Explanation:
Okay, so I need to solve this equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Hmm, the question is asking for the sum of all complex solutions. Alright, let me think about how to approach this.

First, I should note that this equation has rational expressions on the left side and a linear term on the right. The denominators on the left are all quadratic in terms of $x^2$, so maybe I can make a substitution to simplify things. Let me try letting $y = x^2$. Then, the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

Wait, but there's still an $x$ on the right side. Since $y = x^2$, maybe I need to express everything in terms of $y$? However, the right-hand side has $x$, not $y$. That complicates things because $x$ is related to $y$ via $y = x^2$, but unless we can eliminate $x$, this substitution might not help directly. Maybe I need to consider another approach.

Alternatively, perhaps I can note that if I can write the left-hand side as a function of $x^2$, and then relate it to $x$. However, the right-hand side is linear in $x$, which is different from the left-hand side which is in terms of $x^2$. This might mean that the equation is not straightforward to solve algebraically. Maybe there's some symmetry or something else?

Alternatively, perhaps if I consider that the left-hand side is a sum of terms with denominators $x^2 - n$ for $n = 1$ to $4$. If I can combine these fractions, maybe the equation simplifies? Let me try that.

To combine the fractions, I would need a common denominator, which would be $(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)$. That seems very complicated, but let me see. The left side becomes:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = \frac{A(x^2 - 2)(x^2 - 3)(x^2 - 4) + 2(x^2 - 1)(x^2 - 3)(x^2 - 4) + 3(x^2 - 1)(x^2 - 2)(x^2 - 4) + 4(x^2 - 1)(x^2 - 2)(x^2 - 3)}{(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)}.
$$

That's going to be a massive numerator. Maybe this isn't the best approach. Let me think again.

Alternatively, maybe the equation has solutions where $x$ is such that the left-hand side is equal to $2010x - 4$. Since the left-hand side is a function of $x^2$, perhaps I can consider that for each solution $x$, there is a corresponding $-x$, unless $x = 0$? But the right-hand side is linear in $x$, so the equation is not symmetric in $x$ and $-x$. Therefore, the solutions might not come in pairs.

Wait, but if I substitute $x$ with $-x$, then the left-hand side remains the same because $x^2$ is the same. However, the right-hand side becomes $-2010x - 4$. So unless the original equation is equivalent to $ -2010x - 4 $, which is different. Therefore, unless $x$ is a solution, $-x$ is not necessarily a solution unless that equation is satisfied. Therefore, solutions are not symmetric. Hmm. Therefore, maybe we need to consider that each solution is unique in some way?

Alternatively, perhaps considering Vieta's formula? If the equation can be transformed into a polynomial equation, then the sum of the roots can be determined by Vieta's formula.

But given that the equation has denominators, we need to be careful. Let me think. Suppose I denote the left-hand side as a function $f(x)$, then the equation is $f(x) = 2010x - 4$. To solve this, we can rearrange terms:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} - 2010x + 4 = 0.
$$

But this is a complex expression. However, if I can write this as a single rational function, then setting it equal to zero would give me a polynomial equation. However, given the denominators, this would be a rational function with denominator $(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)$, and numerator being some polynomial. Then, setting this equal to zero would involve multiplying both sides by the denominator, leading to a polynomial equation. But this would result in a very high-degree polynomial. For example, the denominator is degree 8, so when multiplied by the left-hand side, the numerator would be degree... let me check.

Each term in the left-hand side is a rational function. For example, $\frac{1}{x^2 - 1}$ multiplied by the denominator would give $(x^2 - 2)(x^2 - 3)(x^2 - 4)$, which is degree 6. Similarly for the others. So adding all those terms, the numerator would be a degree 6 polynomial. Then, subtracting 2010x and adding 4. Wait, actually, the entire equation is:

Multiply both sides by the denominator:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) \cdot (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) - (2010x - 4)(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) = 0.
$$

So the first term is a polynomial of degree 6 (since each term is degree 6 when multiplied out), and the second term is a polynomial of degree 1 + 8 = 9, because 2010x is degree 1 multiplied by the degree 8 denominator. Therefore, the entire equation would be a degree 9 polynomial. However, solving a degree 9 polynomial is not practical manually, especially without knowing specific roots or symmetries.

But the question asks for the sum of all complex solutions. Vieta's formula tells us that for a polynomial equation $a_n x^n + \ldots + a_0 = 0$, the sum of the roots is $-a_{n-1}/a_n$. However, since the original equation is a rational equation, when we multiply both sides by the denominator, we might introduce extraneous roots, but the question says "all complex solutions", which includes the solutions of the original equation. However, we should be careful about the domain. The original equation has denominators $x^2 - 1$, $x^2 - 2$, etc., so $x$ cannot be $\pm1$, $\pm\sqrt{2}$, $\pm\sqrt{3}$, $\pm2$. However, since we are considering complex solutions, those would be excluded, but the polynomial equation we get after multiplying both sides would include those roots as well, so we need to make sure that when we apply Vieta's formula, we don't include those extraneous roots. But maybe they are already excluded? Wait, no. If we multiply both sides by the denominator, the resulting polynomial equation will have all the roots of the original equation, plus the roots of the denominator (i.e., x = ±1, ±√2, etc.), which are not solutions to the original equation. Therefore, the sum of all complex solutions of the original equation would be the sum of all roots of the polynomial equation minus the sum of the extraneous roots. But that seems complicated, unless there's a trick here.

Alternatively, maybe there is a way to find the sum of the solutions without having to find the polynomial. Let me think again.

Alternatively, let me think about the original equation. Since the left side is a function of $x^2$, and the right side is a linear function in x, perhaps we can consider the equation as:

Let me denote $y = x^2$. Then, the left side is a function of y, but the right side is $2010x - 4$. However, since $y = x^2$, maybe I can write x in terms of y? But that's not straightforward. For example, $x = \sqrt{y}$ or $x = -\sqrt{y}$. Therefore, unless there's a way to relate the equation to both y and x, it's not helpful.

Alternatively, suppose we think of x as a variable and consider that each solution corresponds to two solutions for x, except when x=0. But again, the problem with the right side being linear in x complicates this.

Alternatively, maybe think about Vieta's formula on the polynomial equation. Let me try that approach. Let me suppose that after multiplying both sides by the denominator, I get a polynomial equation. Let me call that polynomial P(x) = 0. Then, the sum of all complex solutions of P(x) = 0 is equal to -coefficient of x^8 divided by coefficient of x^9. But if we can compute those coefficients, then we can find the sum. However, the problem is that the polynomial is degree 9, which is too high to compute manually. So unless there's a smarter way.

Alternatively, perhaps consider that the equation is not symmetric in x and -x, so maybe the sum of the roots isn't simply zero or something else. However, maybe there's a trick here. Let me think again.

Wait, the problem asks for the sum of all complex solutions. Maybe there's a way to relate the sum to the coefficients of the original equation? Let me think.

If I consider the original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Let me denote $S = \sum_{k=1}^4 \frac{k}{x^2 - k}$. Then, the equation is S = 2010x - 4. Let me think about taking the sum of all solutions. Suppose that x is a solution, then so is something else? Not sure.

Alternatively, suppose I let x be a solution, then maybe consider substituting x with another value? Not sure.

Alternatively, think about the sum over all solutions. Let me consider Vieta's formula on the polynomial. Let me try to find the sum. So again, if I can get the polynomial equation, then the sum of the roots would be... but how?

Alternatively, let me note that the left-hand side S can be written as:

Let me consider combining them. Let me think of S as:

Let me denote $y = x^2$. Then,

$$
S = \frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}.
$$

So maybe I can find this expression in terms of y, and then set it equal to $2010x - 4$. But since $y = x^2$, unless there is a relation between y and x, which is not straightforward. However, perhaps I can think of this as an equation involving both x and y, where y is dependent on x.

Alternatively, suppose that I rearrange the equation:

$$
S + 4 = 2010x.
$$

So:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} + 4 = 2010x.
$$

But since $y = x^2$, this equation relates x and y. However, without knowing more, I don't see how to proceed.

Alternatively, perhaps think about the function S(y) = $\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}$. Maybe compute S(y) for some y, but not sure.

Alternatively, think about the fact that for any solution, x is such that S(x²) = 2010x - 4. If I can relate this to some polynomial equation. Let me think of S(y) = 2010x - 4, but since y = x², we have S(y) = 2010x - 4. Let me think if there is a way to write this as S(y) + 4 = 2010x, and since y = x², perhaps square both sides?

Wait, if I take S(y) + 4 = 2010x, then squaring both sides would give me something in terms of x and y. However, since y = x², that is:

$$
(S(y) + 4)^2 = (2010x)^2.
$$

But expanding this would result in an equation involving x and y. However, since y = x², we can substitute y into the equation. But this seems like a lot of work and may not lead us anywhere useful. Let me check:

Let me denote A = S(y) + 4. Then, we have A = 2010x, so A² = (2010x)^2. But also, since y = x², then x² = y. So if I can express A in terms of y and x, then maybe...

But A = S(y) + 4. Therefore,

$$
(S(y) + 4)^2 = 4040100x^2 = 4040100y.
$$

So, we have:

$$
(S(y) + 4)^2 = 4040100y.
$$

But S(y) is:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}.
$$

Therefore, we can compute S(y) + 4 and square it, set equal to 4040100y. This would result in a complicated equation, but perhaps it can be simplified? Let me try.

