 71 is a prime factor, we already have 71 as a divisor. Let me check 71: 71*4=284, so yes. Then 71*2=142, which is also a divisor. Then 142*2=284, which is excluded. So the divisors are 1, 2, 4, 71, 142, 284. So the proper divisors are indeed 1, 2, 4, 71, 142. So that seems correct.

Now, the next step is to sum these proper divisors. Let me calculate that. So adding 1 + 2 + 4 + 71 + 142. Let me do this step by step.

1 + 2 = 3

3 + 4 = 7

7 + 71 = 78

78 + 142 = 220

So the sum of the proper divisors of 284 is 220. Wait, is that right? Let me check again.

1 + 2 = 3. Then +4 is 7. Then +71 is 78. Then +142. 78 + 142. Let me compute 78 + 142. 70 + 140 = 210, 8 + 2 = 10, so total 220. Yes, that seems correct.

Okay, so the sum of the proper divisors of 284 is 220. Now, the problem asks for the sum of the proper divisors of this sum, which is 220. So now I need to find the proper divisors of 220, then sum them up.

Alright, let me find the proper divisors of 220. Again, first, maybe find its prime factorization. Let me do that.

220 is even, so divide by 2: 220 ÷ 2 = 110.

110 ÷ 2 = 55.

55 is 5*11. So the prime factors are 2² * 5¹ * 11¹.

So, the prime factorization is 2² × 5 × 11.

Now, to find all divisors, we can take exponents for each prime:

For 2: exponents 0, 1, 2;

For 5: exponents 0, 1;

For 11: exponents 0, 1.

So the number of divisors is (2+1)*(1+1)*(1+1) = 3*2*2=12. So 12 divisors in total. Let me list them:

Start with 2^0 * 5^0 * 11^0 = 1

2^0 * 5^0 * 11^1 = 11

2^0 * 5^1 * 11^0 = 5

2^0 * 5^1 * 11^1 = 5*11=55

2^1 * 5^0 * 11^0 = 2

2^1 * 5^0 * 11^1 = 2*11=22

2^1 * 5^1 * 11^0 = 2*5=10

2^1 * 5^1 * 11^1 = 2*5*11=110

2^2 * 5^0 * 11^0 = 4

2^2 * 5^0 * 11^1 = 4*11=44

2^2 * 5^1 * 11^0 = 4*5=20

2^2 * 5^1 * 11^1 = 4*5*11=220

So the divisors are: 1, 11, 5, 55, 2, 22, 10, 110, 4, 44, 20, 220. Now, proper divisors exclude 220, so the proper divisors are all of these except 220. Let me write them down:

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110.

Let me check if there are 11 of them, which makes sense since total divisors were 12.

Now, let me confirm these by checking if they divide 220.

1 divides 220, yes.

2 divides 220, yes.

4 divides 220 (220 ÷ 4 = 55).

5 divides 220, yes.

10 divides 220, yes.

11 divides 220 (220 ÷ 11 = 20), yes.

20 divides 220 (220 ÷ 20 = 11), yes.

22 divides 220 (220 ÷ 22 = 10), yes.

44 divides 220 (220 ÷ 44 = 5), yes.

55 divides 220 (220 ÷ 55 = 4), yes.

110 divides 220 (220 ÷ 110 = 2), yes.

So all of these are indeed proper divisors. Now, I need to sum them up. Let me do that step by step.

Adding them one by one:

Start with 1 + 2 = 3.

3 + 4 = 7.

7 + 5 = 12.

12 + 10 = 22.

22 + 11 = 33.

33 + 20 = 53.

53 + 22 = 75.

75 + 44 = 119.

119 + 55 = 174.

174 + 110 = 284.

Wait, so the sum of the proper divisors of 220 is 284?

But hold on, that seems interesting because 284 is the original number we started with. Is this a coincidence? Wait, 284 and 220 are known as amicable numbers. Amicable numbers are pairs of numbers where each is the sum of the proper divisors of the other. So 220 and 284 form an amicable pair. That's why the sum of the proper divisors of 284 is 220, and then the sum of the proper divisors of 220 is back to 284. That's pretty cool.

But regardless, the problem is just asking for the sum of the proper divisors of the sum of the proper divisors of 284, which we found to be 284. Therefore, the answer should be 284. Let me double-check my addition steps to make sure I didn't make an arithmetic error.

Let me write the list of proper divisors again: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110.

Let me add them in pairs or groups to check.

Group 1: 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110.

Alternatively, perhaps group them as follows:

First, group the smaller ones:

1 + 2 = 3

4 + 5 = 9

10 + 11 = 21

20 + 22 = 42

44 + 55 = 99

And then the remaining is 110.

So total sum would be 3 + 9 + 21 + 42 + 99 + 110.

Calculating step by step:

3 + 9 = 12

12 + 21 = 33

33 + 42 = 75

75 + 99 = 174

174 + 110 = 284. Yep, same result. So the sum is indeed 284.

