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 the original expression.

#### 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 the original expression.

#### 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 (3*20000=60000), with a remainder of 6200. So 6200/20000 = 0.31. Therefore, 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:

(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

Subtracting 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

Using the quadratic formula:

r = [ -3 ± sqrt( (3)^2 - 4*1*(-0.31) ) ] / (2*1)

Compute discriminant:

D = 9 - 4*1*(-0.31) = 9 + 1.24 = 10.24

sqrt(10.24) is 3.2

So:

r = [ -3 ± 3.2 ] / 2

So two solutions:

r = (-3 + 3.2)/2 = 0.2 / 2 = 0.1

r = (-3 - 3.2)/2 = -6.2 / 2 = -3.1

Since interest rate can't be negative, the only valid solution is r = 0.1, which is 10%.

Wait, but hold on. Let me check if this is correct. Because if we plug r = 0.1 into the original equation, does it satisfy?

Let me check:

Original equation after simplifying was:

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

Let me compute left side with r = 0.1:

( (1.1)^3 - 1 ) / 0.1

Calculate 1.1^3: 1.1 * 1.1 = 1.21, then *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 yes, r = 0.1 or 10% is the solution.

Wait, but that seems straightforward. But why did we get a quadratic equation? Because when we expanded the equation, we ended up with a cubic equation which factored into r times a quadratic. But the quadratic gave us the solution. So, that seems correct.

But let me think again. The original problem is about compound interest with annual deposits. So, the future value of the annuity. So, if the interest rate is 10%, then let's compute the future value step by step to confirm.

First year: deposit $20,000 at the end of year 1. It earns interest for 2 years.

Second year: deposit $20,000 at the end of year 2. It earns interest for 1 year.

Third year: deposit $20,000 at the end of year 3. It earns no interest.

So total amount after 3 years:

First deposit: 20,000*(1 + r)^2

Second deposit: 20,000*(1 + r)^1

Third deposit: 20,000*(1 + r)^0 = 20,000

So total FV = 20,000*( (1 + r)^2 + (1 + r) + 1 )

Which is the same as the formula for the future value of an ordinary annuity. So, if r = 10%, then:

(1.1)^2 + 1.1 + 1 = 1.21 + 1.1 + 1 = 3.31

Multiply by 20,000: 20,000 * 3.31 = 66,200. Exactly the amount needed. So that checks out.

Therefore, the minimal interest rate is 10%.

But wait, the problem says "the minimal compound interest rate that the savings account must provide". So, 10% is the answer? But let me think again. Is there 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, I think it's safe to assume that the interest is compounded annually. Otherwise, if it was compounded more frequently, the required rate would be different. But since the problem doesn't specify, we can assume annual compounding.

Therefore, the answer is 10%.

But let me check if there's another way to approach this. Maybe using present value? Wait, no, since we are dealing with future value. Alternatively, maybe I can think of each deposit and calculate their future values.

First deposit: end of year 1, so it has 2 years to earn interest.

Second deposit: end of year 2, so 1 year.

Third deposit: end of year 3, no interest.

So, as above, total FV is 20,000*(1 + r)^2 + 20,000*(1 + r) + 20,000 = 66,200.

Which is the same as before. So solving that equation gives us r = 10%.

Alternatively, maybe I can use logarithms or something else? But since we already have an exact solution, and verified it, maybe that's sufficient.

But let me check if there's a possibility that maybe the interest is compounded more frequently, but the problem says "compound interest rate", which usually refers to the annual rate. If it was compounded monthly, for example, the effective annual rate would be different. However, since the problem doesn't specify compounding frequency, and the deposits are made annually, it's standard to assume annual compounding. Therefore, 10% is correct.

Alternatively, if the problem had said "compounded monthly", we would have to adjust, but since it's not specified, 10% is the answer.

Therefore, the minimal compound interest rate 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 **after the third deposit**.
- The interest is **compounded annually** (standard assumption unless otherwise stated).
- We are to find the **minimum annual interest rate** $ r $ that satisfies this condition.

---

### **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:

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

$$
\sqrt{10.24} = 3.2
$$

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

$$
r = \frac{-3 - 3.2}{2} = \frac{-6.2}{2} = -3.1 \quad \text{(discard, negative rate)}
$$

So, the only valid solution is:

$$
r = 0.1 = 10\%
$$

---

### **Step 4: Verify the Solution**

Let’s compute the future value with $ r = 10\% $:

$$
FV = 20000 \cdot \left( \frac{(1 + 0.1)^3 - 1}{0.1} \right)
$$

$$
(1.1)^3 = 1.331 \Rightarrow 1.331 - 1 = 0.331
$$

$$
\frac{0.331}{0.1} = 3.31 \Rightarrow 20000 \cdot 3.31 = 66200
$$

This matches the required amount, confirming that **10%** is the correct interest rate.