First, compute S(y):

Let me compute S(y) + 4:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} + 4.
$$

Let me combine these terms. Let me compute each term:

Alternatively, maybe compute S(y):

Let me combine the terms step by step. Let me first combine $\frac{1}{y - 1} + \frac{2}{y - 2}$. Let me find a common denominator:

$$
\frac{(y - 2) + 2(y - 1)}{(y - 1)(y - 2)} = \frac{y - 2 + 2y - 2}{(y - 1)(y - 2)} = \frac{3y - 4}{(y - 1)(y - 2)}.
$$

Then, add $\frac{3}{y - 3}$:

$$
\frac{3y - 4}{(y - 1)(y - 2)} + \frac{3}{y - 3} = \frac{(3y - 4)(y - 3) + 3(y - 1)(y - 2)}{(y - 1)(y - 2)(y - 3)}.
$$

Let me compute the numerator:

First term: (3y - 4)(y - 3) = 3y(y - 3) - 4(y - 3) = 3y² - 9y - 4y + 12 = 3y² - 13y + 12.

Second term: 3(y - 1)(y - 2) = 3[y² - 3y + 2] = 3y² - 9y + 6.

Adding them together: 3y² -13y +12 + 3y² -9y +6 = 6y² - 22y + 18.

Therefore, the expression becomes:

$$
\frac{6y² - 22y + 18}{(y - 1)(y - 2)(y - 3)}.
$$

Now, add $\frac{4}{y - 4}$:

$$
\frac{6y² - 22y + 18}{(y - 1)(y - 2)(y - 3)} + \frac{4}{y - 4} = \frac{(6y² - 22y + 18)(y - 4) + 4(y - 1)(y - 2)(y - 3)}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Compute the numerator:

First term: (6y² - 22y + 18)(y - 4) = 6y²(y - 4) - 22y(y - 4) + 18(y - 4).

Let me compute each part:

- 6y²(y - 4) = 6y³ - 24y²

- -22y(y - 4) = -22y² + 88y

- 18(y - 4) = 18y - 72

Adding these together:

6y³ -24y² -22y² +88y +18y -72 = 6y³ - (24 + 22)y² + (88 + 18)y -72 = 6y³ - 46y² + 106y -72.

Second term: 4(y - 1)(y - 2)(y - 3). Let me compute that:

First, compute (y - 1)(y - 2) = y² - 3y + 2. Then multiply by (y - 3):

(y² - 3y + 2)(y - 3) = y³ - 3y² - 3y² + 9y + 2y - 6 = y³ - 6y² + 11y - 6.

Multiply by 4: 4y³ - 24y² + 44y - 24.

Now, adding the first term and this second term:

First term: 6y³ - 46y² + 106y -72

Second term: +4y³ -24y² +44y -24

Total: (6y³ + 4y³) + (-46y² -24y²) + (106y + 44y) + (-72 -24) = 10y³ -70y² + 150y -96.

Therefore, the entire numerator is 10y³ -70y² + 150y -96, so S(y) is:

$$
\frac{10y³ -70y² + 150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Therefore, S(y) + 4 is:

$$
\frac{10y³ -70y² + 150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)} + 4.
$$

Let me combine these terms:

$$
\frac{10y³ -70y² + 150y -96 + 4(y - 1)(y - 2)(y - 3)(y - 4)}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Now, compute the numerator:

First, compute 4(y - 1)(y - 2)(y - 3)(y - 4). Let me note that (y - 1)(y - 4) = y² -5y +4 and (y - 2)(y - 3) = y² -5y +6. Therefore, multiplying these two:

(y² -5y +4)(y² -5y +6) = Let me denote z = y² -5y, then it becomes (z +4)(z +6) = z² + 10z +24.

So expanding:

(z +4)(z +6) = z² + 10z +24 = (y² -5y)^2 + 10(y² -5y) +24.

Compute this:

First term: (y² -5y)^2 = y^4 -10y³ +25y²

Second term: 10(y² -5y) = 10y² -50y

Third term: +24

Total: y^4 -10y³ +25y² +10y² -50y +24 = y^4 -10y³ +35y² -50y +24.

Multiply by 4: 4y^4 -40y³ +140y² -200y +96.

Therefore, the numerator of S(y) + 4 is:

Original numerator: 10y³ -70y² + 150y -96 + [4y^4 -40y³ +140y² -200y +96] =

Let me compute term by term:

- The 4y^4 term: 4y^4

- The y³ terms: 10y³ -40y³ = -30y³

- The y² terms: -70y² +140y² = 70y²

- The y terms: 150y -200y = -50y

- Constants: -96 +96 = 0.

Therefore, the numerator simplifies to:

$$
4y^4 -30y³ +70y² -50y.
$$

So S(y) + 4 = [4y^4 -30y³ +70y² -50y] / [(y - 1)(y - 2)(y - 3)(y - 4)].

Therefore, the original equation S(y) + 4 = 2010x is:

$$
\frac{4y^4 -30y³ +70y² -50y}{(y - 1)(y - 2)(y - 3)(y - 4)} = 2010x,
$$

but since y = x², we can write:

$$
\frac{4x^8 -30x^6 +70x^4 -50x²}{(x² - 1)(x² - 2)(x² - 3)(x² - 4)} = 2010x.
$$

Let me note that the denominator is (x² - 1)(x² - 2)(x² - 3)(x² - 4) which is the same as before. Therefore, multiplying both sides by the denominator:

$$
4x^8 -30x^6 +70x^4 -50x² = 2010x \cdot (x² - 1)(x² - 2)(x² - 3)(x² - 4).
$$

So, bringing all terms to one side:

$$
4x^8 -30x^6 +70x^4 -50x² - 2010x \cdot (x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.
$$

This is the polynomial equation we get. Now, to find the sum of the roots of this equation. However, as mentioned earlier, this equation is a degree 9 polynomial because:

The term 2010x multiplied by (x² - 1)(x² - 2)(x² - 3)(x² - 4) is a degree 1 + 8 = 9. So the entire equation is a ninth-degree polynomial. So, according to Vieta's formula, the sum of the roots is equal to - (coefficient of x^8) / (coefficient of x^9). However, we need to calculate the coefficients of x^9 and x^8.

But let me note that the original equation was:

Left side: 4x^8 -30x^6 +70x^4 -50x²

Right side: 2010x * (x² - 1)(x² - 2)(x² - 3)(x² - 4)

So let me compute the right-hand side. Let me first compute (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me note that this is equal to [(x² - 1)(x² - 4)][(x² - 2)(x² - 3)].

As before, (x² - 1)(x² - 4) = x^4 -5x² +4

And (x² - 2)(x² - 3) = x^4 -5x² +6

Therefore, multiplying them: (x^4 -5x² +4)(x^4 -5x² +6).

Let me let z = x^4 -5x², then the product is (z + 4)(z + 6) = z² + 10z +24.

Expanding z²:

(z +4)(z +6) = z² + 10z +24 = (x^4 -5x²)^2 +10(x^4 -5x²) +24.

Compute this:

(x^4 -5x²)^2 = x^8 -10x^6 +25x^4

10(x^4 -5x²) = 10x^4 -50x²

Adding them together with +24:

x^8 -10x^6 +25x^4 +10x^4 -50x² +24 = x^8 -10x^6 +35x^4 -50x² +24.

Therefore, the right-hand side is 2010x multiplied by that:

2010x * [x^8 -10x^6 +35x^4 -50x² +24] = 2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x.

Therefore, the entire equation is:

Left side: 4x^8 -30x^6 +70x^4 -50x²

Minus Right side: - (2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x) = 0

Thus, moving all terms to left side:

4x^8 -30x^6 +70x^4 -50x² -2010x^9 + 20100x^7 -70350x^5 + 100500x^3 -48240x = 0

Now, combine like terms:

Start with the highest degree term:

-2010x^9

Next term: +20100x^7

Then: +4x^8

Then: -30x^6

Then: -70350x^5

Then: +70x^4

Then: +100500x^3

Then: -50x²

Then: -48240x.

So arranging in descending order:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0

Therefore, the polynomial is:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0.

Now, Vieta's formula says that the sum of the roots of the polynomial is - (coefficient of x^8) / (coefficient of x^9). Wait, but the coefficient of x^9 is -2010, and the coefficient of x^8 is +4. Therefore, the sum of the roots is - (4) / (-2010) = 4/2010 = 2/1005.

Wait, but hold on, the polynomial we derived is equal to zero. However, this includes all roots of the original equation, but also the roots of the denominator (from prior steps). Wait, but when we multiplied both sides by the denominator, we included the roots where the denominator is zero. But earlier, the original equation had denominators (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the roots x = ±1, ±√2, ±√3, ±2 are excluded from the original equation. Therefore, the polynomial we have now includes these roots as well. However, Vieta's formula counts all roots, including those extraneous ones. Therefore, the sum we just calculated includes those roots. However, we need to adjust for that to find the sum of the original solutions.

But wait, the original equation is defined for all x except x = ±1, ±√2, ±√3, ±2. However, when we multiplied both sides by the denominator, we obtained a polynomial which has those roots as roots as well. However, those roots are not solutions to the original equation. Therefore, the sum we need is the sum of all roots of the polynomial minus the sum of the extraneous roots.

Therefore, we need to find the sum of all roots of the polynomial, then subtract the sum of the roots of the denominator. However, the denominator is (x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0, which gives the roots x = ±1, ±√2, ±√3, ±2. So there are 8 roots from the denominator. However, when we multiplied both sides by the denominator, those roots are now roots of the polynomial equation. However, they are not solutions of the original equation. Therefore, if we can find the sum of those 8 roots, then subtract that from the total sum of the roots of the polynomial equation, we will get the sum of the solutions of the original equation.

Therefore, first, the total sum of all roots of the polynomial is 2/1005 as calculated earlier. Then, let's compute the sum of the roots of the denominator, which are x = ±1, ±√2, ±√3, ±2.

Let me compute that sum:

Sum = (1 + (-1)) + (√2 + (-√2)) + (√3 + (-√3)) + (2 + (-2)) = 0 + 0 + 0 + 0 = 0.

