Explanation:
Okay, so I need to solve the equation $ x = \sqrt{11 - 2x} + 4 $. Hmm, let me think about how to approach this. I remember that when there's a square root in an equation, one common method is to isolate the square root and then square both sides to eliminate it. But I have to be careful because squaring both sides can sometimes introduce extraneous solutions. Let me try that step by step.

First, let me rewrite the equation to make it clearer:

$ x = \sqrt{11 - 2x} + 4 $

I need to isolate the square root term. Let me subtract 4 from both sides to get the square root by itself. So:

$ x - 4 = \sqrt{11 - 2x} $

Okay, now the square root is isolated. The next step is to square both sides of the equation to get rid of the square root. Let me do that:

$ (x - 4)^2 = (\sqrt{11 - 2x})^2 $

Simplifying both sides:

Left side: $ (x - 4)^2 = x^2 - 8x + 16 $

Right side: $ (\sqrt{11 - 2x})^2 = 11 - 2x $

So now the equation becomes:

$ x^2 - 8x + 16 = 11 - 2x $

Now, let me bring all terms to one side to form a quadratic equation. Let me subtract $ 11 - 2x $ from both sides:

$ x^2 - 8x + 16 - 11 + 2x = 0 $

Simplify the terms:

Combine the $ -8x + 2x $ to get $ -6x $

Combine the constants $ 16 - 11 $ to get $ 5 $

So the equation becomes:

$ x^2 - 6x + 5 = 0 $

Now, I need to solve this quadratic equation. Let me try factoring first. Looking for two numbers that multiply to 5 and add up to -6. Well, 5 factors into 1 and 5. Since the middle term is -6x, and the constant term is +5, both numbers should be negative. Let me check:

$ (x - 1)(x - 5) = x^2 - 6x + 5 $. Yes, that works!

So, the factored form is:

$ (x - 1)(x - 5) = 0 $

Setting each factor equal to zero gives the solutions:

$ x - 1 = 0 $ => $ x = 1 $

$ x - 5 = 0 $ => $ x = 5 $

But wait, since I squared both sides of the equation earlier, I need to check these solutions in the original equation to make sure they are not extraneous. Let me do that.

First, check $ x = 1 $:

Original equation: $ x = \sqrt{11 - 2x} + 4 $

Left side: 1

Right side: $ \sqrt{11 - 2(1)} + 4 = \sqrt{11 - 2} + 4 = \sqrt{9} + 4 = 3 + 4 = 7 $

So, 1 vs 7. Not equal. Therefore, $ x = 1 $ is an extraneous solution and must be rejected.

Now check $ x = 5 $:

Left side: 5

Right side: $ \sqrt{11 - 2(5)} + 4 = \sqrt{11 - 10} + 4 = \sqrt{1} + 4 = 1 + 4 = 5 $

So, 5 equals 5. That works. Therefore, $ x = 5 $ is a valid solution.

Wait, but let me just make sure I didn't make any mistakes in my algebra steps. Let me go through them again quickly.

Original equation: $ x = \sqrt{11 - 2x} + 4 $

Subtract 4: $ x - 4 = \sqrt{11 - 2x} $

Square both sides: $ (x - 4)^2 = 11 - 2x $

Expand left side: $ x^2 - 8x + 16 = 11 - 2x $

Bring all terms to left: $ x^2 - 8x + 16 - 11 + 2x = x^2 - 6x + 5 = 0 $

Factor: $ (x - 1)(x - 5) = 0 $. Correct.

Solutions 1 and 5. Check them, 1 doesn't work, 5 does. So only 5 is the solution.

But let me think again: when I squared both sides, could there be another solution? Or maybe I missed something?

Alternatively, maybe I can check the domain of the original equation. The expression under the square root must be non-negative. So:

$ 11 - 2x \geq 0 $

Which implies:

$ -2x \geq -11 $

Divide both sides by -2 (remember to reverse inequality sign):

$ x \leq 11/2 $

So x must be less than or equal to 5.5. Both solutions, 1 and 5, are less than 5.5, so they are in the domain. But even though they are in the domain, one of them didn't satisfy the original equation. So that's why we have to check.

So, conclusion: only x = 5 is the valid solution.

Wait, but let me think again. When I squared both sides, sometimes you can introduce solutions that don't satisfy the original equation because squaring is not a reversible step. So even though x=1 is in the domain, when we squared, we might have introduced it. So checking is essential.

Therefore, the only solution is x=5.

But let me try another approach just to confirm. Let me suppose that x = sqrt(11 - 2x) + 4. Let me denote sqrt(11 - 2x) as y. Then, the equation becomes x = y + 4. Also, from the definition of y, we have y = sqrt(11 - 2x). Therefore, squaring both sides, y² = 11 - 2x. But since x = y + 4, substitute into this equation:

y² = 11 - 2(y + 4)

Simplify:

y² = 11 - 2y - 8

Which is:

y² = 3 - 2y

Bring all terms to left:

y² + 2y - 3 = 0

Factor:

Looking for two numbers that multiply to -3 and add to 2. Those numbers are 3 and -1.

So, (y + 3)(y - 1) = 0

Therefore, y = -3 or y = 1.

But y is sqrt(11 - 2x), which is always non-negative. Therefore, y = -3 is invalid. So only y = 1 is valid.

Therefore, y = 1. Then, since x = y + 4, x = 1 + 4 = 5. So again, x=5 is the solution. This method also gives x=5. So that confirms it.

Therefore, the only solution is 5. So I think that's solid.

But just to make sure, let me plug x=5 back into the original equation again:

Left side: 5

Right side: sqrt(11 - 2*5) + 4 = sqrt(11 - 10) + 4 = sqrt(1) + 4 = 1 + 4 = 5. Yep, same.

And x=1, as before, gives 1 vs 7. So definitely not.

Therefore, the answer is x=5.

I think that's thorough. I considered different methods, checked for extraneous solutions, verified the domain, and confirmed through substitution. So I feel confident that x=5 is the only solution.

**Final Answer**
The value of $ x $ that satisfies the equation is \boxed{5}.

To solve the equation:

$$
x = \sqrt{11 - 2x} + 4
$$

we begin by isolating the square root term. Subtract 4 from both sides:

$$
x - 4 = \sqrt{11 - 2x}
$$

Next, square both sides to eliminate the square root:

$$
(x - 4)^2 = (\sqrt{11 - 2x})^2
$$

Simplify both sides:

$$
x^2 - 8x + 16 = 11 - 2x
$$

Now, bring all terms to one side to form a quadratic equation:

$$
x^2 - 8x + 16 - 11 + 2x = 0
$$

$$
x^2 - 6x + 5 = 0
$$

Factor the quadratic:

$$
(x - 1)(x - 5) = 0
$$

This gives two potential solutions:

$$
x = 1 \quad \text{or} \quad x = 5
$$

---

### Step 3: Check for Extraneous Solutions

Since we squared both sides of the equation, we must verify each solution in the original equation to ensure it satisfies it.

**Check $ x = 1 $:**

$$
\text{Left side: } x = 1 \\
\text{Right side: } \sqrt{11 - 2(1)} + 4 = \sqrt{9} + 4 = 3 + 4 = 7
$$

Since $ 1 \neq 7 $, $ x = 1 $ is **not** a valid solution.

**Check $ x = 5 $:**

$$
\text{Left side: } x = 5 \\
\text{Right side: } \sqrt{11 - 2(5)} + 4 = \sqrt{1} + 4 = 1 + 4 = 5
$$

Since $ 5 = 5 $, $ x = 5 $ is a valid solution.

---

### Step 4: Domain Consideration

The expression under the square root must be non-negative:

$$
11 - 2x \geq 0 \Rightarrow x \leq \frac{11}{2} = 5.5
$$

Both $ x = 1 $ and $ x = 5 $ satisfy this condition, but only $ x = 5 $ satisfies the original equation.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $ x = \sqrt{11 - 2x} + 4 $. Hmm, let me think about how to approach this. I remember that when there's a square root in an equation, one common method is to isolate the square root and then square both sides to eliminate it. But I have to be careful because squaring both sides can sometimes introduce extraneous solutions. Let me try that step by step.

First, let me rewrite the equation to make it clearer:

$ x = \sqrt{11 - 2x} + 4 $

I need to isolate the square root term. Let me subtract 4 from both sides to get the square root by itself. So:

$ x - 4 = \sqrt{11 - 2x} $

Okay, now the square root is isolated. The next step is to square both sides of the equation to get rid of the square root. Let me do that:

$ (x - 4)^2 = (\sqrt{11 - 2x})^2 $

Simplifying both sides:

Left side: $ (x - 4)^2 = x^2 - 8x + 16 $

Right side: $ (\sqrt{11 - 2x})^2 = 11 - 2x $

So now the equation becomes:

$ x^2 - 8x + 16 = 11 - 2x $

Now, let me bring all terms to one side to form a quadratic equation. Let me subtract $ 11 - 2x $ from both sides:

$ x^2 - 8x + 16 - 11 + 2x = 0 $

Simplify the terms:

Combine the $ -8x + 2x $ to get $ -6x $

Combine the constants $ 16 - 11 $ to get $ 5 $

So the equation becomes:

$ x^2 - 6x + 5 = 0 $

Now, I need to solve this quadratic equation. Let me try factoring first. Looking for two numbers that multiply to 5 and add up to -6. Well, 5 factors into 1 and 5. Since the middle term is -6x, and the constant term is +5, both numbers should be negative. Let me check:

$ (x - 1)(x - 5) = x^2 - 6x + 5 $. Yes, that works!

So, the factored form is:

$ (x - 1)(x - 5) = 0 $

Setting each factor equal to zero gives the solutions:

$ x - 1 = 0 $ => $ x = 1 $

$ x - 5 = 0 $ => $ x = 5 $

But wait, since I squared both sides of the equation earlier, I need to check these solutions in the original equation to make sure they are not extraneous. Let me do that.

First, check $ x = 1 $:

Original equation: $ x = \sqrt{11 - 2x} + 4 $

Left side: 1

Right side: $ \sqrt{11 - 2(1)} + 4 = \sqrt{11 - 2} + 4 = \sqrt{9} + 4 = 3 + 4 = 7 $

So, 1 vs 7. Not equal. Therefore, $ x = 1 $ is an extraneous solution and must be rejected.