Therefore, the answer is 284. But let me just make sure I didn't skip any steps or make a mistake in the initial sum.

Earlier, when I summed the proper divisors of 284, I had 1 + 2 + 4 + 71 + 142 = 220. Let me check that again.

1 + 2 = 3

3 + 4 = 7

7 + 71 = 78

78 + 142 = 220. Correct. Then the proper divisors of 220 sum to 284. So the answer is 284. That seems correct.

Alternatively, maybe there's another way to approach this problem? Let me think. For instance, knowing that 284 and 220 are amicable numbers. But since the problem is presented in a way that requires computation, maybe I should have done the same steps. However, given that the problem is from a math competition or something similar, knowing about amicable numbers might be helpful. But since I didn't know that initially, I just did the calculations step by step.

But just to make sure, let me confirm once more that the divisors of 220 are indeed as I listed. Let me list them again:

Divisors of 220:

1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220. So removing 220, the proper divisors are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110. So that's 11 numbers. Their sum is 284. So that's correct. Therefore, the answer is 284.

But let me check once again the sum of these numbers. Maybe another way. For example, using the formula for the sum of divisors. The sum of all divisors of a number is given by multiplying (p^(k+1) - 1)/(p - 1) for each prime factor p with exponent k. Then subtract the number itself to get the sum of proper divisors.

So for 220, prime factorization is 2² * 5¹ * 11¹.

Sum of all divisors is (2³ - 1)/(2 - 1) * (5² - 1)/(5 - 1) * (11² - 1)/(11 - 1).

Calculating each part:

For 2²: (2³ - 1)/1 = (8 - 1)/1 = 7.

For 5¹: (5² - 1)/4 = (25 - 1)/4 = 24/4 = 6.

For 11¹: (11² - 1)/10 = (121 - 1)/10 = 120/10 = 12.

Multiply them all together: 7 * 6 * 12. Let's compute that:

7 * 6 = 42. 42 * 12 = 504. Therefore, the sum of all divisors of 220 is 504. Therefore, the sum of proper divisors is 504 - 220 = 284. Which matches what I got earlier. Therefore, that confirms that the sum is indeed 284.

So the answer is 284. Therefore, the problem's answer is 284, which I can put in a box.

**Final Answer**
\boxed{284}

To solve the problem, we need to determine the **sum of the proper divisors of the sum of the proper divisors of 284**. Let's approach it step by step.

---

### Step 1: Find the Proper Divisors of 284

The **proper divisors** of a number $ N $ are all the positive integers that divide $ N $ without a remainder, **excluding $ N $** itself.

We begin by factoring 284 into prime factors:

$$
284 = 2^2 \times 71
$$

Using the prime factorization, we generate all divisors:

- $2^0 = 1$
- $2^1 = 2$
- $2^2 = 4$
- $71^1 = 71$
- $2 \times 71 = 142$
- $4 \times 71 = 284$

So all the divisors of 284 are:
$$
1, 2, 4, 71, 142, 284
$$

Excluding 284 (as it's not a proper divisor), the **proper divisors** of 284 are:
$$
1, 2, 4, 71, 142
$$

**Sum of proper divisors of 284:**

$$
1 + 2 + 4 + 71 + 142 = 220
$$

---

### Step 2: Find the Proper Divisors of 220

Using the same method, we find the prime factorization of 220:

$$
220 = 2^2 \times 5 \times 11
$$

Now, list all divisors of 220 (excluding the number itself for proper divisors):

$$
1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110
$$

**Sum of these divisors:**

$$
1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284
$$

---

### Conclusion

We've found that:

- The sum of the proper divisors of 284 is $ \boxed{220} $
- The sum of the proper divisors of 220 is $ \boxed{284} $

Thus, the **final answer** is:

$$
\boxed{284}
$$

Solution:
Prime factorize $284=2^2\cdot71$. The sum of the proper divisors of $284$ is
\begin{align*}
1+2+2^2+71+2 \cdot 71 &= (1+2+2^2)(1+71)-284 \\
&= 220 \\
&= 2^2\cdot5\cdot11.
\end{align*}Here we have used the observation that multiplying out $(1+2+2^2)(1+71)$ by distributing yields an expression which is the sum of all $6$ factors of $284.$ Applying this observation again, we find that the sum of the proper divisors of $220$ is $$(1+2+2^2)(1+5)(1+11)-220=7\cdot 6\cdot 12-220=\boxed{284}.$$


Explanation:
Okay, so I need to find the complex number w, which is the result of rotating z around c by π/4 counter-clockwise. Hmm, let me recall how rotation of complex numbers works.

First, I remember that to rotate a complex number around another point, you can translate the complex plane so that the center of rotation becomes the origin, perform the rotation, and then translate back. So, in formula terms, if I have a complex number z and I want to rotate it around point c by an angle θ, the formula should be something like:

w = (z - c) * e^{iθ} + c

Is that right? Let me check. If I subtract c, that moves the center of rotation to the origin. Then multiplying by e^{iθ} rotates it by θ, and then adding c brings it back. Yeah, that makes sense. So θ here is π/4, so e^{iπ/4} would be cos(π/4) + i sin(π/4), which is √2/2 + i√2/2.