Wait, that's zero! Therefore, the sum of all extraneous roots is zero. Therefore, the sum of the original solutions is equal to the total sum of the polynomial's roots, which is 2/1005, minus the sum of the extraneous roots (which is zero). Therefore, the answer is 2/1005.

But let me verify this conclusion carefully. The polynomial equation we derived includes all solutions of the original equation plus the roots of the denominator. Since the roots of the denominator cancel out in their sum, then the sum of the original solutions is just the total sum of the polynomial's roots. Therefore, the answer is 2/1005.

But to make sure, let me check again. If the sum of all roots of the polynomial is 4 / (-(-2010))? Wait, let me recall Vieta's formula. For a polynomial of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = 0, the sum of the roots is -a_{n-1}/a_n.

In our case, the polynomial is:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0.

So, written as:

(-2010)x^9 + 4x^8 + ... = 0.

Therefore, leading coefficient a_n = -2010, and coefficient of x^{n-1} (which is x^8) is 4. Therefore, the sum of all roots is - (4)/(-2010) = 4/2010 = 2/1005. Yes, as before.

But since the sum of the extraneous roots is zero, then the sum of the original solutions is 2/1005. Therefore, the answer is 2/1005. Let me check if this reduces further. 2 and 1005: 1005 factors into 5 * 201 = 5 * 3 * 67. 2 is prime, so no common factors. Therefore, 2/1005 is in simplest terms.

But let me check if I made a mistake in the Vieta's formula. Let me double-check:

Given polynomial equation: P(x) = 0, with leading term -2010x^9 + ... So, write as:

P(x) = (-2010)x^9 + 4x^8 + ... So, according to Vieta, the sum of the roots is - (coefficient of x^8)/ (coefficient of x^9) = - (4)/(-2010) = 4/2010 = 2/1005. Correct.

Therefore, unless there's an error in calculation of the polynomial or in reasoning about the extraneous roots, this should be the answer.

But let me think again about the extraneous roots. The original equation had denominators which are zero when x^2 = 1, 2, 3, 4. Therefore, x = ±1, ±√2, ±√3, ±2. There are 8 roots. The sum of these roots is zero, as I said before. Therefore, yes, if the total sum of all roots of the polynomial is 2/1005, then subtracting zero gives the same value. Therefore, the sum of all complex solutions is 2/1005.

But let me check with an example. Suppose the equation had only the extraneous roots; then the sum would be zero, and if there are other roots, their sum would be 2/1005. Therefore, yes, the conclusion seems valid.

Therefore, the answer is 2/1005, which can be written as \boxed{\dfrac{2}{1005}}.

**Final Answer**
\boxed{\dfrac{2}{1005}}

We are asked to find the **sum of all complex solutions** to the equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

---

### Step 1: Understanding the Structure of the Equation

The left-hand side consists of rational expressions depending on $ x^2 $, while the right-hand side is linear in $ x $. This suggests that the equation can be treated as a function of $ x $, making it suitable for transformation into a polynomial, which will allow us to use **Vieta's formulas** to find the **sum of all complex solutions**.

To proceed, we first consider the denominator of the left-hand side, which is:

$$
(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Let us denote this product by $ D(x) $, and denote the left-hand side as $ S(x) $, so the equation becomes:

$$
S(x) = 2010x - 4.
$$

---

### Step 2: Clearing Denominators via Multiplication

To convert this into a polynomial equation, we multiply both sides of the equation by $ D(x) $, which results in:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) \cdot D(x) = (2010x - 4) \cdot D(x)
$$

This transformation yields a single rational expression on the left-hand side, and a polynomial on the right-hand side.

After algebraic manipulations, we simplify the left-hand side to a polynomial in $ x $, resulting in the expression:

$$
4x^8 - 30x^6 + 70x^4 - 50x^2
$$

On the right-hand side, we expand:

$$
2010x \cdot (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Let’s denote the denominator as:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Then we expand $ D(x) $ as a polynomial:

$$
D(x) = x^8 - 10x^6 + 35x^4 - 50x^2 + 24
$$

Hence, multiplying by $ 2010x $, we get:

$$
2010x \cdot D(x) = 2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x
$$

Subtracting this from the left-hand polynomial gives the final polynomial equation:

$$
\begin{aligned}
& -2010x^9 + 4x^8 + 20100x^7 - 30x^6 - 70350x^5 + 70x^4 \\
& + 100500x^3 - 50x^2 - 48240x = 0
\end{aligned}
$$

---

### Step 3: Applying Vieta’s Formula

The polynomial is of degree 9, and the sum of all complex roots of a polynomial is given by:

$$
-\frac{\text{coefficient of } x^8}{\text{coefficient of } x^9}
$$

From the equation, the coefficient of $ x^8 $ is **+4**, and the coefficient of $ x^9 $ is **-2010**. Thus, the sum of all roots of the polynomial is:

$$
-\frac{4}{-2010} = \frac{4}{2010} = \frac{2}{1005}
$$

---

### Step 4: Eliminating Extraneous Roots

However, note that multiplying both sides of the original equation by the denominator $ D(x) $ introduces **extraneous roots** — those at the values of $ x $ that make the original denominators zero. These are:

$$
x = \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2
$$

There are 8 such values, and summing them gives:

$$
(\pm1) + (\pm\sqrt{2}) + (\pm\sqrt{3}) + (\pm2) = 0
$$

So the **sum of these extraneous roots is 0**.

Therefore, since the sum of all **roots** (including the extraneous ones) is $ \frac{2}{1005} $, and the sum of the **extraneous** roots is 0, the **sum of all complex solutions** to the original equation is also:

$$
\frac{2}{1005}
$$

---

### Final Answer

$$
\boxed{\dfrac{2}{1005}}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Hmm, the question is asking for the sum of all complex solutions. Alright, let me think about how to approach this.

First, I should note that this equation has rational expressions on the left side and a linear term on the right. The denominators on the left are all quadratic in terms of $x^2$, so maybe I can make a substitution to simplify things. Let me try letting $y = x^2$. Then, the equation becomes:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} = 2010x - 4.
$$

Wait, but there's still an $x$ on the right side. Since $y = x^2$, maybe I need to express everything in terms of $y$? However, the right-hand side has $x$, not $y$. That complicates things because $x$ is related to $y$ via $y = x^2$, but unless we can eliminate $x$, this substitution might not help directly. Maybe I need to consider another approach.

Alternatively, perhaps I can note that if I can write the left-hand side as a function of $x^2$, and then relate it to $x$. However, the right-hand side is linear in $x$, which is different from the left-hand side which is in terms of $x^2$. This might mean that the equation is not straightforward to solve algebraically. Maybe there's some symmetry or something else?

Alternatively, perhaps if I consider that the left-hand side is a sum of terms with denominators $x^2 - n$ for $n = 1$ to $4$. If I can combine these fractions, maybe the equation simplifies? Let me try that.

To combine the fractions, I would need a common denominator, which would be $(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)$. That seems very complicated, but let me see. The left side becomes:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = \frac{A(x^2 - 2)(x^2 - 3)(x^2 - 4) + 2(x^2 - 1)(x^2 - 3)(x^2 - 4) + 3(x^2 - 1)(x^2 - 2)(x^2 - 4) + 4(x^2 - 1)(x^2 - 2)(x^2 - 3)}{(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)}.
$$

That's going to be a massive numerator. Maybe this isn't the best approach. Let me think again.

Alternatively, maybe the equation has solutions where $x$ is such that the left-hand side is equal to $2010x - 4$. Since the left-hand side is a function of $x^2$, perhaps I can consider that for each solution $x$, there is a corresponding $-x$, unless $x = 0$? But the right-hand side is linear in $x$, so the equation is not symmetric in $x$ and $-x$. Therefore, the solutions might not come in pairs.

Wait, but if I substitute $x$ with $-x$, then the left-hand side remains the same because $x^2$ is the same. However, the right-hand side becomes $-2010x - 4$. So unless the original equation is equivalent to $ -2010x - 4 $, which is different. Therefore, unless $x$ is a solution, $-x$ is not necessarily a solution unless that equation is satisfied. Therefore, solutions are not symmetric. Hmm. Therefore, maybe we need to consider that each solution is unique in some way?

Alternatively, perhaps considering Vieta's formula? If the equation can be transformed into a polynomial equation, then the sum of the roots can be determined by Vieta's formula.

But given that the equation has denominators, we need to be careful. Let me think. Suppose I denote the left-hand side as a function $f(x)$, then the equation is $f(x) = 2010x - 4$. To solve this, we can rearrange terms:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} - 2010x + 4 = 0.
$$

But this is a complex expression. However, if I can write this as a single rational function, then setting it equal to zero would give me a polynomial equation. However, given the denominators, this would be a rational function with denominator $(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)$, and numerator being some polynomial. Then, setting this equal to zero would involve multiplying both sides by the denominator, leading to a polynomial equation. But this would result in a very high-degree polynomial. For example, the denominator is degree 8, so when multiplied by the left-hand side, the numerator would be degree... let me check.

Each term in the left-hand side is a rational function. For example, $\frac{1}{x^2 - 1}$ multiplied by the denominator would give $(x^2 - 2)(x^2 - 3)(x^2 - 4)$, which is degree 6. Similarly for the others. So adding all those terms, the numerator would be a degree 6 polynomial. Then, subtracting 2010x and adding 4. Wait, actually, the entire equation is:

Multiply both sides by the denominator:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) \cdot (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) - (2010x - 4)(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4) = 0.
$$

So the first term is a polynomial of degree 6 (since each term is degree 6 when multiplied out), and the second term is a polynomial of degree 1 + 8 = 9, because 2010x is degree 1 multiplied by the degree 8 denominator. Therefore, the entire equation would be a degree 9 polynomial. However, solving a degree 9 polynomial is not practical manually, especially without knowing specific roots or symmetries.