Now check $ x = 5 $:

Left side: 5

Right side: $ \sqrt{11 - 2(5)} + 4 = \sqrt{11 - 10} + 4 = \sqrt{1} + 4 = 1 + 4 = 5 $

So, 5 equals 5. That works. Therefore, $ x = 5 $ is a valid solution.

Wait, but let me just make sure I didn't make any mistakes in my algebra steps. Let me go through them again quickly.

Original equation: $ x = \sqrt{11 - 2x} + 4 $

Subtract 4: $ x - 4 = \sqrt{11 - 2x} $

Square both sides: $ (x - 4)^2 = 11 - 2x $

Expand left side: $ x^2 - 8x + 16 = 11 - 2x $

Bring all terms to left: $ x^2 - 8x + 16 - 11 + 2x = x^2 - 6x + 5 = 0 $

Factor: $ (x - 1)(x - 5) = 0 $. Correct.

Solutions 1 and 5. Check them, 1 doesn't work, 5 does. So only 5 is the solution.

But let me think again: when I squared both sides, could there be another solution? Or maybe I missed something?

Alternatively, maybe I can check the domain of the original equation. The expression under the square root must be non-negative. So:

$ 11 - 2x \geq 0 $

Which implies:

$ -2x \geq -11 $

Divide both sides by -2 (remember to reverse inequality sign):

$ x \leq 11/2 $

So x must be less than or equal to 5.5. Both solutions, 1 and 5, are less than 5.5, so they are in the domain. But even though they are in the domain, one of them didn't satisfy the original equation. So that's why we have to check.

So, conclusion: only x = 5 is the valid solution.

Wait, but let me think again. When I squared both sides, sometimes you can introduce solutions that don't satisfy the original equation because squaring is not a reversible step. So even though x=1 is in the domain, when we squared, we might have introduced it. So checking is essential.

Therefore, the only solution is x=5.

But let me try another approach just to confirm. Let me suppose that x = sqrt(11 - 2x) + 4. Let me denote sqrt(11 - 2x) as y. Then, the equation becomes x = y + 4. Also, from the definition of y, we have y = sqrt(11 - 2x). Therefore, squaring both sides, y² = 11 - 2x. But since x = y + 4, substitute into this equation:

y² = 11 - 2(y + 4)

Simplify:

y² = 11 - 2y - 8

Which is:

y² = 3 - 2y

Bring all terms to left:

y² + 2y - 3 = 0

Factor:

Looking for two numbers that multiply to -3 and add to 2. Those numbers are 3 and -1.

So, (y + 3)(y - 1) = 0

Therefore, y = -3 or y = 1.

But y is sqrt(11 - 2x), which is always non-negative. Therefore, y = -3 is invalid. So only y = 1 is valid.

Therefore, y = 1. Then, since x = y + 4, x = 1 + 4 = 5. So again, x=5 is the solution. This method also gives x=5. So that confirms it.

Therefore, the only solution is 5. So I think that's solid.

But just to make sure, let me plug x=5 back into the original equation again:

Left side: 5

Right side: sqrt(11 - 2*5) + 4 = sqrt(11 - 10) + 4 = sqrt(1) + 4 = 1 + 4 = 5. Yep, same.

And x=1, as before, gives 1 vs 7. So definitely not.

Therefore, the answer is x=5.

I think that's thorough. I considered different methods, checked for extraneous solutions, verified the domain, and confirmed through substitution. So I feel confident that x=5 is the only solution.

**Final Answer**
The value of $ x $ that satisfies the equation is \boxed{5}.

To solve the equation:

$$
x = \sqrt{11 - 2x} + 4
$$

we begin by isolating the square root term. Subtract 4 from both sides:

$$
x - 4 = \sqrt{11 - 2x}
$$

Next, square both sides to eliminate the square root:

$$
(x - 4)^2 = (\sqrt{11 - 2x})^2
$$

Simplify both sides:

$$
x^2 - 8x + 16 = 11 - 2x
$$

Now, bring all terms to one side to form a quadratic equation:

$$
x^2 - 8x + 16 - 11 + 2x = 0
$$

$$
x^2 - 6x + 5 = 0
$$

Factor the quadratic:

$$
(x - 1)(x - 5) = 0
$$

This gives two potential solutions:

$$
x = 1 \quad \text{or} \quad x = 5
$$

---

### Step 3: Check for Extraneous Solutions

Since we squared both sides of the equation, we must verify each solution in the original equation to ensure it satisfies it.

**Check $ x = 1 $:**

$$
\text{Left side: } x = 1 \\
\text{Right side: } \sqrt{11 - 2(1)} + 4 = \sqrt{9} + 4 = 3 + 4 = 7
$$

Since $ 1 \neq 7 $, $ x = 1 $ is **not** a valid solution.

**Check $ x = 5 $:**

$$
\text{Left side: } x = 5 \\
\text{Right side: } \sqrt{11 - 2(5)} + 4 = \sqrt{1} + 4 = 1 + 4 = 5
$$

Since $ 5 = 5 $, $ x = 5 $ is a valid solution.

---

### Step 4: Domain Consideration

The expression under the square root must be non-negative:

$$
11 - 2x \geq 0 \Rightarrow x \leq \frac{11}{2} = 5.5
$$

Both $ x = 1 $ and $ x = 5 $ satisfy this condition, but only $ x = 5 $ satisfies the original equation.

---

### Final Answer

$$
\boxed{5}
$$

Solution: We first isolate the square root, so we can then square both sides to get rid of it.  Subtracting 4 from both sides gives $x-4 = \!\sqrt{11-2x}$.  Squaring both sides gives $x^2 - 8x + 16 = 11-2x$, or $x^2 -6x + 5=0$.  Factoring gives $(x-5)(x-1)=0$, so $x=5$ or $x=1$.  Because we squared the equation, we must  check if our solutions are extraneous. For $x=5$, the equation reads $5 = \!\sqrt{11-10} + 4$, which is true. If $x=1$, we have $1 = \!\sqrt{11-2} + 4$, which is not true, so $x=1$ is extraneous. Therefore, our only solution is $\boxed{x=5}$.


Explanation:
Okay, so there's this worker who gets $20,000 every year and deposits it into a savings account at the end of each year. After three years, he wants to have at least $66,200 in the account to buy a house. The question is asking for the minimal compound interest rate that the savings account must provide. Hmm, let me think about how to approach this.

First, I remember that compound interest means that each year, the interest is calculated on the current amount in the account, including previously earned interest. Since the worker is making deposits at the end of each year, this sounds like an annuity problem where payments are made at the end of each period. So, maybe I need to use the future value of an ordinary annuity formula?

Let me recall the formula for the future value of an ordinary annuity. The formula is:

FV = P * [( (1 + r)^n - 1 ) / r ]

Where:
- FV is the future value
- P is the annual payment (deposit)
- r is the annual interest rate
- n is the number of periods (years)

In this case, the worker is making three deposits of $20,000 each, so P is $20,000, n is 3 years, and FV needs to be at least $66,200. We need to solve for r.

So plugging the numbers into the formula:

66,200 = 20,000 * [ ( (1 + r)^3 - 1 ) / r ]

Let me write that equation out:

66200 = 20000 * [ ( (1 + r)^3 - 1 ) / r ]

First, I can divide both sides by 20,000 to simplify:

66200 / 20000 = [ ( (1 + r)^3 - 1 ) / r ]

Calculating the left side:

66200 divided by 20000. Let me compute that. 20000 goes into 66200 three times (since 20000*3=60000), and then there's 6200 left. 6200 / 20000 = 0.31. So total is 3.31.

So:

3.31 = [ ( (1 + r)^3 - 1 ) / r ]

So now, we have the equation:

( (1 + r)^3 - 1 ) / r = 3.31

We need to solve for r here. This seems like a nonlinear equation and might not have an algebraic solution, so maybe I need to use trial and error or some numerical method. Alternatively, maybe I can rearrange the equation?

Let me write the equation again:

( (1 + r)^3 - 1 ) / r = 3.31

Multiply both sides by r:

(1 + r)^3 - 1 = 3.31r

Then:

(1 + r)^3 = 3.31r + 1

Expanding the left side:

(1 + r)^3 = 1 + 3r + 3r^2 + r^3

So:

1 + 3r + 3r^2 + r^3 = 3.31r + 1

Subtract 3.31r + 1 from both sides:

1 + 3r + 3r^2 + r^3 - 3.31r - 1 = 0

Simplify:

(1 - 1) + (3r - 3.31r) + 3r^2 + r^3 = 0

Which is:

0 - 0.31r + 3r^2 + r^3 = 0

So:

r^3 + 3r^2 - 0.31r = 0

Factor out an r:

r(r^2 + 3r - 0.31) = 0

So, the solutions are r = 0 or solutions to r^2 + 3r - 0.31 = 0.

But r = 0 is not a valid interest rate here, so we need to solve the quadratic equation:

r^2 + 3r - 0.31 = 0

Wait, hold on. Let me check my algebra again because this seems conflicting. Let me go back step by step.

Original equation after expanding:

(1 + r)^3 = 3.31r + 1

Expanding left side:

1 + 3r + 3r^2 + r^3 = 3.31r + 1

Subtracting 3.31r + 1 from both sides:

Left side minus right side:

(1 + 3r + 3r^2 + r^3) - (3.31r + 1) = 0

Which is:

1 - 1 + 3r - 3.31r + 3r^2 + r^3 = 0

Simplify:

0 - 0.31r + 3r^2 + r^3 = 0

Which is:

r^3 + 3r^2 - 0.31r = 0

Factor out r:

r(r^2 + 3r - 0.31) = 0

So, yes, that's correct. So the quadratic equation is r^2 + 3r - 0.31 = 0.

Wait, but solving this quadratic equation would give us the roots. Let me compute the discriminant:

Discriminant D = b^2 - 4ac = (3)^2 - 4*(1)*(-0.31) = 9 + 1.24 = 10.24

Then, the roots are:

r = [ -b ± sqrt(D) ] / 2a = [ -3 ± sqrt(10.24) ] / 2

sqrt(10.24) is 3.2, since 3.2^2 = 10.24.