---

### **Conclusion**

The **minimal compound interest rate** required for the worker to have at least \$66,200 in his 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 (3*20000=60000), with a remainder of 6200. So 6200/20000 = 0.31. Therefore, 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:

(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

Subtracting 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

Using the quadratic formula:

r = [ -3 ± sqrt( (3)^2 - 4*1*(-0.31) ) ] / (2*1)

Compute discriminant:

D = 9 - 4*1*(-0.31) = 9 + 1.24 = 10.24

sqrt(10.24) is 3.2

So:

r = [ -3 ± 3.2 ] / 2

So two solutions:

r = (-3 + 3.2)/2 = 0.2 / 2 = 0.1

r = (-3 - 3.2)/2 = -6.2 / 2 = -3.1

Since interest rate can't be negative, the only valid solution is r = 0.1, which is 10%.

Wait, but hold on. Let me check if this is correct. Because if we plug r = 0.1 into the original equation, does it satisfy?

Let me check:

Original equation after simplifying was:

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

Let me compute left side with r = 0.1:

( (1.1)^3 - 1 ) / 0.1

Calculate 1.1^3: 1.1 * 1.1 = 1.21, then *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 yes, r = 0.1 or 10% is the solution.

Wait, but that seems straightforward. But why did we get a quadratic equation? Because when we expanded the equation, we ended up with a cubic equation which factored into r times a quadratic. But the quadratic gave us the solution. So, that seems correct.

But let me think again. The original problem is about compound interest with annual deposits. So, the future value of the annuity. So, if the interest rate is 10%, then let's compute the future value step by step to confirm.

First year: deposit $20,000 at the end of year 1. It earns interest for 2 years.

Second year: deposit $20,000 at the end of year 2. It earns interest for 1 year.

Third year: deposit $20,000 at the end of year 3. It earns no interest.

So total amount after 3 years:

First deposit: 20,000*(1 + r)^2

Second deposit: 20,000*(1 + r)^1

Third deposit: 20,000*(1 + r)^0 = 20,000

So total FV = 20,000*( (1 + r)^2 + (1 + r) + 1 )

Which is the same as the formula for the future value of an ordinary annuity. So, if r = 10%, then:

(1.1)^2 + 1.1 + 1 = 1.21 + 1.1 + 1 = 3.31

Multiply by 20,000: 20,000 * 3.31 = 66,200. Exactly the amount needed. So that checks out.

Therefore, the minimal interest rate is 10%.

But wait, the problem says "the minimal compound interest rate that the savings account must provide". So, 10% is the answer? But let me think again. Is there 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, I think it's safe to assume that the interest is compounded annually. Otherwise, if it was compounded more frequently, the required rate would be different. But since the problem doesn't specify, we can assume annual compounding.

Therefore, the answer is 10%.

But let me check if there's another way to approach this. Maybe using present value? Wait, no, since we are dealing with future value. Alternatively, maybe I can think of each deposit and calculate their future values.

First deposit: end of year 1, so it has 2 years to earn interest.

Second deposit: end of year 2, so 1 year.

Third deposit: end of year 3, no interest.

So, as above, total FV is 20,000*(1 + r)^2 + 20,000*(1 + r) + 20,000 = 66,200.

Which is the same as before. So solving that equation gives us r = 10%.

Alternatively, maybe I can use logarithms or something else? But since we already have an exact solution, and verified it, maybe that's sufficient.

But let me check if there's a possibility that maybe the interest is compounded more frequently, but the problem says "compound interest rate", which usually refers to the annual rate. If it was compounded monthly, for example, the effective annual rate would be different. However, since the problem doesn't specify compounding frequency, and the deposits are made annually, it's standard to assume annual compounding. Therefore, 10% is correct.

Alternatively, if the problem had said "compounded monthly", we would have to adjust, but since it's not specified, 10% is the answer.

Therefore, the minimal compound interest rate 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 **after the third deposit**.
- The interest is **compounded annually** (standard assumption unless otherwise stated).
- We are to find the **minimum annual interest rate** $ r $ that satisfies this condition.