But the question asks for the sum of all complex solutions. Vieta's formula tells us that for a polynomial equation $a_n x^n + \ldots + a_0 = 0$, the sum of the roots is $-a_{n-1}/a_n$. However, since the original equation is a rational equation, when we multiply both sides by the denominator, we might introduce extraneous roots, but the question says "all complex solutions", which includes the solutions of the original equation. However, we should be careful about the domain. The original equation has denominators $x^2 - 1$, $x^2 - 2$, etc., so $x$ cannot be $\pm1$, $\pm\sqrt{2}$, $\pm\sqrt{3}$, $\pm2$. However, since we are considering complex solutions, those would be excluded, but the polynomial equation we get after multiplying both sides would include those roots as well, so we need to make sure that when we apply Vieta's formula, we don't include those extraneous roots. But maybe they are already excluded? Wait, no. If we multiply both sides by the denominator, the resulting polynomial equation will have all the roots of the original equation, plus the roots of the denominator (i.e., x = ±1, ±√2, etc.), which are not solutions to the original equation. Therefore, the sum of all complex solutions of the original equation would be the sum of all roots of the polynomial equation minus the sum of the extraneous roots. But that seems complicated, unless there's a trick here.

Alternatively, maybe there is a way to find the sum of the solutions without having to find the polynomial. Let me think again.

Alternatively, let me think about the original equation. Since the left side is a function of $x^2$, and the right side is a linear function in x, perhaps we can consider the equation as:

Let me denote $y = x^2$. Then, the left side is a function of y, but the right side is $2010x - 4$. However, since $y = x^2$, maybe I can write x in terms of y? But that's not straightforward. For example, $x = \sqrt{y}$ or $x = -\sqrt{y}$. Therefore, unless there's a way to relate the equation to both y and x, it's not helpful.

Alternatively, suppose we think of x as a variable and consider that each solution corresponds to two solutions for x, except when x=0. But again, the problem with the right side being linear in x complicates this.

Alternatively, maybe think about Vieta's formula on the polynomial equation. Let me try that approach. Let me suppose that after multiplying both sides by the denominator, I get a polynomial equation. Let me call that polynomial P(x) = 0. Then, the sum of all complex solutions of P(x) = 0 is equal to -coefficient of x^8 divided by coefficient of x^9. But if we can compute those coefficients, then we can find the sum. However, the problem is that the polynomial is degree 9, which is too high to compute manually. So unless there's a smarter way.

Alternatively, perhaps consider that the equation is not symmetric in x and -x, so maybe the sum of the roots isn't simply zero or something else. However, maybe there's a trick here. Let me think again.

Wait, the problem asks for the sum of all complex solutions. Maybe there's a way to relate the sum to the coefficients of the original equation? Let me think.

If I consider the original equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

Let me denote $S = \sum_{k=1}^4 \frac{k}{x^2 - k}$. Then, the equation is S = 2010x - 4. Let me think about taking the sum of all solutions. Suppose that x is a solution, then so is something else? Not sure.

Alternatively, suppose I let x be a solution, then maybe consider substituting x with another value? Not sure.

Alternatively, think about the sum over all solutions. Let me consider Vieta's formula on the polynomial. Let me try to find the sum. So again, if I can get the polynomial equation, then the sum of the roots would be... but how?

Alternatively, let me note that the left-hand side S can be written as:

Let me consider combining them. Let me think of S as:

Let me denote $y = x^2$. Then,

$$
S = \frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}.
$$

So maybe I can find this expression in terms of y, and then set it equal to $2010x - 4$. But since $y = x^2$, unless there is a relation between y and x, which is not straightforward. However, perhaps I can think of this as an equation involving both x and y, where y is dependent on x.

Alternatively, suppose that I rearrange the equation:

$$
S + 4 = 2010x.
$$

So:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} + 4 = 2010x.
$$

But since $y = x^2$, this equation relates x and y. However, without knowing more, I don't see how to proceed.

Alternatively, perhaps think about the function S(y) = $\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}$. Maybe compute S(y) for some y, but not sure.

Alternatively, think about the fact that for any solution, x is such that S(x²) = 2010x - 4. If I can relate this to some polynomial equation. Let me think of S(y) = 2010x - 4, but since y = x², we have S(y) = 2010x - 4. Let me think if there is a way to write this as S(y) + 4 = 2010x, and since y = x², perhaps square both sides?

Wait, if I take S(y) + 4 = 2010x, then squaring both sides would give me something in terms of x and y. However, since y = x², that is:

$$
(S(y) + 4)^2 = (2010x)^2.
$$

But expanding this would result in an equation involving x and y. However, since y = x², we can substitute y into the equation. But this seems like a lot of work and may not lead us anywhere useful. Let me check:

Let me denote A = S(y) + 4. Then, we have A = 2010x, so A² = (2010x)^2. But also, since y = x², then x² = y. So if I can express A in terms of y and x, then maybe...

But A = S(y) + 4. Therefore,

$$
(S(y) + 4)^2 = 4040100x^2 = 4040100y.
$$

So, we have:

$$
(S(y) + 4)^2 = 4040100y.
$$

But S(y) is:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4}.
$$

Therefore, we can compute S(y) + 4 and square it, set equal to 4040100y. This would result in a complicated equation, but perhaps it can be simplified? Let me try.

First, compute S(y):

Let me compute S(y) + 4:

$$
\frac{1}{y - 1} + \frac{2}{y - 2} + \frac{3}{y - 3} + \frac{4}{y - 4} + 4.
$$

Let me combine these terms. Let me compute each term:

Alternatively, maybe compute S(y):

Let me combine the terms step by step. Let me first combine $\frac{1}{y - 1} + \frac{2}{y - 2}$. Let me find a common denominator:

$$
\frac{(y - 2) + 2(y - 1)}{(y - 1)(y - 2)} = \frac{y - 2 + 2y - 2}{(y - 1)(y - 2)} = \frac{3y - 4}{(y - 1)(y - 2)}.
$$

Then, add $\frac{3}{y - 3}$:

$$
\frac{3y - 4}{(y - 1)(y - 2)} + \frac{3}{y - 3} = \frac{(3y - 4)(y - 3) + 3(y - 1)(y - 2)}{(y - 1)(y - 2)(y - 3)}.
$$

Let me compute the numerator:

First term: (3y - 4)(y - 3) = 3y(y - 3) - 4(y - 3) = 3y² - 9y - 4y + 12 = 3y² - 13y + 12.

Second term: 3(y - 1)(y - 2) = 3[y² - 3y + 2] = 3y² - 9y + 6.

Adding them together: 3y² -13y +12 + 3y² -9y +6 = 6y² - 22y + 18.

Therefore, the expression becomes:

$$
\frac{6y² - 22y + 18}{(y - 1)(y - 2)(y - 3)}.
$$

Now, add $\frac{4}{y - 4}$:

$$
\frac{6y² - 22y + 18}{(y - 1)(y - 2)(y - 3)} + \frac{4}{y - 4} = \frac{(6y² - 22y + 18)(y - 4) + 4(y - 1)(y - 2)(y - 3)}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Compute the numerator:

First term: (6y² - 22y + 18)(y - 4) = 6y²(y - 4) - 22y(y - 4) + 18(y - 4).

Let me compute each part:

- 6y²(y - 4) = 6y³ - 24y²

- -22y(y - 4) = -22y² + 88y

- 18(y - 4) = 18y - 72

Adding these together:

6y³ -24y² -22y² +88y +18y -72 = 6y³ - (24 + 22)y² + (88 + 18)y -72 = 6y³ - 46y² + 106y -72.

Second term: 4(y - 1)(y - 2)(y - 3). Let me compute that:

First, compute (y - 1)(y - 2) = y² - 3y + 2. Then multiply by (y - 3):

(y² - 3y + 2)(y - 3) = y³ - 3y² - 3y² + 9y + 2y - 6 = y³ - 6y² + 11y - 6.

Multiply by 4: 4y³ - 24y² + 44y - 24.

Now, adding the first term and this second term:

First term: 6y³ - 46y² + 106y -72

Second term: +4y³ -24y² +44y -24

Total: (6y³ + 4y³) + (-46y² -24y²) + (106y + 44y) + (-72 -24) = 10y³ -70y² + 150y -96.

Therefore, the entire numerator is 10y³ -70y² + 150y -96, so S(y) is:

$$
\frac{10y³ -70y² + 150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Therefore, S(y) + 4 is:

$$
\frac{10y³ -70y² + 150y -96}{(y - 1)(y - 2)(y - 3)(y - 4)} + 4.
$$

Let me combine these terms:

$$
\frac{10y³ -70y² + 150y -96 + 4(y - 1)(y - 2)(y - 3)(y - 4)}{(y - 1)(y - 2)(y - 3)(y - 4)}.
$$

Now, compute the numerator:

First, compute 4(y - 1)(y - 2)(y - 3)(y - 4). Let me note that (y - 1)(y - 4) = y² -5y +4 and (y - 2)(y - 3) = y² -5y +6. Therefore, multiplying these two:

(y² -5y +4)(y² -5y +6) = Let me denote z = y² -5y, then it becomes (z +4)(z +6) = z² + 10z +24.

So expanding:

(z +4)(z +6) = z² + 10z +24 = (y² -5y)^2 + 10(y² -5y) +24.

Compute this:

First term: (y² -5y)^2 = y^4 -10y³ +25y²

Second term: 10(y² -5y) = 10y² -50y

Third term: +24

Total: y^4 -10y³ +25y² +10y² -50y +24 = y^4 -10y³ +35y² -50y +24.

Multiply by 4: 4y^4 -40y³ +140y² -200y +96.