Therefore:

r = [ -3 ± 3.2 ] / 2

So two solutions:

First solution: [ -3 + 3.2 ] / 2 = 0.2 / 2 = 0.1

Second solution: [ -3 - 3.2 ] / 2 = -6.2 / 2 = -3.1

So the solutions are r = 0.1 or r = -3.1. Since interest rate can't be negative, we take r = 0.1, which is 10%.

Wait, but hold on. If we plug r = 0.1 into the original equation, does it satisfy?

Let me check:

Original equation after dividing by 20,000 was:

( (1 + r)^3 - 1 ) / r = 3.31

Let me compute left side with r = 0.1:

( (1 + 0.1)^3 - 1 ) / 0.1 = ( (1.1)^3 - 1 ) / 0.1

Calculate 1.1^3: 1.1 * 1.1 = 1.21, then 1.21 * 1.1 = 1.331

So 1.331 - 1 = 0.331

Divide by 0.1: 0.331 / 0.1 = 3.31

Which matches the right side. So r = 0.1 is indeed the solution. So 10%?

But wait, the problem says "the minimal compound interest rate that the savings account must provide". So 10%? But let me think again. Because when I set up the equation, I assumed that the future value is exactly 66,200. But the problem says "at least 66,200", so maybe there's a possibility that a lower rate could also result in 66,200? But according to the equation, there's only one positive solution, which is 10%. So that must be the minimal rate. Wait, but maybe I made a mistake in setting up the equation?

Wait, let me check again. The formula for future value of an ordinary annuity is indeed FV = P * [ ((1 + r)^n - 1)/r ].

But let me verify with r = 10% and see if the future value is indeed 66,200.

So, deposits are made at the end of each year. So first deposit is at the end of year 1, which earns interest for 2 years. Second deposit at end of year 2, earns interest for 1 year. Third deposit at end of year 3, earns no interest.

So let's compute manually:

First deposit: $20,000 at end of year 1. After 2 years, it becomes 20,000*(1 + 0.1)^2 = 20,000*1.21 = 24,200

Second deposit: $20,000 at end of year 2. After 1 year, it becomes 20,000*(1 + 0.1) = 22,000

Third deposit: $20,000 at end of year 3. No interest, so remains 20,000

Total FV = 24,200 + 22,000 + 20,000 = 66,200. Exactly. So that's correct. Therefore, 10% is the exact rate. Therefore, the minimal rate is 10%.

But the problem says "the minimal compound interest rate that the savings account must provide". So 10% is the answer? But let me check if maybe there's a lower rate? Wait, according to the equation, there's only one positive solution. So if you have a lower rate, then the future value would be less than 66,200. Let me test with, say, 9%.

Let me compute FV with r = 9% (0.09):

First deposit: 20,000*(1.09)^2 = 20,000*1.1881 = 23,762

Second deposit: 20,000*(1.09) = 21,800

Third deposit: 20,000

Total FV = 23,762 + 21,800 + 20,000 = 65,562, which is less than 66,200. So 9% is insufficient.

What about 10% gives exactly 66,200. Therefore, 10% is the minimal rate. Therefore, the answer is 10.

But let me check if there's a possibility that maybe the interest is compounded more frequently? Wait, the problem says "compound interest rate", but it doesn't specify the compounding frequency. However, since the deposits are made annually, and the question is about the annual interest rate, it's standard to assume that the interest is compounded annually. So the answer is 10%.

But let me check again the equation. When I set up the equation, I used the future value of an ordinary annuity formula, which is correct for annual deposits and annual compounding. Therefore, solving gives 10%, which works. Therefore, the answer is 10.

But let me think again. The problem says "the minimal compound interest rate". So maybe there's a different way to model it? For example, if the interest is compounded more frequently, but the deposits are made annually. However, the question says "compound interest rate", which is usually expressed as an annual rate. If it's compounded more frequently, the effective annual rate would be different. However, since the problem doesn't specify the compounding frequency, it's standard to assume annual compounding. Therefore, 10% is correct.

Alternatively, if the problem had said "the account compounds monthly", then we would have to adjust the rate accordingly. But since it's not specified, we assume annual compounding. Therefore, 10% is the answer.

Wait, but let me check once more. Let me think if there's another approach. For example, maybe using the formula for each deposit:

First deposit: end of year 1, so it earns interest for 2 years.

Second deposit: end of year 2, earns interest for 1 year.

Third deposit: end of year 3, earns no interest.

So total amount is:

20,000*(1 + r)^2 + 20,000*(1 + r) + 20,000 = 66,200

Which is the same as:

20,000[(1 + r)^2 + (1 + r) + 1] = 66,200

Divide both sides by 20,000:

(1 + r)^2 + (1 + r) + 1 = 3.31

Let me compute that:

Let me denote x = 1 + r, so the equation becomes:

x^2 + x + 1 = 3.31

Which simplifies to:

x^2 + x + 1 - 3.31 = 0 => x^2 + x - 2.31 = 0

Solving for x:

x = [ -1 ± sqrt(1 + 4*2.31) ] / 2

Compute discriminant:

1 + 9.24 = 10.24, sqrt(10.24) = 3.2

Therefore:

x = [ -1 + 3.2 ] / 2 = 2.2 / 2 = 1.1

Or x = [ -1 - 3.2 ] / 2 = negative value, which we can ignore.

Therefore, x = 1.1, so 1 + r = 1.1 => r = 0.1, which is 10%. Same result. So that confirms it again.

Therefore, the minimal interest rate is 10%.

But let me check once more with another rate, say 10.5%, just to see if maybe there's a lower rate? Wait, no, since 10% gives exactly 66,200, and lower rates give less. Therefore, 10% is the minimal rate required. Therefore, the answer is 10.

**Final Answer**
\boxed{10}

To determine the **minimal compound interest rate** that allows the worker to accumulate at least \$66,200 in his savings account after three years, we analyze the situation using the **future value of an ordinary annuity**.

---

### **Understanding the Problem**

- The worker deposits **\$20,000** at the **end of each year** for **3 years**.
- The goal is to have **at least \$66,200** in the account **by the end of the third year**.
- The interest is **compounded annually**, and we are to find the **minimum annual interest rate** that achieves this.

---

### **Step 1: Use the Future Value of an Ordinary Annuity Formula**

The formula for the future value of an ordinary annuity is:

$$
FV = P \cdot \left( \frac{(1 + r)^n - 1}{r} \right)
$$

Where:
- $ FV $ = future value = \$66,200
- $ P $ = annual payment = \$20,000
- $ r $ = annual interest rate (to be found)
- $ n $ = number of years = 3

Substitute the known values:

$$
66200 = 20000 \cdot \left( \frac{(1 + r)^3 - 1}{r} \right)
$$

---

### **Step 2: Simplify the Equation**

Divide both sides by 20,000:

$$
\frac{66200}{20000} = \frac{(1 + r)^3 - 1}{r}
$$

$$
3.31 = \frac{(1 + r)^3 - 1}{r}
$$

Multiply both sides by $ r $:

$$
3.31r = (1 + r)^3 - 1
$$

Now expand $ (1 + r)^3 $:

$$
(1 + r)^3 = 1 + 3r + 3r^2 + r^3
$$

Substitute back:

$$
3.31r = 1 + 3r + 3r^2 + r^3 - 1
$$

$$
3.31r = 3r + 3r^2 + r^3
$$

Bring all terms to one side:

$$
r^3 + 3r^2 - 0.31r = 0
$$

Factor out $ r $:

$$
r(r^2 + 3r - 0.31) = 0
$$

So, the solutions are:

- $ r = 0 $ (not valid, as it implies no interest)
- Solve the quadratic: $ r^2 + 3r - 0.31 = 0 $

---

### **Step 3: Solve the Quadratic Equation**

Use the quadratic formula:

$$
r = \frac{-3 \pm \sqrt{3^2 - 4(1)(-0.31)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 1.24}}{2} = \frac{-3 \pm \sqrt{10.24}}{2}
$$

$$
\sqrt{10.24} = 3.2
$$

$$
r = \frac{-3 + 3.2}{2} = \frac{0.2}{2} = 0.1
$$

$$
r = \frac{-3 - 3.2}{2} = -3.1 \quad \text{(discard, negative rate)}
$$

So, the only valid solution is:

$$
r = 0.1 = 10\%
$$

---

### **Step 4: Verify the Result**

Let’s verify that with $ r = 10\% $, the future value is indeed \$66,200.

- First deposit: \$20,000 at end of year 1 → earns interest for 2 years:
  $ 20,000 \cdot (1.1)^2 = 20,000 \cdot 1.21 = 24,200 $

- Second deposit: \$20,000 at end of year 2 → earns interest for 1 year:
  $ 20,000 \cdot 1.1 = 22,000 $

- Third deposit: \$20,000 at end of year 3 → no interest:
  $ 20,000 $

Total:

$$
24,200 + 22,000 + 20,000 = 66,200
$$

✅ The calculation checks out.

---

### **Conclusion**

The **minimum compound interest rate** required to accumulate at least \$66,200 in the account after three years of annual deposits of \$20,000 is:

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so there's this worker who gets $20,000 every year and deposits it into a savings account at the end of each year. After three years, he wants to have at least $66,200 in the account to buy a house. The question is asking for the minimal compound interest rate that the savings account must provide. Hmm, let me think about how to approach this.

First, I remember that compound interest means that each year, the interest is calculated on the current amount in the account, including previously earned interest. Since the worker is making deposits at the end of each year, this sounds like an annuity problem where payments are made at the end of each period. So, maybe I need to use the future value of an ordinary annuity formula?

Let me recall the formula for the future value of an ordinary annuity. The formula is:

FV = P * [( (1 + r)^n - 1 ) / r ]

Where:
- FV is the future value
- P is the annual payment (deposit)
- r is the annual interest rate
- n is the number of periods (years)

In this case, the worker is making three deposits of $20,000 each, so P is $20,000, n is 3 years, and FV needs to be at least $66,200. We need to solve for r.

So plugging the numbers into the formula:

66,200 = 20,000 * [ ( (1 + r)^3 - 1 ) / r ]

Let me write that equation out:

66200 = 20000 * [ ( (1 + r)^3 - 1 ) / r ]

First, I can divide both sides by 20,000 to simplify:

66200 / 20000 = [ ( (1 + r)^3 - 1 ) / r ]

Calculating the left side:

66200 divided by 20000. Let me compute that. 20000 goes into 66200 three times (since 20000*3=60000), and then there's 6200 left. 6200 / 20000 = 0.31. So total is 3.31.