---

### **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:

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

$$
\sqrt{10.24} = 3.2
$$

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

$$
r = \frac{-3 - 3.2}{2} = \frac{-6.2}{2} = -3.1 \quad \text{(discard, negative rate)}
$$

So, the only valid solution is:

$$
r = 0.1 = 10\%
$$

---

### **Step 4: Verify the Solution**

Let’s compute the future value with $ r = 10\% $:

$$
FV = 20000 \cdot \left( \frac{(1 + 0.1)^3 - 1}{0.1} \right)
$$

$$
(1.1)^3 = 1.331 \Rightarrow 1.331 - 1 = 0.331
$$

$$
\frac{0.331}{0.1} = 3.31 \Rightarrow 20000 \cdot 3.31 = 66200
$$

This matches the required amount, confirming that **10%** is the correct interest rate.

---

### **Conclusion**

The **minimal compound interest rate** required for the worker to have at least \$66,200 in his 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's assume 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 we 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:

f(x + y) - xy - 1 = 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 combine like terms:

For x² and y² terms: a x² + a y².

For the xy terms: 2a xy - xy = (2a - 1)xy.

For the linear terms: b x + b y.

Constants: c - 1.

So, putting it all together, f(x + y) - xy - 1 equals:

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

Now, according to the functional equation, this should equal f(x) + f(y), which is:

a(x² + y²) + b(x + y) + 2c.

Therefore, setting them equal:

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

Subtracting the right-hand side from both sides:

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

Simplifying term by term:

a(x² + y²) - a(x² + y²) cancels out.

(2a - 1)xy remains.

b(x + y) - b(x + y) cancels out.

(c - 1) - 2c = -c - 1.

Therefore, the equation reduces to:

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

This 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 equation must hold for all x and y, the coefficient of xy must be zero, and the constant term must also be zero.

So, setting coefficients equal to zero:

For the xy term: 2a - 1 = 0 ⇒ 2a = 1 ⇒ a = 1/2.

For the constant term: -c - 1 = 0 ⇒ -c = 1 ⇒ c = -1.

So, we have found a = 1/2 and c = -1. The coefficient b hasn't been determined yet. Wait, but in the equation above, after subtracting, the only terms left were (2a - 1)xy - c - 1. There was no term involving b, because both sides had b(x + y). Therefore, the value of b is not determined by this equation? That seems odd. Maybe I need to check my steps again.

Wait, let me check again. The original equation after substitution was:

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

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

Therefore, equating them:

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

Subtracting the left side from both sides:

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

Which simplifies to:

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

So, yes, that's correct. Therefore, the equation reduces to (2a - 1)xy - c - 1 = 0 for all x, y. Therefore, since this must hold for all x and y, the coefficients of xy and the constant term must be zero. Therefore, 2a - 1 = 0 and -c - 1 = 0. Therefore, a = 1/2 and c = -1. The value of b is not determined here. Wait, but how?

But in the original functional equation, there is no term involving b? Wait, no. Wait, when we equated f(x) + f(y) and f(x + y) - xy - 1, the terms involving b(x + y) canceled out. Therefore, the value of b is arbitrary? But that can't be. Wait, but maybe there is another condition? Because we have f(1) = 1 given. So, maybe we can use that to find b?

Yes! Since we have 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 we are told that f(1) = 1. Therefore:

b - 1/2 = 1 ⇒ b = 1 + 1/2 = 3/2.

Therefore, b = 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 side equals right side. Good.

Another check: Let me take x = 0 and y = 0. Then:

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

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

Therefore, 2f(0) = f(0) - 1 ⇒ f(0) = -1. Let's check with our function: f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. Correct.

Another check: Let me take x = 1, y = 0. Then:

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

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

Another check: Let me take x = 2, y = 3. Let's compute both sides.

First, compute f(2) = 4 + 3 - 1 = 6? Wait, earlier I had f(2) = 4? Wait, let me recalculate:

Wait, f(x) = (1/2)x² + (3/2)x - 1. So for x = 2:

(1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Yes, that's correct. Similarly, f(3) = (1/2)(9) + (3/2)(3) - 1 = 4.5 + 4.5 - 1 = 8. So f(3) = 8.

Left side: f(2) + f(3) = 4 + 8 = 12.

Right side: f(2 + 3) - (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. Correct. So that works.

Therefore, the function f(x) = (1/2)x² + (3/2)x - 1 seems to satisfy the functional equation and the given condition f(1) = 1. Great, so that's our function.

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 the left side:

n² + 3n - 2 - 2n = n² + n - 2 = 0.

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

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

n = [-1 ± √9]/2 = [-1 ± 3]/2.

Therefore, the solutions are:

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) = 1, which is given. Correct.

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

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Yes, that equals -2. So, f(-2) = -2. Correct.