Therefore, the numerator of S(y) + 4 is:

Original numerator: 10y³ -70y² + 150y -96 + [4y^4 -40y³ +140y² -200y +96] =

Let me compute term by term:

- The 4y^4 term: 4y^4

- The y³ terms: 10y³ -40y³ = -30y³

- The y² terms: -70y² +140y² = 70y²

- The y terms: 150y -200y = -50y

- Constants: -96 +96 = 0.

Therefore, the numerator simplifies to:

$$
4y^4 -30y³ +70y² -50y.
$$

So S(y) + 4 = [4y^4 -30y³ +70y² -50y] / [(y - 1)(y - 2)(y - 3)(y - 4)].

Therefore, the original equation S(y) + 4 = 2010x is:

$$
\frac{4y^4 -30y³ +70y² -50y}{(y - 1)(y - 2)(y - 3)(y - 4)} = 2010x,
$$

but since y = x², we can write:

$$
\frac{4x^8 -30x^6 +70x^4 -50x²}{(x² - 1)(x² - 2)(x² - 3)(x² - 4)} = 2010x.
$$

Let me note that the denominator is (x² - 1)(x² - 2)(x² - 3)(x² - 4) which is the same as before. Therefore, multiplying both sides by the denominator:

$$
4x^8 -30x^6 +70x^4 -50x² = 2010x \cdot (x² - 1)(x² - 2)(x² - 3)(x² - 4).
$$

So, bringing all terms to one side:

$$
4x^8 -30x^6 +70x^4 -50x² - 2010x \cdot (x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0.
$$

This is the polynomial equation we get. Now, to find the sum of the roots of this equation. However, as mentioned earlier, this equation is a degree 9 polynomial because:

The term 2010x multiplied by (x² - 1)(x² - 2)(x² - 3)(x² - 4) is a degree 1 + 8 = 9. So the entire equation is a ninth-degree polynomial. So, according to Vieta's formula, the sum of the roots is equal to - (coefficient of x^8) / (coefficient of x^9). However, we need to calculate the coefficients of x^9 and x^8.

But let me note that the original equation was:

Left side: 4x^8 -30x^6 +70x^4 -50x²

Right side: 2010x * (x² - 1)(x² - 2)(x² - 3)(x² - 4)

So let me compute the right-hand side. Let me first compute (x² - 1)(x² - 2)(x² - 3)(x² - 4). Let me note that this is equal to [(x² - 1)(x² - 4)][(x² - 2)(x² - 3)].

As before, (x² - 1)(x² - 4) = x^4 -5x² +4

And (x² - 2)(x² - 3) = x^4 -5x² +6

Therefore, multiplying them: (x^4 -5x² +4)(x^4 -5x² +6).

Let me let z = x^4 -5x², then the product is (z + 4)(z + 6) = z² + 10z +24.

Expanding z²:

(z +4)(z +6) = z² + 10z +24 = (x^4 -5x²)^2 +10(x^4 -5x²) +24.

Compute this:

(x^4 -5x²)^2 = x^8 -10x^6 +25x^4

10(x^4 -5x²) = 10x^4 -50x²

Adding them together with +24:

x^8 -10x^6 +25x^4 +10x^4 -50x² +24 = x^8 -10x^6 +35x^4 -50x² +24.

Therefore, the right-hand side is 2010x multiplied by that:

2010x * [x^8 -10x^6 +35x^4 -50x² +24] = 2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x.

Therefore, the entire equation is:

Left side: 4x^8 -30x^6 +70x^4 -50x²

Minus Right side: - (2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x) = 0

Thus, moving all terms to left side:

4x^8 -30x^6 +70x^4 -50x² -2010x^9 + 20100x^7 -70350x^5 + 100500x^3 -48240x = 0

Now, combine like terms:

Start with the highest degree term:

-2010x^9

Next term: +20100x^7

Then: +4x^8

Then: -30x^6

Then: -70350x^5

Then: +70x^4

Then: +100500x^3

Then: -50x²

Then: -48240x.

So arranging in descending order:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0

Therefore, the polynomial is:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0.

Now, Vieta's formula says that the sum of the roots of the polynomial is - (coefficient of x^8) / (coefficient of x^9). Wait, but the coefficient of x^9 is -2010, and the coefficient of x^8 is +4. Therefore, the sum of the roots is - (4) / (-2010) = 4/2010 = 2/1005.

Wait, but hold on, the polynomial we derived is equal to zero. However, this includes all roots of the original equation, but also the roots of the denominator (from prior steps). Wait, but when we multiplied both sides by the denominator, we included the roots where the denominator is zero. But earlier, the original equation had denominators (x² - 1)(x² - 2)(x² - 3)(x² - 4). Therefore, the roots x = ±1, ±√2, ±√3, ±2 are excluded from the original equation. Therefore, the polynomial we have now includes these roots as well. However, Vieta's formula counts all roots, including those extraneous ones. Therefore, the sum we just calculated includes those roots. However, we need to adjust for that to find the sum of the original solutions.

But wait, the original equation is defined for all x except x = ±1, ±√2, ±√3, ±2. However, when we multiplied both sides by the denominator, we obtained a polynomial which has those roots as roots as well. However, those roots are not solutions to the original equation. Therefore, the sum we need is the sum of all roots of the polynomial minus the sum of the extraneous roots.

Therefore, we need to find the sum of all roots of the polynomial, then subtract the sum of the roots of the denominator. However, the denominator is (x² - 1)(x² - 2)(x² - 3)(x² - 4) = 0, which gives the roots x = ±1, ±√2, ±√3, ±2. So there are 8 roots from the denominator. However, when we multiplied both sides by the denominator, those roots are now roots of the polynomial equation. However, they are not solutions of the original equation. Therefore, if we can find the sum of those 8 roots, then subtract that from the total sum of the roots of the polynomial equation, we will get the sum of the solutions of the original equation.

Therefore, first, the total sum of all roots of the polynomial is 2/1005 as calculated earlier. Then, let's compute the sum of the roots of the denominator, which are x = ±1, ±√2, ±√3, ±2.

Let me compute that sum:

Sum = (1 + (-1)) + (√2 + (-√2)) + (√3 + (-√3)) + (2 + (-2)) = 0 + 0 + 0 + 0 = 0.

Wait, that's zero! Therefore, the sum of all extraneous roots is zero. Therefore, the sum of the original solutions is equal to the total sum of the polynomial's roots, which is 2/1005, minus the sum of the extraneous roots (which is zero). Therefore, the answer is 2/1005.

But let me verify this conclusion carefully. The polynomial equation we derived includes all solutions of the original equation plus the roots of the denominator. Since the roots of the denominator cancel out in their sum, then the sum of the original solutions is just the total sum of the polynomial's roots. Therefore, the answer is 2/1005.

But to make sure, let me check again. If the sum of all roots of the polynomial is 4 / (-(-2010))? Wait, let me recall Vieta's formula. For a polynomial of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_0 = 0, the sum of the roots is -a_{n-1}/a_n.

In our case, the polynomial is:

-2010x^9 + 4x^8 + 20100x^7 -30x^6 -70350x^5 +70x^4 +100500x^3 -50x² -48240x = 0.

So, written as:

(-2010)x^9 + 4x^8 + ... = 0.

Therefore, leading coefficient a_n = -2010, and coefficient of x^{n-1} (which is x^8) is 4. Therefore, the sum of all roots is - (4)/(-2010) = 4/2010 = 2/1005. Yes, as before.

But since the sum of the extraneous roots is zero, then the sum of the original solutions is 2/1005. Therefore, the answer is 2/1005. Let me check if this reduces further. 2 and 1005: 1005 factors into 5 * 201 = 5 * 3 * 67. 2 is prime, so no common factors. Therefore, 2/1005 is in simplest terms.

But let me check if I made a mistake in the Vieta's formula. Let me double-check:

Given polynomial equation: P(x) = 0, with leading term -2010x^9 + ... So, write as:

P(x) = (-2010)x^9 + 4x^8 + ... So, according to Vieta, the sum of the roots is - (coefficient of x^8)/ (coefficient of x^9) = - (4)/(-2010) = 4/2010 = 2/1005. Correct.

Therefore, unless there's an error in calculation of the polynomial or in reasoning about the extraneous roots, this should be the answer.

But let me think again about the extraneous roots. The original equation had denominators which are zero when x^2 = 1, 2, 3, 4. Therefore, x = ±1, ±√2, ±√3, ±2. There are 8 roots. The sum of these roots is zero, as I said before. Therefore, yes, if the total sum of all roots of the polynomial is 2/1005, then subtracting zero gives the same value. Therefore, the sum of all complex solutions is 2/1005.

But let me check with an example. Suppose the equation had only the extraneous roots; then the sum would be zero, and if there are other roots, their sum would be 2/1005. Therefore, yes, the conclusion seems valid.

Therefore, the answer is 2/1005, which can be written as \boxed{\dfrac{2}{1005}}.

**Final Answer**
\boxed{\dfrac{2}{1005}}

We are asked to find the **sum of all complex solutions** to the equation:

$$
\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.
$$

---

### Step 1: Understanding the Structure of the Equation

The left-hand side consists of rational expressions depending on $ x^2 $, while the right-hand side is linear in $ x $. This suggests that the equation can be treated as a function of $ x $, making it suitable for transformation into a polynomial, which will allow us to use **Vieta's formulas** to find the **sum of all complex solutions**.

To proceed, we first consider the denominator of the left-hand side, which is:

$$
(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Let us denote this product by $ D(x) $, and denote the left-hand side as $ S(x) $, so the equation becomes:

$$
S(x) = 2010x - 4.
$$

---

### Step 2: Clearing Denominators via Multiplication

To convert this into a polynomial equation, we multiply both sides of the equation by $ D(x) $, which results in:

$$
\left( \frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} \right) \cdot D(x) = (2010x - 4) \cdot D(x)
$$

This transformation yields a single rational expression on the left-hand side, and a polynomial on the right-hand side.