So let me write down z and c first.

Given z = 2 + √2 - (3 + 3√2)i. Let me parse that. The real part is 2 + √2, and the imaginary part is -(3 + 3√2). Wait, so writing it as a complex number, z = (2 + √2) - (3 + 3√2)i.

And c is 2 - 3i.

So first step is to compute z - c. Let me do that.

z - c = [ (2 + √2) - 2 ] + [ - (3 + 3√2)i - (-3i) ]

Wait, let me compute real and imaginary parts separately.

Real part: (2 + √2) - 2 = √2.

Imaginary part: - (3 + 3√2)i - (-3i) = - (3 + 3√2)i + 3i = [ -3 - 3√2 + 3 ] i = -3√2 i.

So z - c = √2 - 3√2 i.

Hmm, that seems right. Let me verify again.

Original z: real part 2 + √2, imaginary part -(3 + 3√2).

Subtract c: which is 2 - 3i. So subtract 2 from the real part, and subtract (-3i) which is adding 3i to the imaginary part.

So real part: 2 + √2 - 2 = √2.

Imaginary part: -(3 + 3√2)i + 3i = -3i - 3√2 i + 3i = -3√2 i. Yep, that's correct.

So z - c is √2 - 3√2 i.

Now, I need to multiply this by e^{iπ/4} which is (√2/2 + i√2/2).

So let me compute (√2 - 3√2 i) * (√2/2 + i√2/2).

First, maybe factor out √2 from both terms? Let me see.

Let me write the first term as √2*(1 - 3i).

Then multiplying by √2/2 + i√2/2.

So that would be √2*(1 - 3i) * (√2/2 + i√2/2).

Let me compute that.

First, note that √2*(√2/2) = (2)/2 = 1. Similarly, √2*(i√2/2) = i*(2)/2 = i.

So maybe expanding the multiplication:

Alternatively, let me just multiply the two complex numbers:

(√2 - 3√2 i) * (√2/2 + i√2/2).

Let me compute this step by step.

Let me denote A = √2 and B = -3√2 i.

Multiply A*(√2/2) + A*(i√2/2) + B*(√2/2) + B*(i√2/2).

Compute term by term:

First term: √2 * √2/2 = (2)/2 = 1.

Second term: √2 * i√2/2 = i*(√2 * √2)/2 = i*(2)/2 = i.

Third term: -3√2 i * √2 / 2 = -3 * (√2 * √2) * i / 2 = -3*(2)/2 * i = -3i.

Fourth term: -3√2 i * i√2 / 2 = -3 * (√2 * √2) * i^2 / 2.

Compute that: √2 * √2 is 2, i^2 is -1.

So this term is -3 * 2 * (-1)/2 = -3 * (-2)/2 = 3.

Wait, let me check again.

Fourth term: -3√2 i * i√2 / 2

First, multiply constants:

-3 * √2 * √2 * i * i / 2

Which is -3 * 2 * (i^2) / 2

Which is -3 * 2 * (-1) / 2

Which is (-3 * -2) / 2

Which is 6 / 2 = 3.

So yes, the fourth term is 3.

So adding all four terms:

First term: 1

Second term: i

Third term: -3i

Fourth term: 3

So total is (1 + 3) + (i - 3i) = 4 - 2i.

So the result of multiplying (z - c) by e^{iπ/4} is 4 - 2i.

Then, to get w, we need to add c back.

So w = (z - c) rotated + c = 4 - 2i + c.

But c is 2 - 3i.

So adding 2 - 3i to 4 - 2i:

Real parts: 4 + 2 = 6

Imaginary parts: -2i + (-3i) = -5i

Therefore, w = 6 - 5i.

Wait, is that correct? Let me check again.

Wait, the rotation step gave us 4 - 2i, and then adding c which is 2 - 3i.

So adding 4 + 2 = 6, and -2i -3i = -5i. So yes, w = 6 -5i.

But let me check all the steps again because sometimes when dealing with complex numbers, there can be a mistake in signs or arithmetic.

First, z - c:

Original z is (2 + √2) - (3 + 3√2)i.

c is 2 - 3i.

So z - c is:

Real part: 2 + √2 - 2 = √2.

Imaginary part: [ - (3 + 3√2) - (-3) ] i ?

Wait, no. Wait, the imaginary part of z is -(3 + 3√2), and the imaginary part of c is -3.

So when subtracting c, which is (2 - 3i), it's equivalent to subtracting 2 and adding 3i.

So z - c has imaginary part: [ - (3 + 3√2) ] - (-3) = -3 - 3√2 + 3 = -3√2. So that's correct.

So z - c is √2 - 3√2 i.

Then multiplying by e^{iπ/4} which is √2/2 + i√2/2.