So:

3.31 = [ ( (1 + r)^3 - 1 ) / r ]

So now, we have the equation:

( (1 + r)^3 - 1 ) / r = 3.31

We need to solve for r here. This seems like a nonlinear equation and might not have an algebraic solution, so maybe I need to use trial and error or some numerical method. Alternatively, maybe I can rearrange the equation?

Let me write the equation again:

( (1 + r)^3 - 1 ) / r = 3.31

Multiply both sides by r:

(1 + r)^3 - 1 = 3.31r

Then:

(1 + r)^3 = 3.31r + 1

Expanding the left side:

(1 + r)^3 = 1 + 3r + 3r^2 + r^3

So:

1 + 3r + 3r^2 + r^3 = 3.31r + 1

Subtract 3.31r + 1 from both sides:

1 + 3r + 3r^2 + r^3 - 3.31r - 1 = 0

Simplify:

(1 - 1) + (3r - 3.31r) + 3r^2 + r^3 = 0

Which is:

0 - 0.31r + 3r^2 + r^3 = 0

So:

r^3 + 3r^2 - 0.31r = 0

Factor out an r:

r(r^2 + 3r - 0.31) = 0

So, the solutions are r = 0 or solutions to r^2 + 3r - 0.31 = 0.

But r = 0 is not a valid interest rate here, so we need to solve the quadratic equation:

r^2 + 3r - 0.31 = 0

Wait, hold on. Let me check my algebra again because this seems conflicting. Let me go back step by step.

Original equation after expanding:

(1 + r)^3 = 3.31r + 1

Expanding left side:

1 + 3r + 3r^2 + r^3 = 3.31r + 1

Subtracting 3.31r + 1 from both sides:

Left side minus right side:

(1 + 3r + 3r^2 + r^3) - (3.31r + 1) = 0

Which is:

1 - 1 + 3r - 3.31r + 3r^2 + r^3 = 0

Simplify:

0 - 0.31r + 3r^2 + r^3 = 0

Which is:

r^3 + 3r^2 - 0.31r = 0

Factor out r:

r(r^2 + 3r - 0.31) = 0

So, yes, that's correct. So the quadratic equation is r^2 + 3r - 0.31 = 0.

Wait, but solving this quadratic equation would give us the roots. Let me compute the discriminant:

Discriminant D = b^2 - 4ac = (3)^2 - 4*(1)*(-0.31) = 9 + 1.24 = 10.24

Then, the roots are:

r = [ -b ± sqrt(D) ] / 2a = [ -3 ± sqrt(10.24) ] / 2

sqrt(10.24) is 3.2, since 3.2^2 = 10.24.

Therefore:

r = [ -3 ± 3.2 ] / 2

So two solutions:

First solution: [ -3 + 3.2 ] / 2 = 0.2 / 2 = 0.1

Second solution: [ -3 - 3.2 ] / 2 = -6.2 / 2 = -3.1

So the solutions are r = 0.1 or r = -3.1. Since interest rate can't be negative, we take r = 0.1, which is 10%.

Wait, but hold on. If we plug r = 0.1 into the original equation, does it satisfy?

Let me check:

Original equation after dividing by 20,000 was:

( (1 + r)^3 - 1 ) / r = 3.31

Let me compute left side with r = 0.1:

( (1 + 0.1)^3 - 1 ) / 0.1 = ( (1.1)^3 - 1 ) / 0.1

Calculate 1.1^3: 1.1 * 1.1 = 1.21, then 1.21 * 1.1 = 1.331

So 1.331 - 1 = 0.331

Divide by 0.1: 0.331 / 0.1 = 3.31

Which matches the right side. So r = 0.1 is indeed the solution. So 10%?

But wait, the problem says "the minimal compound interest rate that the savings account must provide". So 10%? But let me think again. Because when I set up the equation, I assumed that the future value is exactly 66,200. But the problem says "at least 66,200", so maybe there's a possibility that a lower rate could also result in 66,200? But according to the equation, there's only one positive solution, which is 10%. So that must be the minimal rate. Wait, but maybe I made a mistake in setting up the equation?

Wait, let me check again. The formula for future value of an ordinary annuity is indeed FV = P * [ ((1 + r)^n - 1)/r ].

But let me verify with r = 10% and see if the future value is indeed 66,200.

So, deposits are made at the end of each year. So first deposit is at the end of year 1, which earns interest for 2 years. Second deposit at end of year 2, earns interest for 1 year. Third deposit at end of year 3, earns no interest.

So let's compute manually:

First deposit: $20,000 at end of year 1. After 2 years, it becomes 20,000*(1 + 0.1)^2 = 20,000*1.21 = 24,200

Second deposit: $20,000 at end of year 2. After 1 year, it becomes 20,000*(1 + 0.1) = 22,000

Third deposit: $20,000 at end of year 3. No interest, so remains 20,000

Total FV = 24,200 + 22,000 + 20,000 = 66,200. Exactly. So that's correct. Therefore, 10% is the exact rate. Therefore, the minimal rate is 10%.

But the problem says "the minimal compound interest rate that the savings account must provide". So 10% is the answer? But let me check if maybe there's a lower rate? Wait, according to the equation, there's only one positive solution. So if you have a lower rate, then the future value would be less than 66,200. Let me test with, say, 9%.

Let me compute FV with r = 9% (0.09):

First deposit: 20,000*(1.09)^2 = 20,000*1.1881 = 23,762

Second deposit: 20,000*(1.09) = 21,800

Third deposit: 20,000

Total FV = 23,762 + 21,800 + 20,000 = 65,562, which is less than 66,200. So 9% is insufficient.

What about 10% gives exactly 66,200. Therefore, 10% is the minimal rate. Therefore, the answer is 10.

But let me check if there's a possibility that maybe the interest is compounded more frequently? Wait, the problem says "compound interest rate", but it doesn't specify the compounding frequency. However, since the deposits are made annually, and the question is about the annual interest rate, it's standard to assume that the interest is compounded annually. So the answer is 10%.

But let me check again the equation. When I set up the equation, I used the future value of an ordinary annuity formula, which is correct for annual deposits and annual compounding. Therefore, solving gives 10%, which works. Therefore, the answer is 10.

But let me think again. The problem says "the minimal compound interest rate". So maybe there's a different way to model it? For example, if the interest is compounded more frequently, but the deposits are made annually. However, the question says "compound interest rate", which is usually expressed as an annual rate. If it's compounded more frequently, the effective annual rate would be different. However, since the problem doesn't specify the compounding frequency, it's standard to assume annual compounding. Therefore, 10% is correct.

Alternatively, if the problem had said "the account compounds monthly", then we would have to adjust the rate accordingly. But since it's not specified, we assume annual compounding. Therefore, 10% is the answer.

Wait, but let me check once more. Let me think if there's another approach. For example, maybe using the formula for each deposit:

First deposit: end of year 1, so it earns interest for 2 years.

Second deposit: end of year 2, earns interest for 1 year.

Third deposit: end of year 3, earns no interest.

So total amount is:

20,000*(1 + r)^2 + 20,000*(1 + r) + 20,000 = 66,200

Which is the same as:

20,000[(1 + r)^2 + (1 + r) + 1] = 66,200

Divide both sides by 20,000:

(1 + r)^2 + (1 + r) + 1 = 3.31

Let me compute that:

Let me denote x = 1 + r, so the equation becomes:

x^2 + x + 1 = 3.31

Which simplifies to:

x^2 + x + 1 - 3.31 = 0 => x^2 + x - 2.31 = 0

Solving for x:

x = [ -1 ± sqrt(1 + 4*2.31) ] / 2

Compute discriminant:

1 + 9.24 = 10.24, sqrt(10.24) = 3.2

Therefore:

x = [ -1 + 3.2 ] / 2 = 2.2 / 2 = 1.1

Or x = [ -1 - 3.2 ] / 2 = negative value, which we can ignore.

Therefore, x = 1.1, so 1 + r = 1.1 => r = 0.1, which is 10%. Same result. So that confirms it again.

Therefore, the minimal interest rate is 10%.

But let me check once more with another rate, say 10.5%, just to see if maybe there's a lower rate? Wait, no, since 10% gives exactly 66,200, and lower rates give less. Therefore, 10% is the minimal rate required. Therefore, the answer is 10.

**Final Answer**
\boxed{10}

To determine the **minimal compound interest rate** that allows the worker to accumulate at least \$66,200 in his savings account after three years, we analyze the situation using the **future value of an ordinary annuity**.

---

### **Understanding the Problem**

- The worker deposits **\$20,000** at the **end of each year** for **3 years**.
- The goal is to have **at least \$66,200** in the account **by the end of the third year**.
- The interest is **compounded annually**, and we are to find the **minimum annual interest rate** that achieves this.

---

### **Step 1: Use the Future Value of an Ordinary Annuity Formula**

The formula for the future value of an ordinary annuity is:

$$
FV = P \cdot \left( \frac{(1 + r)^n - 1}{r} \right)
$$

Where:
- $ FV $ = future value = \$66,200
- $ P $ = annual payment = \$20,000
- $ r $ = annual interest rate (to be found)
- $ n $ = number of years = 3

Substitute the known values:

$$
66200 = 20000 \cdot \left( \frac{(1 + r)^3 - 1}{r} \right)
$$

---

### **Step 2: Simplify the Equation**

Divide both sides by 20,000:

$$
\frac{66200}{20000} = \frac{(1 + r)^3 - 1}{r}
$$

$$
3.31 = \frac{(1 + r)^3 - 1}{r}
$$

Multiply both sides by $ r $:

$$
3.31r = (1 + r)^3 - 1
$$

Now expand $ (1 + r)^3 $:

$$
(1 + r)^3 = 1 + 3r + 3r^2 + r^3
$$

Substitute back:

$$
3.31r = 1 + 3r + 3r^2 + r^3 - 1
$$

$$
3.31r = 3r + 3r^2 + r^3
$$

Bring all terms to one side:

$$
r^3 + 3r^2 - 0.31r = 0
$$

Factor out $ r $:

$$
r(r^2 + 3r - 0.31) = 0
$$

So, the solutions are:

- $ r = 0 $ (not valid, as it implies no interest)
- Solve the quadratic: $ r^2 + 3r - 0.31 = 0 $

---

### **Step 3: Solve the Quadratic Equation**

Use the quadratic formula:

$$
r = \frac{-3 \pm \sqrt{3^2 - 4(1)(-0.31)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 1.24}}{2} = \frac{-3 \pm \sqrt{10.24}}{2}
$$

$$
\sqrt{10.24} = 3.2
$$

$$
r = \frac{-3 + 3.2}{2} = \frac{0.2}{2} = 0.1
$$

$$
r = \frac{-3 - 3.2}{2} = -3.1 \quad \text{(discard, negative rate)}
$$

So, the only valid solution is:

$$
r = 0.1 = 10\%
$$

---

### **Step 4: Verify the Result**

Let’s verify that with $ r = 10\% $, the future value is indeed \$66,200.