After algebraic manipulations, we simplify the left-hand side to a polynomial in $ x $, resulting in the expression:

$$
4x^8 - 30x^6 + 70x^4 - 50x^2
$$

On the right-hand side, we expand:

$$
2010x \cdot (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Let’s denote the denominator as:

$$
D(x) = (x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4)
$$

Then we expand $ D(x) $ as a polynomial:

$$
D(x) = x^8 - 10x^6 + 35x^4 - 50x^2 + 24
$$

Hence, multiplying by $ 2010x $, we get:

$$
2010x \cdot D(x) = 2010x^9 - 20100x^7 + 70350x^5 - 100500x^3 + 48240x
$$

Subtracting this from the left-hand polynomial gives the final polynomial equation:

$$
\begin{aligned}
& -2010x^9 + 4x^8 + 20100x^7 - 30x^6 - 70350x^5 + 70x^4 \\
& + 100500x^3 - 50x^2 - 48240x = 0
\end{aligned}
$$

---

### Step 3: Applying Vieta’s Formula

The polynomial is of degree 9, and the sum of all complex roots of a polynomial is given by:

$$
-\frac{\text{coefficient of } x^8}{\text{coefficient of } x^9}
$$

From the equation, the coefficient of $ x^8 $ is **+4**, and the coefficient of $ x^9 $ is **-2010**. Thus, the sum of all roots of the polynomial is:

$$
-\frac{4}{-2010} = \frac{4}{2010} = \frac{2}{1005}
$$

---

### Step 4: Eliminating Extraneous Roots

However, note that multiplying both sides of the original equation by the denominator $ D(x) $ introduces **extraneous roots** — those at the values of $ x $ that make the original denominators zero. These are:

$$
x = \pm1, \pm\sqrt{2}, \pm\sqrt{3}, \pm2
$$

There are 8 such values, and summing them gives:

$$
(\pm1) + (\pm\sqrt{2}) + (\pm\sqrt{3}) + (\pm2) = 0
$$

So the **sum of these extraneous roots is 0**.

Therefore, since the sum of all **roots** (including the extraneous ones) is $ \frac{2}{1005} $, and the sum of the **extraneous** roots is 0, the **sum of all complex solutions** to the original equation is also:

$$
\frac{2}{1005}
$$

---

### Final Answer

$$
\boxed{\dfrac{2}{1005}}
$$

Solution:
We can add 4 to both sides to get
\[\frac{1}{x^2 - 1} + 1 + \frac{2}{x^2 - 2} + 1 + \frac{3}{x^2 - 3} + 1 + \frac{4}{x^2 - 4} + 1 = 2010x.\]This simplifies to
\[\frac{x^2}{x^2 - 1} + \frac{x^2}{x^2 - 2} + \frac{x^2}{x^2 - 3} + \frac{x^2}{x^2 - 4} = 2010x.\]We see that $x = 0$ is a solution (which does not affect our sum).  Otherwise, we can divide both sides by $x$:
\[\frac{x}{x^2 - 1} + \frac{x}{x^2 - 2} + \frac{x}{x^2 - 3} + \frac{x}{x^2 - 4} = 2010.\]Clearing the denominators, we get
\begin{align*}
&x(x^2 - 2)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 3) \\
&\quad = 2010(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
\end{align*}This expands to
\[4x^7 + \dotsb = 2010x^8 + \dotsb,\]where only terms that have degree 7 or greater are shown.  Then
\[2010x^8 - 4x^7 + \dotsb = 0,\]so by Vieta's formulas, the sum of the roots is $\frac{4}{2010} = \boxed{\frac{2}{1005}}.$


Explanation:
Okay, so I need to figure out for which prime numbers p the congruence 8x ≡ 1 mod p has no solutions. Then, sum all those primes. Hmm, let me start by recalling some number theory concepts.

First, the equation 8x ≡ 1 mod p is a linear congruence. I remember that a linear congruence ax ≡ b mod m has a solution if and only if gcd(a, m) divides b. In this case, a is 8, b is 1, and m is p, which is a prime. So, the condition for the congruence to have a solution is that gcd(8, p) divides 1.

Now, since p is a prime, the gcd(8, p) can only be 1 or p. Because if p divides 8, then gcd(8, p) would be p, otherwise it's 1. Wait, but p is a prime. Let me think: If p divides 8, then p must be one of the prime factors of 8. The prime factors of 8 are just 2, since 8 is 2 cubed. Therefore, gcd(8, p) is p if p = 2, otherwise it's 1.

So, the condition gcd(8, p) divides 1. Well, since gcd(8, p) is either 1 or 2. Then, 1 divides 1, which is always true. However, if gcd(8, p) is 2, then we need 2 divides 1? Which is not true. Therefore, the congruence 8x ≡ 1 mod p has no solutions if and only if gcd(8, p) does not divide 1, which happens when gcd(8, p) is 2. Which, as we saw, occurs only when p divides 8. Since p is prime, p must be 2.

Wait, so the only prime p where the congruence 8x ≡ 1 mod p has no solutions is p=2? So, then the sum of all such primes is just 2? But the problem says "there are finitely many primes p", and asks to determine the sum. Maybe there's a mistake here?

Wait, let me check again. Let me test p=2. The congruence is 8x ≡ 1 mod 2. But 8 mod 2 is 0, so 0x ≡ 1 mod 2, which is 0 ≡ 1 mod 2, which is false. So indeed no solution. So p=2 is one such prime.

What about other primes? Let me check p=3. Then the congruence is 8x ≡ 1 mod 3. Since 8 mod 3 is 2, so 2x ≡ 1 mod 3. Then multiplying both sides by inverse of 2 mod 3. The inverse of 2 mod 3 is 2, since 2*2=4≡1 mod3. So x ≡ 2*1=2 mod3. So x=2 is a solution. Therefore, for p=3, there is a solution. Similarly for p=5: 8x≡1 mod5. 8 mod5 is 3, so 3x≡1 mod5. Inverse of 3 mod5 is 2, because 3*2=6≡1 mod5. So x≡2 mod5. So solution exists.

What about p=2? As we saw, no solution. What about p=7? 8 mod7 is 1, so 1x≡1 mod7, so x≡1 mod7. Solution exists.

So seems like only p=2 is the prime where there's no solution. So the answer is sum of all such p is 2. But the problem says "finitely many primes p", which is true here, since only p=2. So is the answer 2?

Wait, but maybe I made a mistake? Let me think again. The general theory says that the equation ax ≡ b mod p has a solution if and only if gcd(a, p) divides b. Since p is prime, gcd(a, p) is 1 if p does not divide a, and p if p divides a. So if p divides a, then the equation becomes 0x ≡ b mod p, which has a solution only if b ≡0 mod p. In our case, b=1. Therefore, if p divides a=8, then the equation becomes 0x≡1 mod p, which has no solution. Therefore, primes p dividing 8. Since 8=2^3, the primes dividing 8 are just 2. Hence, only p=2. Therefore, sum is 2.

So is the answer 2? But maybe there is something else. Wait, the question says "there are finitely many primes p", which is true, but maybe the question is designed to trick us? Let me check again.

Wait, maybe if p divides 8, which is p=2, then there's no solution. However, for other primes, since gcd(8,p)=1, which divides 1, so solutions exist. So only p=2. So the sum is 2. Therefore, answer is 2.

But I feel like maybe the problem is more complicated? Maybe I need to check if 8 is invertible modulo p? Which is equivalent to gcd(8,p)=1. So, if gcd(8,p)=1, then 8 is invertible, so equation has solution. If gcd(8,p)=p, which is only when p divides 8, i.e., p=2, then 8 is not invertible, and the equation 8x≡1 mod p becomes 0x≡1 mod p which is impossible. Therefore, only p=2.

Therefore, the sum is 2. Therefore, the answer is \boxed{2}.

Wait, but the problem says "finitely many primes p", but maybe there are more? For example, if the equation is 8x ≡ 1 mod p. Suppose p divides 8, which is only 2. So only p=2.

Alternatively, maybe there are other primes where 8x ≡ 1 mod p has no solutions? But for any prime p ≠ 2, gcd(8,p)=1, so 8 is invertible mod p, so there exists x such that 8x ≡ 1 mod p. Hence, only p=2.

Alternatively, maybe I need to check if 8 is a unit mod p? Which is true if p does not divide 8, so p≠2. Hence, only p=2 is the prime where 8 is not invertible. Therefore, only p=2.

Therefore, the sum of all such p is 2.

Alternatively, maybe the problem is in the translation? The original problem says "the congruence 8x ≡ 1 mod p has no solutions x". So, for which primes p does this happen? Only p=2. Hence, the answer is 2.

Wait, but maybe the problem is in some contexts where p divides 8, but maybe also other primes? Wait, 8 is 2^3. So the only prime divisor is 2.

Alternatively, maybe the problem is referring to primes for which the equation has no solution, but maybe if p divides 1? Which is impossible.

Alternatively, is there a mistake here? Let me check again. For example, take p=2. 8x ≡1 mod 2. Since 8 is 0 mod 2, so 0x ≡1 mod2. Which is 0≡1 mod2, which is false. So no solution.

Take p=3. 8x≡1 mod3 is 2x≡1 mod3. Multiply both sides by inverse of 2 mod3, which is 2. So x≡2*1=2 mod3. So x=2 is solution.

p=5: 8x≡1 mod5 is 3x≡1 mod5. Inverse of 3 mod5 is 2, since 3*2=6≡1 mod5. So x≡2 mod5.

p=7: 8x≡1 mod7 is 1x≡1 mod7, so x≡1 mod7.

p=11: 8x≡1 mod11. Let me find inverse of 8 mod11. 8*? ≡1 mod11. 8* 7=56≡1 mod11 (since 56-55=1). So x≡7 mod11.