Therefore, the integers are 1 and -2. Wait, but the problem says "all integers n", so maybe there are more? Wait, but the quadratic equation only has two roots. So, unless there are other solutions?

Wait, but the equation f(n) = n reduces to a quadratic equation, which can have at most two real roots. Since we found two integer roots, 1 and -2, those are the only solutions. Therefore, the answer should be n = -2 and 1.

But let me check if there are other integer solutions. Suppose someone might think maybe there are more? Let me check for n = 0: f(0) = -1 ≠ 0. n = 2: f(2) = 4 ≠ 2. n = -1: f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 0.5 - 1.5 - 1 = -2 ≠ -1. n = 3: f(3) = 8 ≠ 3. n = -3: f(-3) = (1/2)(9) + (3/2)(-3) - 1 = 4.5 - 4.5 - 1 = -1 ≠ -3. So, no other integers satisfy f(n) = n. Therefore, only -2 and 1.

But let me check again the quadratic equation. The equation was n² + n - 2 = 0. So, factors: (n + 2)(n - 1) = 0. Yes, so roots at n = -2 and 1. Therefore, only those two integers. Therefore, the answer is -2 and 1.

But let me just make sure that my function is correct. Let me check another integer. For example, n = 2: f(2) = 4, which is not equal to 2. n = -2: f(-2) = -2. Correct. So, seems okay.

Wait, but let me check if there's a possibility that the function isn't quadratic? Wait, the functional equation is similar to Cauchy's equation but with extra terms. However, since we were able to find a quadratic function that satisfies the equation and the given condition, and since the functional equation is of degree 2 (because of the xy term), it's likely that the function is indeed quadratic. Also, since we derived the function by assuming it's quadratic and found a solution that works, and the problem gives f(1) = 1, which allowed us to find the coefficients, so I think that's solid.

Alternatively, maybe there's another approach to solving the functional equation without assuming the form of f. Let me try that as a check.

Another approach: Let me try to find f(0). Let me set x = 0 and y = 0 in the functional equation:

f(0) + f(0) = f(0 + 0) - 0*0 - 1 ⇒ 2f(0) = f(0) - 1 ⇒ f(0) = -1. Which matches our previous result.

Now, let me try to find f(-x) in terms of f(x). Let me set y = -x in the functional equation:

f(x) + f(-x) = f(x + (-x)) - x*(-x) - 1 ⇒ f(x) + f(-x) = f(0) + x² - 1 ⇒ f(x) + f(-x) = -1 + x² - 1 = x² - 2.

Therefore, f(-x) = x² - 2 - f(x).

Hmm, interesting. Maybe we can use this relation later.

Alternatively, let me try to find a recursive formula. Let me set y = 1. Then, the functional 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.

So, this gives a recursive formula for f(x + 1) in terms of f(x). Since we know f(1) = 1, maybe we can compute f(n) for integer n?

Let me try that. Let me compute f(0) = -1, as before.

Then, for x = 0: f(0 + 1) = f(0) + 0 + 2 ⇒ f(1) = -1 + 0 + 2 = 1. Which matches the given value.

For x = 1: f(1 + 1) = f(1) + 1 + 2 ⇒ f(2) = 1 + 1 + 2 = 4. Which matches our previous result.

For x = 2: f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which matches.

For x = 3: f(4) = 8 + 3 + 2 = 13.

Similarly, for x = -1: Let me see. Wait, the recursive formula is for f(x + 1) in terms of f(x). If I want to compute f(-1), maybe I need another approach. Let me set x = -1 and y = 1 in the original equation:

f(-1) + f(1) = f(0) - (-1)(1) - 1 ⇒ f(-1) + 1 = -1 + 1 - 1 ⇒ f(-1) + 1 = -1 ⇒ f(-1) = -2. Which matches our previous calculation for f(-2) when x = -2? Wait, no. Wait, f(-1) = -2? Wait, earlier when I calculated f(-1), I had:

f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 0.5 - 1.5 - 1 = -2. Yes, correct. So f(-1) = -2.

Alternatively, using the relation f(-x) = x² - 2 - f(x). Let me check for x = 1: f(-1) = (1)^2 - 2 - f(1) = 1 - 2 - 1 = -2. Correct. Similarly, for x = 2: f(-2) = (2)^2 - 2 - f(2) = 4 - 2 - 4 = -2. Which matches our previous result. So that works.

So, using the recursive formula f(x + 1) = f(x) + x + 2, we can compute f(n) for integer n. Let me try to find a general formula for f(n) where n is an integer.

Let me consider n as an integer. Let me compute f(n) using the recursive formula. Starting from f(0) = -1.