- First deposit: \$20,000 at end of year 1 → earns interest for 2 years:
  $ 20,000 \cdot (1.1)^2 = 20,000 \cdot 1.21 = 24,200 $

- Second deposit: \$20,000 at end of year 2 → earns interest for 1 year:
  $ 20,000 \cdot 1.1 = 22,000 $

- Third deposit: \$20,000 at end of year 3 → no interest:
  $ 20,000 $

Total:

$$
24,200 + 22,000 + 20,000 = 66,200
$$

✅ The calculation checks out.

---

### **Conclusion**

The **minimum compound interest rate** required to accumulate at least \$66,200 in the account after three years of annual deposits of \$20,000 is:

$$
\boxed{10}
$$

Solution: If the interest rate is $r$, it follows that $$20000(1+r)^2 + 20000(1+r) + 20000 \ge 66200.$$ If we set $x = 1+r$ and divide through the inequality by $200$, it follows that $$100x^2 + 100x - 231 \ge 0.$$ Since $231 = 11 \cdot 21$, we can factor the quadratic as $(10x - 11)(10x + 21) \ge 0$, so it follows that $x \ge \frac {11}{10}$ or $x \le \frac{-21}{10}$. Since we are looking for an interest rate percentage, it follows that $x \ge \frac{11}{10} = 1.1$, and $r = x - 1 = \boxed{10}\%$.


Explanation:
Okay, so I need to solve this functional equation problem. The function f satisfies the equation f(x) + f(y) = f(x + y) - xy - 1 for all real numbers x and y, and we're given that f(1) = 1. The task is to find all integers n such that f(n) = n. Alright, let's start by understanding the problem and figuring out a strategy.

First, functional equations often require finding the form of the function f. Since the equation holds for all real numbers x and y, maybe I can assume a particular form for f, like a quadratic function, because the equation involves terms like xy, which suggests that f might have a quadratic component. Let me try that approach.

Let me suppose that f(x) is a quadratic function. Let me write f(x) = ax² + bx + c, where a, b, c are constants to be determined. Then, let's plug this into the functional equation and see if I can find a, b, c.

So, let's compute f(x) + f(y):

f(x) + f(y) = a x² + b x + c + a y² + b y + c = a(x² + y²) + b(x + y) + 2c.

On the other hand, f(x + y) - xy - 1 is:

a(x + y)² + b(x + y) + c - xy - 1.

Let me expand that:

a(x² + 2xy + y²) + b(x + y) + c - xy - 1 = a x² + 2a xy + a y² + b x + b y + c - xy - 1.

Now, let's set the two expressions equal:

Left side: a(x² + y²) + b(x + y) + 2c.

Right side: a x² + 2a xy + a y² + b x + b y + c - xy - 1.

So, equate left and right:

a(x² + y²) + b(x + y) + 2c = a x² + 2a xy + a y² + b x + b y + c - xy - 1.

Let me subtract the left side from both sides to bring everything to one side:

0 = [a x² + 2a xy + a y² + b x + b y + c - xy - 1] - [a(x² + y²) + b(x + y) + 2c]

Simplify term by term:

First, a x² - a x² = 0, same with a y² - a y² = 0.

Then, 2a xy - xy = (2a - 1)xy.

Then, b x - b x = 0, same with b y - b y = 0.

Then, c - 2c = -c.

Then, -1 remains.

So overall:

0 = (2a - 1)xy - c - 1.

But this equation must hold for all real numbers x and y. Therefore, the coefficients of the terms involving x and y must be zero. However, here we have a term with xy and a constant term. Since the left side is zero for all x and y, the coefficients of xy and the constant term must each be zero.

Therefore:

Coefficient of xy: 2a - 1 = 0 ⇒ 2a = 1 ⇒ a = 1/2.

Constant term: -c - 1 = 0 ⇒ -c = 1 ⇒ c = -1.

So, we have a = 1/2, c = -1. The coefficient b hasn't been determined yet. Wait, but in the above equation, there was no term with b, because when we subtracted the left side, the b terms canceled out. So, does that mean that b can be any value? But that can't be, because the original functional equation must hold for all x and y. Wait, but we already used the equation to find a and c. However, the original equation didn't involve b? Wait, let me check again.

Wait, when I subtracted the left side from the right side, the terms with b canceled out. Therefore, the equation reduces to 0 = (2a - 1)xy - c - 1. Therefore, the value of b doesn't affect this equation. Therefore, does that mean that b can be arbitrary? But that seems odd. However, we also have the condition that f(1) = 1. Let me use that to find b.

So, f(x) = (1/2)x² + b x - 1. Then f(1) = (1/2)(1)^2 + b(1) - 1 = 1/2 + b - 1 = b - 1/2. And this is given to be 1. Therefore:

b - 1/2 = 1 ⇒ b = 1 + 1/2 = 3/2.

Therefore, the function is f(x) = (1/2)x² + (3/2)x - 1.

Let me check if this function satisfies the original functional equation. Let me verify with some values.

First, let me check with x = 1 and y = 1. Then:

Left side: f(1) + f(1) = 1 + 1 = 2.

Right side: f(1 + 1) - (1)(1) - 1 = f(2) - 1 - 1 = f(2) - 2.

Compute f(2): (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Therefore, right side is 4 - 2 = 2. So left equals right. Good.

Another check: x = 0, y = 0.

Left side: f(0) + f(0) = 2f(0).

Compute f(0): (1/2)(0) + (3/2)(0) - 1 = -1. Therefore, left side is 2*(-1) = -2.

Right side: f(0 + 0) - 0*0 - 1 = f(0) - 0 - 1 = -1 - 1 = -2. So that works.

Another check: x = 1, y = 0.

Left side: f(1) + f(0) = 1 + (-1) = 0.

Right side: f(1 + 0) - (1)(0) - 1 = f(1) - 0 - 1 = 1 - 1 = 0. Good.

Another check: x = 2, y = 3.

Left side: f(2) + f(3). f(2) is 4 as before. f(3) = (1/2)(9) + (3/2)(3) - 1 = 4.5 + 4.5 - 1 = 8. So left side is 4 + 8 = 12.

Right side: f(5) - (2)(3) - 1 = f(5) - 6 - 1 = f(5) - 7.

Compute f(5): (1/2)(25) + (3/2)(5) - 1 = 12.5 + 7.5 - 1 = 19. Therefore, right side is 19 - 7 = 12. Which matches. So seems like the function works.

Therefore, the function is f(x) = (1/2)x² + (3/2)x - 1. Now, the problem asks for all integers n such that f(n) = n. So, we need to solve the equation f(n) = n for integer n.

So, let's write the equation:

(1/2)n² + (3/2)n - 1 = n.

Multiply both sides by 2 to eliminate denominators:

n² + 3n - 2 = 2n.

Bring all terms to left side:

n² + 3n - 2 - 2n = n² + n - 2 = 0.

So, quadratic equation: n² + n - 2 = 0.

Let me solve this quadratic equation. The discriminant is D = 1² - 4*1*(-2) = 1 + 8 = 9. Therefore, solutions are:

n = [-1 ± √9]/2 = [-1 ± 3]/2.

So, two solutions:

n = (-1 + 3)/2 = 2/2 = 1,

and

n = (-1 - 3)/2 = -4/2 = -2.

Therefore, the integer solutions are n = 1 and n = -2.

But wait, the problem says "find all integers n such that f(n) = n." So, we need to check if these are indeed solutions. Let me verify.

First, n = 1: f(1) is given as 1, so that's correct.

n = -2: Let's compute f(-2):

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Yes, that's correct. So f(-2) = -2.

Therefore, the integers are 1 and -2.

But wait, the problem says "all integers n such that f(n) = n." So, are there any other integers? Let me check if the quadratic equation n² + n - 2 = 0 has only these two roots. Since it's a quadratic, there are only two roots, so only n = 1 and n = -2. Therefore, these are the only integer solutions.

But let me just make sure that there are no other integer solutions. For example, suppose someone might think maybe there are more solutions? But since the equation reduces to a quadratic, which only has two roots, so unless there's some mistake in my derivation, these are the only solutions.

Wait, let me check my steps again. Starting from f(n) = n:

(1/2)n² + (3/2)n - 1 = n.

Multiply by 2: n² + 3n - 2 = 2n ⇒ n² + n - 2 = 0. Correct.

Solutions: n = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2. So 1 and -2. Correct.

Therefore, the answer should be integers 1 and -2.

But let me check if there's another way to approach this problem, maybe without assuming f is quadratic. Maybe the functional equation can be manipulated to find f(n) directly?

Let me try that approach as a verification.

Given the functional equation:

f(x) + f(y) = f(x + y) - xy - 1.

Let me rearrange the equation:

f(x + y) = f(x) + f(y) + xy + 1.

This resembles the Cauchy equation but with an extra term. For Cauchy's equation, f(x + y) = f(x) + f(y), but here we have an extra term. Maybe we can find a function g(x) such that f(x) = g(x) + something, to convert the equation into a standard form.

Let me try to find such a function. Let me suppose that f(x) = g(x) + ax² + bx + c. Wait, but earlier I assumed a quadratic function. Alternatively, maybe think of the equation as similar to a quadratic function.

Alternatively, let me consider the equation:

f(x + y) - f(x) - f(y) = xy + 1.