So definitely, for p≠2, there is a solution. So the only prime is 2. Therefore, the answer is 2.

But the problem says "finitely many primes p", which is true, as only 2. So the sum is 2. Therefore, I think the answer is 2. However, I need to check if maybe there's another prime p where 8x≡1 mod p has no solution. Wait, suppose p divides 8, but p is a prime. So only p=2. So no.

Alternatively, maybe the problem is misstated? For instance, if the equation was 8x ≡ 1 mod p has no solutions, meaning that 8 is not invertible. Which is when p divides 8. So again, p=2.

Alternatively, maybe I should think in terms of multiplicative inverses. For the equation 8x ≡1 mod p to have a solution, 8 must have an inverse modulo p. The inverse exists if and only if 8 and p are coprime, i.e., gcd(8, p)=1. Since p is prime, gcd(8,p)=1 unless p divides 8. Therefore, if p divides 8, which is p=2, then the inverse does not exist, so no solution. Therefore, again, only p=2.

Therefore, I think the answer is 2. Maybe the problem is a trick question?

Wait, maybe the question is in the context of modulus p, but maybe p divides 8? But p is prime. So again only p=2.

Alternatively, maybe I made a mistake with the logic. Let me check the general condition again.

The linear congruence ax ≡ b mod m has a solution iff gcd(a, m) divides b.

Here, a=8, b=1, m=p. So gcd(8, p) divides 1.

Since p is prime, gcd(8, p) is either 1 or p.

So if gcd(8, p) divides 1, then either:

Case 1: gcd(8, p) = 1. Then 1 divides 1, which is true. So solutions exist.

Case 2: gcd(8, p) = p. Then p divides 1. But p is a prime, so p divides 1 only if p=1. But 1 is not a prime. Therefore, this case cannot occur.

Wait, hold on! Wait, this suggests that the condition gcd(8, p) divides 1 is always true? Because if gcd(8,p)=p, then p divides 1, which is impossible. Therefore, the congruence 8x≡1 mod p has solutions for all primes p? But this contradicts our earlier analysis where p=2 has no solution.

Wait, what is the problem here? Let me check the general theorem again. The theorem says that ax ≡ b mod m has a solution iff gcd(a, m) divides b.

But in our case, m=p is prime. So if gcd(8, p) divides 1.

But gcd(8, p) is either 1 or p.

So if gcd(8, p) divides 1, then we have two possibilities:

1. If gcd(8, p) = 1: then 1 divides 1, which is true. Therefore, solutions exist.

2. If gcd(8, p) = p: then p divides 1. But p is a prime ≥2, so p divides 1 is false. Therefore, in this case, the congruence has no solution.

Therefore, the congruence has no solution only when gcd(8, p) divides 1 is false, which is when p divides 1? No. Wait, the condition for the congruence to have a solution is that gcd(a, m) divides b. Here, b=1. So the congruence has a solution iff gcd(8, p) divides 1. Which, as above, is true if gcd(8,p)=1 (since 1 divides 1) and false if gcd(8,p)=p (since p divides 1 only if p=1, which is not prime). Therefore, the congruence has no solution iff gcd(8, p) does not divide 1, which is when gcd(8,p)=p. Which happens when p divides 8. Therefore, the primes p dividing 8. Which is only p=2. Therefore, the congruence has no solution only when p=2. Hence, answer is 2.

But then earlier confusion came from misapplying the logic. But no, seems correct. Therefore, the answer is 2.

Alternatively, maybe the problem is in the translation of the question? "There are finitely many primes p for which the congruence... has no solutions". So yes, only p=2, so sum is 2.

Alternatively, maybe the original problem was about equations with different coefficients, but as stated, with 8x≡1 mod p, only p=2.

Alternatively, maybe the question is from a source where they consider p=2 as having solutions? Wait, for p=2, 8x ≡1 mod2 is equivalent to 0x ≡1 mod2, which is 0≡1 mod2, which is false. Therefore, no solution.

Therefore, the conclusion is that the only prime is 2, sum is 2. So answer is \boxed{2}.

**Final Answer**
\boxed{2}

To determine for which prime numbers $ p $ the congruence
$$
8x \equiv 1 \pmod{p}
$$
has **no solutions**, we begin by analyzing a key number theory concept: the solvability of linear congruences.

---

### **Step 1: Solvability Condition for Linear Congruences**

A linear congruence of the form:
$$
ax \equiv b \pmod{m}
$$
has a solution if and only if $ \gcd(a, m) \mid b $.

In our case:
- $ a = 8 $
- $ b = 1 $
- $ m = p $, a prime number

So the congruence has a solution if and only if:
$$
\gcd(8, p) \mid 1
$$

Note that:
- Since $ p $ is prime, $ \gcd(8, p) $ must be either 1 or $ p $.
- The only divisors of 1 are $ \pm1 $, so the condition $ \gcd(8, p) \mid 1 $ holds **if and only if $ \gcd(8, p) = 1 $**.

Let’s analyze the two cases based on the value of $ p $:

---

### **Case 1: $ \gcd(8, p) = 1 $**

This occurs when $ p $ does **not** divide 8. Since the prime divisors of 8 are only $ p = 2 $, for all primes $ p > 2 $, $ \gcd(8, p) = 1 $.

In this case, the congruence has solutions because $ \gcd(8, p) \mid 1 $. For example:
- For $ p = 3 $: $ 8x \equiv 1 \pmod{3} \Rightarrow 2x \equiv 1 \Rightarrow x \equiv 2 \pmod{3} $
- For $ p = 5 $: $ 8x \equiv 1 \pmod{5} \Rightarrow 3x \equiv 1 \Rightarrow x \equiv 2 \pmod{5} $

So the equation is solvable for $ p \neq 2 $.

---

### **Case 2: $ \gcd(8, p) = p $**

This happens only when $ p \mid 8 $, i.e., when $ p = 2 $, since 8's only prime factor is 2.

So let’s test $ p = 2 $:
$$
8x \equiv 1 \pmod{2}
\Rightarrow 0 \cdot x \equiv 1 \pmod{2}
\Rightarrow 0 \equiv 1 \pmod{2}
$$
This is **false**, so the congruence has no solution.

Therefore, **only** when $ p = 2 $ does the congruence $ 8x \equiv 1 \pmod{p} $ have **no solutions**.

---

### **Step 2: Conclusion**

- The only prime $ p $ for which the congruence has **no solutions** is $ p = 2 $.
- Hence, there is **only one** such prime.
- The **sum** of all such primes is:
$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to figure out for which prime numbers p the congruence 8x ≡ 1 mod p has no solutions. Then, sum all those primes. Hmm, let me start by recalling some number theory concepts.

First, the equation 8x ≡ 1 mod p is a linear congruence. I remember that a linear congruence ax ≡ b mod m has a solution if and only if gcd(a, m) divides b. In this case, a is 8, b is 1, and m is p, which is a prime. So, the condition for the congruence to have a solution is that gcd(8, p) divides 1.

Now, since p is a prime, the gcd(8, p) can only be 1 or p. Because if p divides 8, then gcd(8, p) would be p, otherwise it's 1. Wait, but p is a prime. Let me think: If p divides 8, then p must be one of the prime factors of 8. The prime factors of 8 are just 2, since 8 is 2 cubed. Therefore, gcd(8, p) is p if p = 2, otherwise it's 1.

So, the condition gcd(8, p) divides 1. Well, since gcd(8, p) is either 1 or 2. Then, 1 divides 1, which is always true. However, if gcd(8, p) is 2, then we need 2 divides 1? Which is not true. Therefore, the congruence 8x ≡ 1 mod p has no solutions if and only if gcd(8, p) does not divide 1, which happens when gcd(8, p) is 2. Which, as we saw, occurs only when p divides 8. Since p is prime, p must be 2.

Wait, so the only prime p where the congruence 8x ≡ 1 mod p has no solutions is p=2? So, then the sum of all such primes is just 2? But the problem says "there are finitely many primes p", and asks to determine the sum. Maybe there's a mistake here?

Wait, let me check again. Let me test p=2. The congruence is 8x ≡ 1 mod 2. But 8 mod 2 is 0, so 0x ≡ 1 mod 2, which is 0 ≡ 1 mod 2, which is false. So indeed no solution. So p=2 is one such prime.

What about other primes? Let me check p=3. Then the congruence is 8x ≡ 1 mod 3. Since 8 mod 3 is 2, so 2x ≡ 1 mod 3. Then multiplying both sides by inverse of 2 mod 3. The inverse of 2 mod 3 is 2, since 2*2=4≡1 mod3. So x ≡ 2*1=2 mod3. So x=2 is a solution. Therefore, for p=3, there is a solution. Similarly for p=5: 8x≡1 mod5. 8 mod5 is 3, so 3x≡1 mod5. Inverse of 3 mod5 is 2, because 3*2=6≡1 mod5. So x≡2 mod5. So solution exists.

What about p=2? As we saw, no solution. What about p=7? 8 mod7 is 1, so 1x≡1 mod7, so x≡1 mod7. Solution exists.

So seems like only p=2 is the prime where there's no solution. So the answer is sum of all such p is 2. But the problem says "finitely many primes p", which is true here, since only p=2. So is the answer 2?

Wait, but maybe I made a mistake? Let me think again. The general theory says that the equation ax ≡ b mod p has a solution if and only if gcd(a, p) divides b. Since p is prime, gcd(a, p) is 1 if p does not divide a, and p if p divides a. So if p divides a, then the equation becomes 0x ≡ b mod p, which has a solution only if b ≡0 mod p. In our case, b=1. Therefore, if p divides a=8, then the equation becomes 0x≡1 mod p, which has no solution. Therefore, primes p dividing 8. Since 8=2^3, the primes dividing 8 are just 2. Hence, only p=2. Therefore, sum is 2.