Let me write √2 - 3√2 i as √2(1 - 3i).

Multiply by e^{iπ/4} = √2/2 (1 + i).

Therefore, √2(1 - 3i) * √2/2 (1 + i) = (√2 * √2 / 2) * (1 - 3i)(1 + i)

Which is (2 / 2) * [ (1)(1) + 1(i) -3i(1) -3i(i) ]

Simplify: 1 * [ 1 + i -3i -3i² ]

Since i² = -1, so -3i² = 3.

So inside the brackets: 1 + i - 3i + 3 = (1 + 3) + (i - 3i) = 4 - 2i.

Therefore, multiplying by 1: 4 - 2i.

So that is correct. Then adding c, which is 2 - 3i.

So 4 - 2i + 2 - 3i = 6 -5i.

Therefore, w = 6 -5i.

Wait, but let me check with another method.

Alternatively, maybe I can represent all complex numbers as vectors and perform rotation.

But maybe I can check with coordinates.

Let me think.

Original z is 2 + √2 - (3 + 3√2)i.

So in coordinates, that's (2 + √2, -3 - 3√2).

Point c is (2, -3).

So vector from c to z is z - c, which is ( (2 + √2 - 2), (-3 - 3√2 - (-3)) ) = (√2, -3√2).

Which is the same as before.

So rotating this vector by π/4 counter-clockwise.

The rotation matrix for π/4 is [cosθ -sinθ; sinθ cosθ] which is [√2/2 -√2/2; √2/2 √2/2].

So applying this to the vector (√2, -3√2).

So new x-coordinate: √2*(√2/2) - (-3√2)*(√2/2)

Compute that: √2*(√2)/2 + 3√2*(√2)/2

Which is (2)/2 + 3*(2)/2 = 1 + 3 = 4.

New y-coordinate: √2*(√2/2) + (-3√2)*(√2/2)

Which is (2)/2 + (-3*2)/2 = 1 - 3 = -2.

Therefore, the rotated vector is (4, -2).

Then adding back the point c, which is (2, -3), gives w's coordinates as (4 + 2, -2 + (-3)) = (6, -5).

Therefore, w is 6 -5i.

So that matches the previous result.

Therefore, seems that the answer is 6 -5i.

But let me check once again.

Alternatively, maybe I can use complex numbers again.

Let me verify the rotation formula again.

w = (z - c) * e^{iπ/4} + c

We have z - c = √2 - 3√2 i

Multiply by e^{iπ/4} = cos(π/4) + i sin(π/4)

So ( √2 - 3√2 i ) * ( cos(π/4) + i sin(π/4) )

Which is √2 cos(π/4) + i √2 sin(π/4) - 3√2 i cos(π/4) - 3√2 i^2 sin(π/4)

Compute each term:

First term: √2 * √2/2 = 2/2 = 1

Second term: i * √2 * √2/2 = i * 2/2 = i

Third term: -3√2 * √2/2 * i = -3 * 2 / 2 * i = -3i

Fourth term: -3√2 * (-1) * √2 / 2

Wait, because i^2 is -1, so that term is -3√2 * i^2 * sin(π/4) = -3√2 * (-1) * sin(π/4)

Wait, no. Let me redo that. Wait:

Original term is -3√2 i * i sin(π/4) ?

Wait, let me re-express the multiplication properly.

Wait, expanding (a + b)(c + d) = ac + ad + bc + bd.

So in this case, a = √2, b = -3√2 i, c = cos(π/4), d = i sin(π/4).

Therefore, the terms are:

ac = √2 * cos(π/4)

ad = √2 * i sin(π/4)

bc = -3√2 i * cos(π/4)

bd = -3√2 i * i sin(π/4)

So then:

ac = √2 * cos(π/4) = √2 * √2/2 = 1

ad = √2 * i * √2/2 = i * 2 / 2 = i

bc = -3√2 * √2/2 * i = -3 * 2 / 2 * i = -3i

bd = -3√2 i * i * √2/2 = -3√2 * √2/2 * i^2 = -3*(2)/2 * (-1) = -3*1*(-1) = 3

Therefore, adding all terms: 1 + i - 3i + 3 = 4 - 2i.

Same result as before.

Then adding c = 2 - 3i gives 4 -2i + 2 -3i = 6 -5i.

So all steps confirm that.

Therefore, the answer is 6 -5i.

But let me check if maybe the Asymptote code can give some hint?

The Asymptote code defines Z as (2 + sqrt(2), -3 - 3*sqrt(2)), which matches the z we had.

c is (2, -3).

Then W is rotate(45, C)*(Z).

So rotating Z around C by 45 degrees, which is π/4.

So according to the Asymptote code, the coordinates of W should be (6, -5), which is 6 -5i.

Therefore, that seems to confirm.

Therefore, the answer is 6 -5i.

But since the question says "Find w", and the answer is 6 -5i.

Wait, but maybe I need to check once again if there is a miscalculation.