This is similar to the equation for quadratic functions. For example, if f(x) is quadratic, then f(x + y) - f(x) - f(y) would be something like 2a xy + ... Let me check.

Suppose f(x) = ax² + bx + c. Then f(x + y) - f(x) - f(y) = a(x + y)^2 + b(x + y) + c - ax² - bx - c - ay² - by - c = a(x² + 2xy + y²) + b(x + y) + c - ax² - bx - c - ay² - by - c = 2a xy - c. Therefore, according to the functional equation, this should equal xy + 1. Therefore:

2a xy - c = xy + 1. Therefore, equating coefficients:

2a = 1 ⇒ a = 1/2,

and -c = 1 ⇒ c = -1.

Which is exactly what I found earlier. Then, the linear term b is not determined by this equation, but we can find it using the given condition f(1) = 1. Therefore, this approach also leads to the same result. Therefore, the function is indeed f(x) = (1/2)x² + bx - 1, and with f(1) = 1, we find b = 3/2. Therefore, the function is as before.

Therefore, solving f(n) = n gives n = 1 and n = -2. Therefore, the answer is these two integers.

But just to be thorough, let me check if there's another way to approach the problem, maybe by using specific values for x and y to find f(n).

For example, let me set y = 0. Then the functional equation becomes:

f(x) + f(0) = f(x + 0) - x*0 - 1 ⇒ f(x) + f(0) = f(x) - 0 - 1 ⇒ f(0) = -1.

So, f(0) = -1. Which matches with our previous result, since f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. Good.

Another idea: Let me set y = 1. Then the equation becomes:

f(x) + f(1) = f(x + 1) - x*1 - 1 ⇒ f(x) + 1 = f(x + 1) - x - 1 ⇒ f(x + 1) = f(x) + x + 2.

This gives a recursive relation for f(x + 1) in terms of f(x). Maybe we can use this to find f(n) for integer n?

Since we know f(1) = 1, let's compute f(2), f(3), etc., and also f(0), f(-1), etc.

We already know f(0) = -1.

Compute f(1) = 1.

Compute f(2) = f(1 + 1) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Which matches our previous result.

Compute f(3) = f(2 + 1) = f(2) + 2 + 2 = 4 + 2 + 2 = 8. Which matches our previous calculation (f(3) = 8).

Similarly, f(4) = f(3) + 3 + 2 = 8 + 5 = 13. Let me check with the formula: f(4) = (1/2)(16) + (3/2)(4) - 1 = 8 + 6 - 1 = 13. Correct.

Similarly, for negative integers, let's try to compute f(-1). Let me use the equation with x = -1 and y = 1:

f(-1) + f(1) = f(0) - (-1)(1) - 1 ⇒ f(-1) + 1 = -1 + 1 - 1 ⇒ f(-1) + 1 = -1 ⇒ f(-1) = -2.

Alternatively, using the recursive relation. Let me see if I can find a relation for negative numbers. Let me set x = -1 and y = 1, but maybe another approach.

Alternatively, use the equation f(x + 1) = f(x) + x + 2. Let me solve this recurrence relation for integer x. Let me think of x as an integer. Let me consider x as an integer variable. Let me try to find f(n) for integer n.

We have f(n + 1) = f(n) + n + 2. This is a recurrence relation. Let me solve it.

First, let me write the recurrence:

f(n + 1) - f(n) = n + 2.

This is a linear recurrence. To solve it, we can telescope the sum.

For integer n, starting from some base case. Let me assume n is an integer. Let me compute f(n) in terms of f(0).

For example, f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1. Which matches.

f(2) = f(1) + 1 + 2 = 1 + 3 = 4.

Similarly, f(3) = f(2) + 2 + 2 = 4 + 4 = 8.

In general, for n ≥ 0, f(n) = f(0) + sum_{k=0}^{n-1} (k + 2).

Wait, let me check:

Starting from f(0), then f(1) = f(0) + (0 + 2), f(2) = f(1) + (1 + 2), ..., f(n) = f(0) + sum_{k=0}^{n-1} (k + 2).

Therefore, sum_{k=0}^{n-1} (k + 2) = sum_{k=0}^{n-1} k + sum_{k=0}^{n-1} 2 = [ (n - 1)n / 2 ] + 2n.

Therefore, f(n) = -1 + [ (n² - n)/2 + 2n ] = -1 + (n² - n + 4n)/2 = -1 + (n² + 3n)/2 = (n² + 3n)/2 - 1 = (n² + 3n - 2)/2.

Wait, but according to our previous formula, f(n) = (1/2)n² + (3/2)n - 1, which is the same as (n² + 3n - 2)/2. So yes, that's consistent. Therefore, for integer n ≥ 0, f(n) = (n² + 3n - 2)/2.

Similarly, for negative integers, we can use the recurrence relation. Let me try to find f(-1). From earlier, we found f(-1) = -2. Let me check with the formula:

For n = -1, f(-1) = ((-1)^2 + 3*(-1) - 2)/2 = (1 - 3 - 2)/2 = (-4)/2 = -2. Correct.

Similarly, let's compute f(-2). Using the recurrence relation, let me use x = -2 and y = 2:

Wait, maybe better to use the equation f(x + y) = f(x) + f(y) + xy + 1. Let me set x = -1, y = -1:

f(-1 + -1) = f(-1) + f(-1) + (-1)(-1) + 1 ⇒ f(-2) = 2*(-2) + 1 + 1 ⇒ f(-2) = -4 + 2 = -2? Wait, let me compute again:

Wait, f(-2) = f(-1) + f(-1) + (-1)(-1) + 1 = (-2) + (-2) + 1 + 1 = -4 + 2 = -2. Which matches our previous result. Also, using the formula: f(-2) = ((-2)^2 + 3*(-2) - 2)/2 = (4 - 6 - 2)/2 = (-4)/2 = -2. Correct.

Alternatively, using the recurrence relation for negative numbers. Let me think. If I have f(n + 1) = f(n) + n + 2, then for negative n, if I set m = n + 1, then for n = m - 1, f(m) = f(m - 1) + (m - 1) + 2 = f(m - 1) + m + 1. Therefore, for m = 0, f(0) = f(-1) + (-1) + 2 ⇒ -1 = f(-1) + 1 ⇒ f(-1) = -2. Which is correct. Similarly, for m = -1, f(-1) = f(-2) + (-2) + 2 ⇒ -2 = f(-2) + 0 ⇒ f(-2) = -2. Which is correct. So, the formula f(n) = (n² + 3n - 2)/2 holds for all integers n.

Therefore, solving f(n) = n:

(n² + 3n - 2)/2 = n ⇒ n² + 3n - 2 = 2n ⇒ n² + n - 2 = 0 ⇒ same quadratic equation as before, leading to n = 1 and n = -2. Therefore, the same solutions.

Therefore, regardless of the approach, the solutions are n = 1 and n = -2.

But just to be thorough, let me check for other integers. For example, n = 0: f(0) = -1 ≠ 0. n = 2: f(2) = 4 ≠ 2. n = -1: f(-1) = -2 ≠ -1. n = 3: f(3) = 8 ≠ 3. n = -2: f(-2) = -2, which is equal. n = -3: f(-3) = ((-3)^2 + 3*(-3) - 2)/2 = (9 - 9 - 2)/2 = (-2)/2 = -1. So f(-3) = -1 ≠ -3. Similarly, n = 4: f(4) = (16 + 12 - 2)/2 = 26/2 = 13 ≠ 4. So, only n = 1 and n = -2 satisfy f(n) = n.

Therefore, the answer is integers 1 and -2.

**Final Answer**
The integers $ n $ such that $ f(n) = n $ are \boxed{-2} and \boxed{1}.

\boxed{-2}, \boxed{1}

To solve the functional equation:

$$
f(x) + f(y) = f(x + y) - xy - 1
$$

for all real numbers $ x $ and $ y $, and given that $ f(1) = 1 $, we aim to find all **integers** $ n $ such that $ f(n) = n $.

---

### Step 1: Assume a Form for $ f(x) $

Given the structure of the equation, which includes a quadratic term $ xy $, it's reasonable to assume that $ f(x) $ is a **quadratic function**. Let us suppose:

$$
f(x) = ax^2 + bx + c
$$

Substituting into the functional equation:

$$
f(x) + f(y) = a(x^2 + y^2) + b(x + y) + 2c
$$
$$
f(x + y) - xy - 1 = a(x + y)^2 + b(x + y) + c - xy - 1
$$

Expanding the right-hand side:

$$
a(x^2 + 2xy + y^2) + b(x + y) + c - xy - 1 = a x^2 + 2a xy + a y^2 + b x + b y + c - xy - 1
$$

Now equating both sides:

$$
a(x^2 + y^2) + b(x + y) + 2c = a x^2 + 2a xy + a y^2 + b x + b y + c - xy - 1
$$

Subtracting the left-hand side from the right-hand side:

$$
0 = (2a - 1)xy - c - 1
$$

This must hold for all $ x, y $, so the coefficients of $ xy $ and the constant term must be zero:

- $ 2a - 1 = 0 \Rightarrow a = \frac{1}{2} $
- $ -c - 1 = 0 \Rightarrow c = -1 $

Now, we use the given condition $ f(1) = 1 $ to find $ b $:

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

Set this equal to 1:

$$
b - \frac{1}{2} = 1 \Rightarrow b = \frac{3}{2}
$$

Thus, the function is:

$$
f(x) = \frac{1}{2}x^2 + \frac{3}{2}x - 1
$$

---

### Step 2: Solve $ f(n) = n $ for Integer $ n $

We now solve:

$$
\frac{1}{2}n^2 + \frac{3}{2}n - 1 = n
$$

Multiply both sides by 2 to eliminate fractions:

$$
n^2 + 3n - 2 = 2n
$$

Bring all terms to one side:

$$
n^2 + n - 2 = 0
$$

Solve the quadratic:

$$
n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}
$$

So the solutions are:

$$
n = \frac{-1 + 3}{2} = 1, \quad n = \frac{-1 - 3}{2} = -2
$$

---

### Step 3: Verify the Solutions

- $ f(1) = \frac{1}{2}(1)^2 + \frac{3}{2}(1) - 1 = \frac{1}{2} + \frac{3}{2} - 1 = 1 $ ✅
- $ f(-2) = \frac{1}{2}(4) + \frac{3}{2}(-2) - 1 = 2 - 3 - 1 = -2 $ ✅

Thus, both $ n = 1 $ and $ n = -2 $ satisfy $ f(n) = n $.