So is the answer 2? But maybe there is something else. Wait, the question says "there are finitely many primes p", which is true, but maybe the question is designed to trick us? Let me check again.

Wait, maybe if p divides 8, which is p=2, then there's no solution. However, for other primes, since gcd(8,p)=1, which divides 1, so solutions exist. So only p=2. So the sum is 2. Therefore, answer is 2.

But I feel like maybe the problem is more complicated? Maybe I need to check if 8 is invertible modulo p? Which is equivalent to gcd(8,p)=1. So, if gcd(8,p)=1, then 8 is invertible, so equation has solution. If gcd(8,p)=p, which is only when p divides 8, i.e., p=2, then 8 is not invertible, and the equation 8x≡1 mod p becomes 0x≡1 mod p which is impossible. Therefore, only p=2.

Therefore, the sum is 2. Therefore, the answer is \boxed{2}.

Wait, but the problem says "finitely many primes p", but maybe there are more? For example, if the equation is 8x ≡ 1 mod p. Suppose p divides 8, which is only 2. So only p=2.

Alternatively, maybe there are other primes where 8x ≡ 1 mod p has no solutions? But for any prime p ≠ 2, gcd(8,p)=1, so 8 is invertible mod p, so there exists x such that 8x ≡ 1 mod p. Hence, only p=2.

Alternatively, maybe I need to check if 8 is a unit mod p? Which is true if p does not divide 8, so p≠2. Hence, only p=2 is the prime where 8 is not invertible. Therefore, only p=2.

Therefore, the sum of all such p is 2.

Alternatively, maybe the problem is in the translation? The original problem says "the congruence 8x ≡ 1 mod p has no solutions x". So, for which primes p does this happen? Only p=2. Hence, the answer is 2.

Wait, but maybe the problem is in some contexts where p divides 8, but maybe also other primes? Wait, 8 is 2^3. So the only prime divisor is 2.

Alternatively, maybe the problem is referring to primes for which the equation has no solution, but maybe if p divides 1? Which is impossible.

Alternatively, is there a mistake here? Let me check again. For example, take p=2. 8x ≡1 mod 2. Since 8 is 0 mod 2, so 0x ≡1 mod2. Which is 0≡1 mod2, which is false. So no solution.

Take p=3. 8x≡1 mod3 is 2x≡1 mod3. Multiply both sides by inverse of 2 mod3, which is 2. So x≡2*1=2 mod3. So x=2 is solution.

p=5: 8x≡1 mod5 is 3x≡1 mod5. Inverse of 3 mod5 is 2, since 3*2=6≡1 mod5. So x≡2 mod5.

p=7: 8x≡1 mod7 is 1x≡1 mod7, so x≡1 mod7.

p=11: 8x≡1 mod11. Let me find inverse of 8 mod11. 8*? ≡1 mod11. 8* 7=56≡1 mod11 (since 56-55=1). So x≡7 mod11.

So definitely, for p≠2, there is a solution. So the only prime is 2. Therefore, the answer is 2.

But the problem says "finitely many primes p", which is true, as only 2. So the sum is 2. Therefore, I think the answer is 2. However, I need to check if maybe there's another prime p where 8x≡1 mod p has no solution. Wait, suppose p divides 8, but p is a prime. So only p=2. So no.

Alternatively, maybe the problem is misstated? For instance, if the equation was 8x ≡ 1 mod p has no solutions, meaning that 8 is not invertible. Which is when p divides 8. So again, p=2.

Alternatively, maybe I should think in terms of multiplicative inverses. For the equation 8x ≡1 mod p to have a solution, 8 must have an inverse modulo p. The inverse exists if and only if 8 and p are coprime, i.e., gcd(8, p)=1. Since p is prime, gcd(8,p)=1 unless p divides 8. Therefore, if p divides 8, which is p=2, then the inverse does not exist, so no solution. Therefore, again, only p=2.

Therefore, I think the answer is 2. Maybe the problem is a trick question?

Wait, maybe the question is in the context of modulus p, but maybe p divides 8? But p is prime. So again only p=2.

Alternatively, maybe I made a mistake with the logic. Let me check the general condition again.

The linear congruence ax ≡ b mod m has a solution iff gcd(a, m) divides b.

Here, a=8, b=1, m=p. So gcd(8, p) divides 1.

Since p is prime, gcd(8, p) is either 1 or p.

So if gcd(8, p) divides 1, then either:

Case 1: gcd(8, p) = 1. Then 1 divides 1, which is true. So solutions exist.

Case 2: gcd(8, p) = p. Then p divides 1. But p is a prime, so p divides 1 only if p=1. But 1 is not a prime. Therefore, this case cannot occur.

Wait, hold on! Wait, this suggests that the condition gcd(8, p) divides 1 is always true? Because if gcd(8,p)=p, then p divides 1, which is impossible. Therefore, the congruence 8x≡1 mod p has solutions for all primes p? But this contradicts our earlier analysis where p=2 has no solution.

Wait, what is the problem here? Let me check the general theorem again. The theorem says that ax ≡ b mod m has a solution iff gcd(a, m) divides b.

But in our case, m=p is prime. So if gcd(8, p) divides 1.

But gcd(8, p) is either 1 or p.

So if gcd(8, p) divides 1, then we have two possibilities:

1. If gcd(8, p) = 1: then 1 divides 1, which is true. Therefore, solutions exist.

2. If gcd(8, p) = p: then p divides 1. But p is a prime ≥2, so p divides 1 is false. Therefore, in this case, the congruence has no solution.

Therefore, the congruence has no solution only when gcd(8, p) divides 1 is false, which is when p divides 1? No. Wait, the condition for the congruence to have a solution is that gcd(a, m) divides b. Here, b=1. So the congruence has a solution iff gcd(8, p) divides 1. Which, as above, is true if gcd(8,p)=1 (since 1 divides 1) and false if gcd(8,p)=p (since p divides 1 only if p=1, which is not prime). Therefore, the congruence has no solution iff gcd(8, p) does not divide 1, which is when gcd(8,p)=p. Which happens when p divides 8. Therefore, the primes p dividing 8. Which is only p=2. Therefore, the congruence has no solution only when p=2. Hence, answer is 2.

But then earlier confusion came from misapplying the logic. But no, seems correct. Therefore, the answer is 2.

Alternatively, maybe the problem is in the translation of the question? "There are finitely many primes p for which the congruence... has no solutions". So yes, only p=2, so sum is 2.

Alternatively, maybe the original problem was about equations with different coefficients, but as stated, with 8x≡1 mod p, only p=2.

Alternatively, maybe the question is from a source where they consider p=2 as having solutions? Wait, for p=2, 8x ≡1 mod2 is equivalent to 0x ≡1 mod2, which is 0≡1 mod2, which is false. Therefore, no solution.

Therefore, the conclusion is that the only prime is 2, sum is 2. So answer is \boxed{2}.

**Final Answer**
\boxed{2}

To determine for which prime numbers $ p $ the congruence
$$
8x \equiv 1 \pmod{p}
$$
has **no solutions**, we begin by analyzing a key number theory concept: the solvability of linear congruences.

---

### **Step 1: Solvability Condition for Linear Congruences**

A linear congruence of the form:
$$
ax \equiv b \pmod{m}
$$
has a solution if and only if $ \gcd(a, m) \mid b $.

In our case:
- $ a = 8 $
- $ b = 1 $
- $ m = p $, a prime number

So the congruence has a solution if and only if:
$$
\gcd(8, p) \mid 1
$$

Note that:
- Since $ p $ is prime, $ \gcd(8, p) $ must be either 1 or $ p $.
- The only divisors of 1 are $ \pm1 $, so the condition $ \gcd(8, p) \mid 1 $ holds **if and only if $ \gcd(8, p) = 1 $**.

Let’s analyze the two cases based on the value of $ p $:

---

### **Case 1: $ \gcd(8, p) = 1 $**

This occurs when $ p $ does **not** divide 8. Since the prime divisors of 8 are only $ p = 2 $, for all primes $ p > 2 $, $ \gcd(8, p) = 1 $.

In this case, the congruence has solutions because $ \gcd(8, p) \mid 1 $. For example:
- For $ p = 3 $: $ 8x \equiv 1 \pmod{3} \Rightarrow 2x \equiv 1 \Rightarrow x \equiv 2 \pmod{3} $
- For $ p = 5 $: $ 8x \equiv 1 \pmod{5} \Rightarrow 3x \equiv 1 \Rightarrow x \equiv 2 \pmod{5} $

So the equation is solvable for $ p \neq 2 $.

---

### **Case 2: $ \gcd(8, p) = p $**

This happens only when $ p \mid 8 $, i.e., when $ p = 2 $, since 8's only prime factor is 2.

So let’s test $ p = 2 $:
$$
8x \equiv 1 \pmod{2}
\Rightarrow 0 \cdot x \equiv 1 \pmod{2}
\Rightarrow 0 \equiv 1 \pmod{2}
$$
This is **false**, so the congruence has no solution.

Therefore, **only** when $ p = 2 $ does the congruence $ 8x \equiv 1 \pmod{p} $ have **no solutions**.

---

### **Step 2: Conclusion**

- The only prime $ p $ for which the congruence has **no solutions** is $ p = 2 $.
- Hence, there is **only one** such prime.
- The **sum** of all such primes is:
$$
\boxed{2}
$$

Solution: A solution exists if and only if $8$ is invertible modulo $p$. In other words, $\gcd(8,p)=1$. Since $8=2^3$ is a power of $2$, $8$ is invertible modulo $q$ if and only if $q$ is an odd integer. All primes except for $2$ are odd, so the number we are looking for is $\boxed{2}$.