Wait, another way: Let me compute modulus and argument.

Alternatively, maybe not necessary.

But let me think.

Suppose I have z - c = √2 - 3√2 i.

So modulus is sqrt( (√2)^2 + (-3√2)^2 ) = sqrt(2 + 9*2) = sqrt(2 + 18) = sqrt(20) = 2*sqrt(5).

The argument is arctangent of (-3√2 / √2) = arctangent(-3).

So the original angle is arctan(-3).

Then rotating by π/4, so new angle is arctan(-3) + π/4.

But maybe it's not necessary.

Alternatively, compute modulus after rotation is same as before, 2√5.

Then if I compute the new complex number, which is 4 - 2i, modulus is sqrt(16 + 4) = sqrt(20) = 2√5. So that's same modulus.

So that checks out.

Then adding c gives 6 -5i, modulus sqrt(36 + 25) = sqrt(61).

But maybe not necessary.

Alternatively, think about the original coordinates.

Original point z is (2 + √2, -3 - 3√2).

After rotating around c = (2, -3) by π/4, we get w = (6, -5).

Let me check if this is correct.

Let me imagine that.

The vector from c to z is (√2, -3√2).

Rotating this vector by π/4 CCW gives (4, -2) as we found.

Then adding to c gives (2 + 4, -3 + (-2)) = (6, -5).

Yes.

Alternatively, if I think about rotating the vector (√2, -3√2) by π/4.

Let me recall that rotating a vector (x, y) by θ is:

x' = x cosθ - y sinθ

y' = x sinθ + y cosθ

So with θ = π/4, cosθ = sinθ = √2/2.

So for vector (√2, -3√2):

x' = √2*(√2/2) - (-3√2)*(√2/2)

= (2/2) + 3*(2)/2

= 1 + 3 = 4

y' = √2*(√2/2) + (-3√2)*(√2/2)

= 2/2 - 3*2/2

= 1 - 3 = -2

Therefore, yes, (4, -2).

Then adding (2, -3) gives (6, -5).

Therefore, all methods confirm that w = 6 -5i.

So I think this is correct.

Therefore, the answer is \boxed{6 - 5i}.

**Final Answer**
\boxed{6 - 5i}

To find the complex number $ w $, which is the result of rotating $ z $ around $ c $ by $ \frac{\pi}{4} $ counter-clockwise, we can use the standard formula for rotating a complex number about another point:

$$
w = (z - c) \cdot e^{i\theta} + c
$$

where:
- $ z = 2 + \sqrt{2} - (3 + 3\sqrt{2})i $
- $ c = 2 - 3i $
- $ \theta = \frac{\pi}{4} $

---

### Step 1: Compute $ z - c $

We perform the subtraction:

$$
z - c = [2 + \sqrt{2} - 2] + [-(3 + 3\sqrt{2})i + 3i] = \sqrt{2} - 3\sqrt{2}i
$$

So we now have:

$$
z - c = \sqrt{2} - 3\sqrt{2}i
$$

---

### Step 2: Multiply by $ e^{i\frac{\pi}{4}} $

We recall that:

$$
e^{i\frac{\pi}{4}} = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}
$$

So we compute:

$$
(\sqrt{2} - 3\sqrt{2}i) \cdot \left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right)
$$

Distribute:

$$
= \sqrt{2} \cdot \frac{\sqrt{2}}{2} + \sqrt{2} \cdot i\frac{\sqrt{2}}{2} - 3\sqrt{2}i \cdot \frac{\sqrt{2}}{2} - 3\sqrt{2}i \cdot i\frac{\sqrt{2}}{2}
$$

Now compute each term:

- $ \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1 $
- $ \sqrt{2} \cdot i\frac{\sqrt{2}}{2} = i \cdot \frac{2}{2} = i $
- $ -3\sqrt{2}i \cdot \frac{\sqrt{2}}{2} = -3i \cdot \frac{2}{2} = -3i $
- $ -3\sqrt{2}i \cdot i\frac{\sqrt{2}}{2} = -3i^2 \cdot \frac{2}{2} = 3 $

Adding these together:

$$
1 + i - 3i + 3 = 4 - 2i
$$

---

### Step 3: Add back the center point $ c $

Now we add $ c = 2 - 3i $ to the result:

$$
w = 4 - 2i + 2 - 3i = (4 + 2) + (-2i - 3i) = 6 - 5i
$$

---

### Final Answer

$$
\boxed{6 - 5i}
$$

[Assistant]
Explanation:
Okay, so I need to find the complex number w, which is the result of rotating z around c by π/4 counter-clockwise. Hmm, let me recall how rotation of complex numbers works.

First, I remember that to rotate a complex number around another point, you can translate the complex plane so that the center of rotation becomes the origin, perform the rotation, and then translate back. So, in formula terms, if I have a complex number z and I want to rotate it around point c by an angle θ, the formula should be something like:

w = (z - c) * e^{iθ} + c

Is that right? Let me check. If I subtract c, that moves the center of rotation to the origin. Then multiplying by e^{iθ} rotates it by θ, and then adding c brings it back. Yeah, that makes sense. So θ here is π/4, so e^{iπ/4} would be cos(π/4) + i sin(π/4), which is √2/2 + i√2/2.