---

### Final Answer

$$
\boxed{-2}, \boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to solve this functional equation problem. The function f satisfies the equation f(x) + f(y) = f(x + y) - xy - 1 for all real numbers x and y, and we're given that f(1) = 1. The task is to find all integers n such that f(n) = n. Alright, let's start by understanding the problem and figuring out a strategy.

First, functional equations often require finding the form of the function f. Since the equation holds for all real numbers x and y, maybe I can assume a particular form for f, like a quadratic function, because the equation involves terms like xy, which suggests that f might have a quadratic component. Let me try that approach.

Let me suppose that f(x) is a quadratic function. Let me write f(x) = ax² + bx + c, where a, b, c are constants to be determined. Then, let's plug this into the functional equation and see if I can find a, b, c.

So, let's compute f(x) + f(y):

f(x) + f(y) = a x² + b x + c + a y² + b y + c = a(x² + y²) + b(x + y) + 2c.

On the other hand, f(x + y) - xy - 1 is:

a(x + y)² + b(x + y) + c - xy - 1.

Let me expand that:

a(x² + 2xy + y²) + b(x + y) + c - xy - 1 = a x² + 2a xy + a y² + b x + b y + c - xy - 1.

Now, let's set the two expressions equal:

Left side: a(x² + y²) + b(x + y) + 2c.

Right side: a x² + 2a xy + a y² + b x + b y + c - xy - 1.

So, equate left and right:

a(x² + y²) + b(x + y) + 2c = a x² + 2a xy + a y² + b x + b y + c - xy - 1.

Let me subtract the left side from both sides to bring everything to one side:

0 = [a x² + 2a xy + a y² + b x + b y + c - xy - 1] - [a(x² + y²) + b(x + y) + 2c]

Simplify term by term:

First, a x² - a x² = 0, same with a y² - a y² = 0.

Then, 2a xy - xy = (2a - 1)xy.

Then, b x - b x = 0, same with b y - b y = 0.

Then, c - 2c = -c.

Then, -1 remains.

So overall:

0 = (2a - 1)xy - c - 1.

But this equation must hold for all real numbers x and y. Therefore, the coefficients of the terms involving x and y must be zero. However, here we have a term with xy and a constant term. Since the left side is zero for all x and y, the coefficients of xy and the constant term must each be zero.

Therefore:

Coefficient of xy: 2a - 1 = 0 ⇒ 2a = 1 ⇒ a = 1/2.

Constant term: -c - 1 = 0 ⇒ -c = 1 ⇒ c = -1.

So, we have a = 1/2, c = -1. The coefficient b hasn't been determined yet. Wait, but in the above equation, there was no term with b, because when we subtracted the left side, the b terms canceled out. So, does that mean that b can be any value? But that can't be, because the original functional equation must hold for all x and y. Wait, but we already used the equation to find a and c. However, the original equation didn't involve b? Wait, let me check again.

Wait, when I subtracted the left side from the right side, the terms with b canceled out. Therefore, the equation reduces to 0 = (2a - 1)xy - c - 1. Therefore, the value of b doesn't affect this equation. Therefore, does that mean that b can be arbitrary? But that seems odd. However, we also have the condition that f(1) = 1. Let me use that to find b.

So, f(x) = (1/2)x² + b x - 1. Then f(1) = (1/2)(1)^2 + b(1) - 1 = 1/2 + b - 1 = b - 1/2. And this is given to be 1. Therefore:

b - 1/2 = 1 ⇒ b = 1 + 1/2 = 3/2.

Therefore, the function is f(x) = (1/2)x² + (3/2)x - 1.

Let me check if this function satisfies the original functional equation. Let me verify with some values.

First, let me check with x = 1 and y = 1. Then:

Left side: f(1) + f(1) = 1 + 1 = 2.

Right side: f(1 + 1) - (1)(1) - 1 = f(2) - 1 - 1 = f(2) - 2.

Compute f(2): (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Therefore, right side is 4 - 2 = 2. So left equals right. Good.

Another check: x = 0, y = 0.

Left side: f(0) + f(0) = 2f(0).

Compute f(0): (1/2)(0) + (3/2)(0) - 1 = -1. Therefore, left side is 2*(-1) = -2.

Right side: f(0 + 0) - 0*0 - 1 = f(0) - 0 - 1 = -1 - 1 = -2. So that works.

Another check: x = 1, y = 0.

Left side: f(1) + f(0) = 1 + (-1) = 0.

Right side: f(1 + 0) - (1)(0) - 1 = f(1) - 0 - 1 = 1 - 1 = 0. Good.

Another check: x = 2, y = 3.

Left side: f(2) + f(3). f(2) is 4 as before. f(3) = (1/2)(9) + (3/2)(3) - 1 = 4.5 + 4.5 - 1 = 8. So left side is 4 + 8 = 12.

Right side: f(5) - (2)(3) - 1 = f(5) - 6 - 1 = f(5) - 7.

Compute f(5): (1/2)(25) + (3/2)(5) - 1 = 12.5 + 7.5 - 1 = 19. Therefore, right side is 19 - 7 = 12. Which matches. So seems like the function works.

Therefore, the function is f(x) = (1/2)x² + (3/2)x - 1. Now, the problem asks for all integers n such that f(n) = n. So, we need to solve the equation f(n) = n for integer n.

So, let's write the equation:

(1/2)n² + (3/2)n - 1 = n.

Multiply both sides by 2 to eliminate denominators:

n² + 3n - 2 = 2n.

Bring all terms to left side:

n² + 3n - 2 - 2n = n² + n - 2 = 0.

So, quadratic equation: n² + n - 2 = 0.

Let me solve this quadratic equation. The discriminant is D = 1² - 4*1*(-2) = 1 + 8 = 9. Therefore, solutions are:

n = [-1 ± √9]/2 = [-1 ± 3]/2.

So, two solutions:

n = (-1 + 3)/2 = 2/2 = 1,

and

n = (-1 - 3)/2 = -4/2 = -2.

Therefore, the integer solutions are n = 1 and n = -2.

But wait, the problem says "find all integers n such that f(n) = n." So, we need to check if these are indeed solutions. Let me verify.

First, n = 1: f(1) is given as 1, so that's correct.

n = -2: Let's compute f(-2):

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Yes, that's correct. So f(-2) = -2.

Therefore, the integers are 1 and -2.

But wait, the problem says "all integers n such that f(n) = n." So, are there any other integers? Let me check if the quadratic equation n² + n - 2 = 0 has only these two roots. Since it's a quadratic, there are only two roots, so only n = 1 and n = -2. Therefore, these are the only integer solutions.

But let me just make sure that there are no other integer solutions. For example, suppose someone might think maybe there are more solutions? But since the equation reduces to a quadratic, which only has two roots, so unless there's some mistake in my derivation, these are the only solutions.

Wait, let me check my steps again. Starting from f(n) = n:

(1/2)n² + (3/2)n - 1 = n.

Multiply by 2: n² + 3n - 2 = 2n ⇒ n² + n - 2 = 0. Correct.

Solutions: n = [-1 ± sqrt(1 + 8)] / 2 = [-1 ± 3]/2. So 1 and -2. Correct.

Therefore, the answer should be integers 1 and -2.

But let me check if there's another way to approach this problem, maybe without assuming f is quadratic. Maybe the functional equation can be manipulated to find f(n) directly?

Let me try that approach as a verification.

Given the functional equation:

f(x) + f(y) = f(x + y) - xy - 1.

Let me rearrange the equation:

f(x + y) = f(x) + f(y) + xy + 1.

This resembles the Cauchy equation but with an extra term. For Cauchy's equation, f(x + y) = f(x) + f(y), but here we have an extra term. Maybe we can find a function g(x) such that f(x) = g(x) + something, to convert the equation into a standard form.

Let me try to find such a function. Let me suppose that f(x) = g(x) + ax² + bx + c. Wait, but earlier I assumed a quadratic function. Alternatively, maybe think of the equation as similar to a quadratic function.

Alternatively, let me consider the equation:

f(x + y) - f(x) - f(y) = xy + 1.

This is similar to the equation for quadratic functions. For example, if f(x) is quadratic, then f(x + y) - f(x) - f(y) would be something like 2a xy + ... Let me check.

Suppose f(x) = ax² + bx + c. Then f(x + y) - f(x) - f(y) = a(x + y)^2 + b(x + y) + c - ax² - bx - c - ay² - by - c = a(x² + 2xy + y²) + b(x + y) + c - ax² - bx - c - ay² - by - c = 2a xy - c. Therefore, according to the functional equation, this should equal xy + 1. Therefore:

2a xy - c = xy + 1. Therefore, equating coefficients:

2a = 1 ⇒ a = 1/2,

and -c = 1 ⇒ c = -1.

Which is exactly what I found earlier. Then, the linear term b is not determined by this equation, but we can find it using the given condition f(1) = 1. Therefore, this approach also leads to the same result. Therefore, the function is indeed f(x) = (1/2)x² + bx - 1, and with f(1) = 1, we find b = 3/2. Therefore, the function is as before.

Therefore, solving f(n) = n gives n = 1 and n = -2. Therefore, the answer is these two integers.

But just to be thorough, let me check if there's another way to approach the problem, maybe by using specific values for x and y to find f(n).

For example, let me set y = 0. Then the functional equation becomes:

f(x) + f(0) = f(x + 0) - x*0 - 1 ⇒ f(x) + f(0) = f(x) - 0 - 1 ⇒ f(0) = -1.

So, f(0) = -1. Which matches with our previous result, since f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. Good.

Another idea: Let me set y = 1. Then the equation becomes:

f(x) + f(1) = f(x + 1) - x*1 - 1 ⇒ f(x) + 1 = f(x + 1) - x - 1 ⇒ f(x + 1) = f(x) + x + 2.

This gives a recursive relation for f(x + 1) in terms of f(x). Maybe we can use this to find f(n) for integer n?

Since we know f(1) = 1, let's compute f(2), f(3), etc., and also f(0), f(-1), etc.

We already know f(0) = -1.

Compute f(1) = 1.