So let me write down z and c first.

Given z = 2 + √2 - (3 + 3√2)i. Let me parse that. The real part is 2 + √2, and the imaginary part is -(3 + 3√2). Wait, so writing it as a complex number, z = (2 + √2) - (3 + 3√2)i.

And c is 2 - 3i.

So first step is to compute z - c. Let me do that.

z - c = [ (2 + √2) - 2 ] + [ - (3 + 3√2)i - (-3i) ]

Wait, let me compute real and imaginary parts separately.

Real part: (2 + √2) - 2 = √2.

Imaginary part: - (3 + 3√2)i - (-3i) = - (3 + 3√2)i + 3i = [ -3 - 3√2 + 3 ] i = -3√2 i.

So z - c = √2 - 3√2 i.

Hmm, that seems right. Let me verify again.

Original z: real part 2 + √2, imaginary part -(3 + 3√2).

Subtract c: which is 2 - 3i. So subtract 2 from the real part, and subtract (-3i) which is adding 3i to the imaginary part.

So real part: 2 + √2 - 2 = √2.

Imaginary part: -(3 + 3√2)i + 3i = -3i - 3√2 i + 3i = -3√2 i. Yep, that's correct.

So z - c is √2 - 3√2 i.

Now, I need to multiply this by e^{iπ/4} which is (√2/2 + i√2/2).

So let me compute (√2 - 3√2 i) * (√2/2 + i√2/2).

First, maybe factor out √2 from both terms? Let me see.

Let me write the first term as √2*(1 - 3i).

Then multiplying by √2/2 + i√2/2.

So that would be √2*(1 - 3i) * (√2/2 + i√2/2).

Let me compute that.

First, note that √2*(√2/2) = (2)/2 = 1. Similarly, √2*(i√2/2) = i*(2)/2 = i.

So maybe expanding the multiplication:

Alternatively, let me just multiply the two complex numbers:

(√2 - 3√2 i) * (√2/2 + i√2/2).

Let me compute this step by step.

Let me denote A = √2 and B = -3√2 i.

Multiply A*(√2/2) + A*(i√2/2) + B*(√2/2) + B*(i√2/2).

Compute term by term:

First term: √2 * √2/2 = (2)/2 = 1.

Second term: √2 * i√2/2 = i*(√2 * √2)/2 = i*(2)/2 = i.

Third term: -3√2 i * √2 / 2 = -3 * (√2 * √2) * i / 2 = -3*(2)/2 * i = -3i.

Fourth term: -3√2 i * i√2 / 2 = -3 * (√2 * √2) * i^2 / 2.

Compute that: √2 * √2 is 2, i^2 is -1.

So this term is -3 * 2 * (-1)/2 = -3 * (-2)/2 = 3.

Wait, let me check again.

Fourth term: -3√2 i * i√2 / 2

First, multiply constants:

-3 * √2 * √2 * i * i / 2

Which is -3 * 2 * (i^2) / 2

Which is -3 * 2 * (-1) / 2

Which is (-3 * -2) / 2

Which is 6 / 2 = 3.

So yes, the fourth term is 3.

So adding all four terms:

First term: 1

Second term: i

Third term: -3i

Fourth term: 3

So total is (1 + 3) + (i - 3i) = 4 - 2i.

So the result of multiplying (z - c) by e^{iπ/4} is 4 - 2i.

Then, to get w, we need to add c back.

So w = (z - c) rotated + c = 4 - 2i + c.

But c is 2 - 3i.

So adding 2 - 3i to 4 - 2i:

Real parts: 4 + 2 = 6

Imaginary parts: -2i + (-3i) = -5i

Therefore, w = 6 - 5i.

Wait, is that correct? Let me check again.

Wait, the rotation step gave us 4 - 2i, and then adding c which is 2 - 3i.

So adding 4 + 2 = 6, and -2i -3i = -5i. So yes, w = 6 -5i.

But let me check all the steps again because sometimes when dealing with complex numbers, there can be a mistake in signs or arithmetic.

First, z - c:

Original z is (2 + √2) - (3 + 3√2)i.

c is 2 - 3i.

So z - c is:

Real part: 2 + √2 - 2 = √2.

Imaginary part: [ - (3 + 3√2) - (-3) ] i ?

Wait, no. Wait, the imaginary part of z is -(3 + 3√2), and the imaginary part of c is -3.

So when subtracting c, which is (2 - 3i), it's equivalent to subtracting 2 and adding 3i.

So z - c has imaginary part: [ - (3 + 3√2) ] - (-3) = -3 - 3√2 + 3 = -3√2. So that's correct.

So z - c is √2 - 3√2 i.

Then multiplying by e^{iπ/4} which is √2/2 + i√2/2.

Let me write √2 - 3√2 i as √2(1 - 3i).

Multiply by e^{iπ/4} = √2/2 (1 + i).

Therefore, √2(1 - 3i) * √2/2 (1 + i) = (√2 * √2 / 2) * (1 - 3i)(1 + i)

Which is (2 / 2) * [ (1)(1) + 1(i) -3i(1) -3i(i) ]

Simplify: 1 * [ 1 + i -3i -3i² ]

Since i² = -1, so -3i² = 3.

So inside the brackets: 1 + i - 3i + 3 = (1 + 3) + (i - 3i) = 4 - 2i.

Therefore, multiplying by 1: 4 - 2i.

So that is correct. Then adding c, which is 2 - 3i.

So 4 - 2i + 2 - 3i = 6 -5i.

Therefore, w = 6 -5i.

Wait, but let me check with another method.

Alternatively, maybe I can represent all complex numbers as vectors and perform rotation.

But maybe I can check with coordinates.

Let me think.

Original z is 2 + √2 - (3 + 3√2)i.

So in coordinates, that's (2 + √2, -3 - 3√2).

Point c is (2, -3).

So vector from c to z is z - c, which is ( (2 + √2 - 2), (-3 - 3√2 - (-3)) ) = (√2, -3√2).

Which is the same as before.

So rotating this vector by π/4 counter-clockwise.

The rotation matrix for π/4 is [cosθ -sinθ; sinθ cosθ] which is [√2/2 -√2/2; √2/2 √2/2].

So applying this to the vector (√2, -3√2).

So new x-coordinate: √2*(√2/2) - (-3√2)*(√2/2)

Compute that: √2*(√2)/2 + 3√2*(√2)/2

Which is (2)/2 + 3*(2)/2 = 1 + 3 = 4.

New y-coordinate: √2*(√2/2) + (-3√2)*(√2/2)

Which is (2)/2 + (-3*2)/2 = 1 - 3 = -2.

Therefore, the rotated vector is (4, -2).

Then adding back the point c, which is (2, -3), gives w's coordinates as (4 + 2, -2 + (-3)) = (6, -5).

Therefore, w is 6 -5i.

So that matches the previous result.

Therefore, seems that the answer is 6 -5i.

But let me check once again.

Alternatively, maybe I can use complex numbers again.

Let me verify the rotation formula again.

w = (z - c) * e^{iπ/4} + c

We have z - c = √2 - 3√2 i

Multiply by e^{iπ/4} = cos(π/4) + i sin(π/4)

So ( √2 - 3√2 i ) * ( cos(π/4) + i sin(π/4) )

Which is √2 cos(π/4) + i √2 sin(π/4) - 3√2 i cos(π/4) - 3√2 i^2 sin(π/4)

Compute each term:

First term: √2 * √2/2 = 2/2 = 1

Second term: i * √2 * √2/2 = i * 2/2 = i

Third term: -3√2 * √2/2 * i = -3 * 2 / 2 * i = -3i

Fourth term: -3√2 * (-1) * √2 / 2

Wait, because i^2 is -1, so that term is -3√2 * i^2 * sin(π/4) = -3√2 * (-1) * sin(π/4)

Wait, no. Let me redo that. Wait:

Original term is -3√2 i * i sin(π/4) ?

Wait, let me re-express the multiplication properly.

Wait, expanding (a + b)(c + d) = ac + ad + bc + bd.

So in this case, a = √2, b = -3√2 i, c = cos(π/4), d = i sin(π/4).

Therefore, the terms are:

ac = √2 * cos(π/4)

ad = √2 * i sin(π/4)

bc = -3√2 i * cos(π/4)

bd = -3√2 i * i sin(π/4)

So then:

ac = √2 * cos(π/4) = √2 * √2/2 = 1

ad = √2 * i * √2/2 = i * 2 / 2 = i

bc = -3√2 * √2/2 * i = -3 * 2 / 2 * i = -3i

bd = -3√2 i * i * √2/2 = -3√2 * √2/2 * i^2 = -3*(2)/2 * (-1) = -3*1*(-1) = 3

Therefore, adding all terms: 1 + i - 3i + 3 = 4 - 2i.

Same result as before.

Then adding c = 2 - 3i gives 4 -2i + 2 -3i = 6 -5i.

So all steps confirm that.

Therefore, the answer is 6 -5i.

But let me check if maybe the Asymptote code can give some hint?

The Asymptote code defines Z as (2 + sqrt(2), -3 - 3*sqrt(2)), which matches the z we had.

c is (2, -3).

Then W is rotate(45, C)*(Z).

So rotating Z around C by 45 degrees, which is π/4.

So according to the Asymptote code, the coordinates of W should be (6, -5), which is 6 -5i.

Therefore, that seems to confirm.

Therefore, the answer is 6 -5i.

But since the question says "Find w", and the answer is 6 -5i.