Compute f(2) = f(1 + 1) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Which matches our previous result.

Compute f(3) = f(2 + 1) = f(2) + 2 + 2 = 4 + 2 + 2 = 8. Which matches our previous calculation (f(3) = 8).

Similarly, f(4) = f(3) + 3 + 2 = 8 + 5 = 13. Let me check with the formula: f(4) = (1/2)(16) + (3/2)(4) - 1 = 8 + 6 - 1 = 13. Correct.

Similarly, for negative integers, let's try to compute f(-1). Let me use the equation with x = -1 and y = 1:

f(-1) + f(1) = f(0) - (-1)(1) - 1 ⇒ f(-1) + 1 = -1 + 1 - 1 ⇒ f(-1) + 1 = -1 ⇒ f(-1) = -2.

Alternatively, using the recursive relation. Let me see if I can find a relation for negative numbers. Let me set x = -1 and y = 1, but maybe another approach.

Alternatively, use the equation f(x + 1) = f(x) + x + 2. Let me solve this recurrence relation for integer x. Let me think of x as an integer. Let me consider x as an integer variable. Let me try to find f(n) for integer n.

We have f(n + 1) = f(n) + n + 2. This is a recurrence relation. Let me solve it.

First, let me write the recurrence:

f(n + 1) - f(n) = n + 2.

This is a linear recurrence. To solve it, we can telescope the sum.

For integer n, starting from some base case. Let me assume n is an integer. Let me compute f(n) in terms of f(0).

For example, f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1. Which matches.

f(2) = f(1) + 1 + 2 = 1 + 3 = 4.

Similarly, f(3) = f(2) + 2 + 2 = 4 + 4 = 8.

In general, for n ≥ 0, f(n) = f(0) + sum_{k=0}^{n-1} (k + 2).

Wait, let me check:

Starting from f(0), then f(1) = f(0) + (0 + 2), f(2) = f(1) + (1 + 2), ..., f(n) = f(0) + sum_{k=0}^{n-1} (k + 2).

Therefore, sum_{k=0}^{n-1} (k + 2) = sum_{k=0}^{n-1} k + sum_{k=0}^{n-1} 2 = [ (n - 1)n / 2 ] + 2n.

Therefore, f(n) = -1 + [ (n² - n)/2 + 2n ] = -1 + (n² - n + 4n)/2 = -1 + (n² + 3n)/2 = (n² + 3n)/2 - 1 = (n² + 3n - 2)/2.

Wait, but according to our previous formula, f(n) = (1/2)n² + (3/2)n - 1, which is the same as (n² + 3n - 2)/2. So yes, that's consistent. Therefore, for integer n ≥ 0, f(n) = (n² + 3n - 2)/2.

Similarly, for negative integers, we can use the recurrence relation. Let me try to find f(-1). From earlier, we found f(-1) = -2. Let me check with the formula:

For n = -1, f(-1) = ((-1)^2 + 3*(-1) - 2)/2 = (1 - 3 - 2)/2 = (-4)/2 = -2. Correct.

Similarly, let's compute f(-2). Using the recurrence relation, let me use x = -2 and y = 2:

Wait, maybe better to use the equation f(x + y) = f(x) + f(y) + xy + 1. Let me set x = -1, y = -1:

f(-1 + -1) = f(-1) + f(-1) + (-1)(-1) + 1 ⇒ f(-2) = 2*(-2) + 1 + 1 ⇒ f(-2) = -4 + 2 = -2? Wait, let me compute again:

Wait, f(-2) = f(-1) + f(-1) + (-1)(-1) + 1 = (-2) + (-2) + 1 + 1 = -4 + 2 = -2. Which matches our previous result. Also, using the formula: f(-2) = ((-2)^2 + 3*(-2) - 2)/2 = (4 - 6 - 2)/2 = (-4)/2 = -2. Correct.

Alternatively, using the recurrence relation for negative numbers. Let me think. If I have f(n + 1) = f(n) + n + 2, then for negative n, if I set m = n + 1, then for n = m - 1, f(m) = f(m - 1) + (m - 1) + 2 = f(m - 1) + m + 1. Therefore, for m = 0, f(0) = f(-1) + (-1) + 2 ⇒ -1 = f(-1) + 1 ⇒ f(-1) = -2. Which is correct. Similarly, for m = -1, f(-1) = f(-2) + (-2) + 2 ⇒ -2 = f(-2) + 0 ⇒ f(-2) = -2. Which is correct. So, the formula f(n) = (n² + 3n - 2)/2 holds for all integers n.

Therefore, solving f(n) = n:

(n² + 3n - 2)/2 = n ⇒ n² + 3n - 2 = 2n ⇒ n² + n - 2 = 0 ⇒ same quadratic equation as before, leading to n = 1 and n = -2. Therefore, the same solutions.

Therefore, regardless of the approach, the solutions are n = 1 and n = -2.

But just to be thorough, let me check for other integers. For example, n = 0: f(0) = -1 ≠ 0. n = 2: f(2) = 4 ≠ 2. n = -1: f(-1) = -2 ≠ -1. n = 3: f(3) = 8 ≠ 3. n = -2: f(-2) = -2, which is equal. n = -3: f(-3) = ((-3)^2 + 3*(-3) - 2)/2 = (9 - 9 - 2)/2 = (-2)/2 = -1. So f(-3) = -1 ≠ -3. Similarly, n = 4: f(4) = (16 + 12 - 2)/2 = 26/2 = 13 ≠ 4. So, only n = 1 and n = -2 satisfy f(n) = n.

Therefore, the answer is integers 1 and -2.

**Final Answer**
The integers $ n $ such that $ f(n) = n $ are \boxed{-2} and \boxed{1}.

\boxed{-2}, \boxed{1}

To solve the functional equation:

$$
f(x) + f(y) = f(x + y) - xy - 1
$$

for all real numbers $ x $ and $ y $, and given that $ f(1) = 1 $, we aim to find all **integers** $ n $ such that $ f(n) = n $.

---

### Step 1: Assume a Form for $ f(x) $

Given the structure of the equation, which includes a quadratic term $ xy $, it's reasonable to assume that $ f(x) $ is a **quadratic function**. Let us suppose:

$$
f(x) = ax^2 + bx + c
$$

Substituting into the functional equation:

$$
f(x) + f(y) = a(x^2 + y^2) + b(x + y) + 2c
$$
$$
f(x + y) - xy - 1 = a(x + y)^2 + b(x + y) + c - xy - 1
$$

Expanding the right-hand side:

$$
a(x^2 + 2xy + y^2) + b(x + y) + c - xy - 1 = a x^2 + 2a xy + a y^2 + b x + b y + c - xy - 1
$$

Now equating both sides:

$$
a(x^2 + y^2) + b(x + y) + 2c = a x^2 + 2a xy + a y^2 + b x + b y + c - xy - 1
$$

Subtracting the left-hand side from the right-hand side:

$$
0 = (2a - 1)xy - c - 1
$$

This must hold for all $ x, y $, so the coefficients of $ xy $ and the constant term must be zero:

- $ 2a - 1 = 0 \Rightarrow a = \frac{1}{2} $
- $ -c - 1 = 0 \Rightarrow c = -1 $

Now, we use the given condition $ f(1) = 1 $ to find $ b $:

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

Set this equal to 1:

$$
b - \frac{1}{2} = 1 \Rightarrow b = \frac{3}{2}
$$

Thus, the function is:

$$
f(x) = \frac{1}{2}x^2 + \frac{3}{2}x - 1
$$

---

### Step 2: Solve $ f(n) = n $ for Integer $ n $

We now solve:

$$
\frac{1}{2}n^2 + \frac{3}{2}n - 1 = n
$$

Multiply both sides by 2 to eliminate fractions:

$$
n^2 + 3n - 2 = 2n
$$

Bring all terms to one side:

$$
n^2 + n - 2 = 0
$$

Solve the quadratic:

$$
n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2} = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}
$$

So the solutions are:

$$
n = \frac{-1 + 3}{2} = 1, \quad n = \frac{-1 - 3}{2} = -2
$$

---

### Step 3: Verify the Solutions

- $ f(1) = \frac{1}{2}(1)^2 + \frac{3}{2}(1) - 1 = \frac{1}{2} + \frac{3}{2} - 1 = 1 $ ✅
- $ f(-2) = \frac{1}{2}(4) + \frac{3}{2}(-2) - 1 = 2 - 3 - 1 = -2 $ ✅

Thus, both $ n = 1 $ and $ n = -2 $ satisfy $ f(n) = n $.

---

### Final Answer

$$
\boxed{-2}, \boxed{1}
$$

Solution:
Setting $x = y = 0,$ we get
\[2f(0) = f(0) - 1,\]so $f(0) = -1.$

Setting $y = 1,$ we get
\[f(x) + 1 = f(x + 1) - x - 1,\]so
\[f(x + 1) - f(x) = x + 2.\]Thus,
\begin{align*}
f(2) - f(1) &= 1 + 2, \\
f(3) - f(2) &= 2 + 2, \\
f(4) - f(3) &= 3 + 2, \\
&\dots, \\
f(n) - f(n - 1) &= (n - 1) + 2.
\end{align*}Adding all the equations, we get
\[f(n) - f(1) = 1 + 2 + 3 + \dots + (n - 1) + 2(n - 1) = \frac{(n - 1)n}{2} + 2n - 2 = \frac{n^2 + 3n - 4}{2},\]so
\[f(n) = \frac{n^2 + 3n - 2}{2}\]for all positive integers $n.$

Setting $x = -n$ and $y = n,$ where $n$ is a positive integer, we get
\[f(-n) + f(n) = f(0) + n^2 - 1.\]Then
\[f(-n) = n^2 - f(n) + f(0) - 1 = n^2 - \frac{n^2 + 3n - 2}{2} - 2 = \frac{n^2 - 3n - 2}{2}.\]Thus, the formula
\[f(n) = \frac{n^2 + 3n - 2}{2}\]holds for all integers $n.$

We want to solve $f(n) = n,$ or
\[\frac{n^2 + 3n - 2}{2} = n.\]Then $n^2 + 3n - 2 = 2n,$ or $n^2 + n - 2 = 0.$  This factors as $(n - 1)(n + 2) = 0,$ so the solutions are $n = \boxed{1,-2}.$