Wait, but maybe I need to check once again if there is a miscalculation.

Wait, another way: Let me compute modulus and argument.

Alternatively, maybe not necessary.

But let me think.

Suppose I have z - c = √2 - 3√2 i.

So modulus is sqrt( (√2)^2 + (-3√2)^2 ) = sqrt(2 + 9*2) = sqrt(2 + 18) = sqrt(20) = 2*sqrt(5).

The argument is arctangent of (-3√2 / √2) = arctangent(-3).

So the original angle is arctan(-3).

Then rotating by π/4, so new angle is arctan(-3) + π/4.

But maybe it's not necessary.

Alternatively, compute modulus after rotation is same as before, 2√5.

Then if I compute the new complex number, which is 4 - 2i, modulus is sqrt(16 + 4) = sqrt(20) = 2√5. So that's same modulus.

So that checks out.

Then adding c gives 6 -5i, modulus sqrt(36 + 25) = sqrt(61).

But maybe not necessary.

Alternatively, think about the original coordinates.

Original point z is (2 + √2, -3 - 3√2).

After rotating around c = (2, -3) by π/4, we get w = (6, -5).

Let me check if this is correct.

Let me imagine that.

The vector from c to z is (√2, -3√2).

Rotating this vector by π/4 CCW gives (4, -2) as we found.

Then adding to c gives (2 + 4, -3 + (-2)) = (6, -5).

Yes.

Alternatively, if I think about rotating the vector (√2, -3√2) by π/4.

Let me recall that rotating a vector (x, y) by θ is:

x' = x cosθ - y sinθ

y' = x sinθ + y cosθ

So with θ = π/4, cosθ = sinθ = √2/2.

So for vector (√2, -3√2):

x' = √2*(√2/2) - (-3√2)*(√2/2)

= (2/2) + 3*(2)/2

= 1 + 3 = 4

y' = √2*(√2/2) + (-3√2)*(√2/2)

= 2/2 - 3*2/2

= 1 - 3 = -2

Therefore, yes, (4, -2).

Then adding (2, -3) gives (6, -5).

Therefore, all methods confirm that w = 6 -5i.

So I think this is correct.

Therefore, the answer is \boxed{6 - 5i}.

**Final Answer**
\boxed{6 - 5i}

To find the complex number $ w $, which is the result of rotating $ z $ around $ c $ by $ \frac{\pi}{4} $ counter-clockwise, we can use the standard formula for rotating a complex number about another point:

$$
w = (z - c) \cdot e^{i\theta} + c
$$

where:
- $ z = 2 + \sqrt{2} - (3 + 3\sqrt{2})i $
- $ c = 2 - 3i $
- $ \theta = \frac{\pi}{4} $

---

### Step 1: Compute $ z - c $

We perform the subtraction:

$$
z - c = [2 + \sqrt{2} - 2] + [-(3 + 3\sqrt{2})i + 3i] = \sqrt{2} - 3\sqrt{2}i
$$

So we now have:

$$
z - c = \sqrt{2} - 3\sqrt{2}i
$$

---

### Step 2: Multiply by $ e^{i\frac{\pi}{4}} $

We recall that:

$$
e^{i\frac{\pi}{4}} = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}
$$

So we compute:

$$
(\sqrt{2} - 3\sqrt{2}i) \cdot \left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right)
$$

Distribute:

$$
= \sqrt{2} \cdot \frac{\sqrt{2}}{2} + \sqrt{2} \cdot i\frac{\sqrt{2}}{2} - 3\sqrt{2}i \cdot \frac{\sqrt{2}}{2} - 3\sqrt{2}i \cdot i\frac{\sqrt{2}}{2}
$$

Now compute each term:

- $ \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1 $
- $ \sqrt{2} \cdot i\frac{\sqrt{2}}{2} = i \cdot \frac{2}{2} = i $
- $ -3\sqrt{2}i \cdot \frac{\sqrt{2}}{2} = -3i \cdot \frac{2}{2} = -3i $
- $ -3\sqrt{2}i \cdot i\frac{\sqrt{2}}{2} = -3i^2 \cdot \frac{2}{2} = 3 $

Adding these together:

$$
1 + i - 3i + 3 = 4 - 2i
$$

---

### Step 3: Add back the center point $ c $

Now we add $ c = 2 - 3i $ to the result:

$$
w = 4 - 2i + 2 - 3i = (4 + 2) + (-2i - 3i) = 6 - 5i
$$

---

### Final Answer

$$
\boxed{6 - 5i}
$$

Solution:
Rotating by $\frac{\pi}{4}$ counter-clockwise corresponds to the complex number
\[e^{\pi i/4} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}.\]Hence,
\[w - c = \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (z - c),\]so
\begin{align*}
w &= \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (z - c) + c \\
&= \left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \right) (\sqrt{2} - 3i \sqrt{2}) + 2 - 3i \\
&= (4 - 2i) + 2 - 3i \\
&= \boxed{6 - 5i}.
\end{align*}