For n = 0: f(0) = -1.

For n = 1: f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1.

For n = 2: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4.

For n = 3: f(3) = f(2) + 2 + 2 = 4 + 4 = 8.

For n = 4: f(4) = f(3) + 3 + 2 = 8 + 5 = 13.

For n = 5: f(5) = 13 + 4 + 2 = 19.

Wait, but according to our quadratic function, f(n) = (1/2)n² + (3/2)n - 1. Let me check for n = 4: (1/2)(16) + (3/2)(4) - 1 = 8 + 6 - 1 = 13. Correct. For n = 5: (1/2)(25) + (3/2)(5) - 1 = 12.5 + 7.5 - 1 = 19. Correct. So, the recursive formula gives the same results as the quadratic function. Therefore, the general formula for integer n is f(n) = (1/2)n² + (3/2)n - 1.

Alternatively, we can derive this formula from the recursive relation. Let me try that.

The recursive formula is f(n + 1) - f(n) = n + 2. This is a recurrence relation. Let me solve it for integer n.

This is a linear recurrence. Let me write it as:

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

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

For integer n ≥ 0, starting from f(0) = -1.

Then, f(1) = f(0) + 0 + 2.

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

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

...

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

Therefore, summing from k = 0 to k = n - 1:

f(n) = f(0) + Σ_{k=0}^{n - 1} [k + 2].

Compute the sum:

Σ_{k=0}^{n - 1} [k + 2] = Σ_{k=0}^{n - 1} k + Σ_{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 quadratic function, f(n) = (1/2)n² + (3/2)n - 1, which is the same as (n² + 3n - 2)/2. Yes, exactly. Therefore, the formula is correct.

Therefore, for integer n, f(n) = (n² + 3n - 2)/2. Therefore, solving f(n) = n:

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

Therefore, the answer is \boxed{-2}, \boxed{1}.

But the problem says "Enter all such integers, separated by commas." So, probably they want both numbers in a box separated by commas. But the instructions say "put your final answer within \boxed{}". Maybe as multiple boxed numbers? Wait, the original problem says "find all integers n such that f(n) = n. Enter all such integers, separated by commas."

But the standard practice for such problems is to put all answers in a single box, separated by commas. For example, \boxed{-2}, \boxed{1} if multiple answers, but maybe here they want \boxed{-2}, \boxed{1} or \boxed{-2, 1}? Wait, the exact instruction says "put your final answer within \boxed{}". So maybe they want all answers inside a single box, separated by commas. Let me check previous problems. For example, if the answer is multiple numbers, they are separated by commas inside the box. For example, \boxed{-2, 1}.

But to confirm, let me check the problem statement again: "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So, the answer is two integers: -2 and 1. Therefore, the final answer should be \boxed{-2}, \boxed{1}? Or \boxed{-2, 1}? Depending on the convention. In many Olympiad problems, multiple answers are put in a single box separated by commas. For example, on AoPS or similar, they might expect \boxed{-2}, \boxed{1} if separate, but since the user instruction says "put your final answer within \boxed{}", maybe they want all in one box. Let me check the original problem's exact wording: "Enter all such integers, separated by commas." So, likely, they want something like \boxed{-2}, \boxed{1} if multiple boxes, but since the user says "put your final answer within \boxed{}", maybe they want \boxed{-2, 1}.

But to be safe, maybe check if the system accepts multiple boxes. However, since the user hasn't specified, but given that in many cases for multiple answers, they are put in one box separated by commas. For example, previous problems on this platform often have multiple answers in one box. So, I think \boxed{-2, 1} is acceptable.

But let me check once again. If I write \boxed{-2}, \boxed{1}, that would be two boxes. However, the instruction says "put your final answer within \boxed{}", which suggests a single box. Therefore, I should write \boxed{-2, 1}.

But to make sure, let me check the problem again. The original problem says "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So, the answer is two integers, so they should be separated by commas. Therefore, the final answer is \boxed{-2}, \boxed{1} if multiple boxes are allowed, but if only one box is allowed, then \boxed{-2, 1}.

But since the user hasn't specified, but given that in the problem statement they say "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will go with \boxed{-2, 1}.

But let me check once again if there are other integer solutions. Suppose n is an integer, and f(n) = n. We have the equation n² + n - 2 = 0, which factors as (n + 2)(n - 1) = 0. Therefore, only solutions are n = -2 and n = 1. Therefore, no other integers. Therefore, the answer is definitely -2 and 1.

Therefore, I think the answer is \boxed{-2} and \boxed{1}, but since the user might expect comma-separated in one box, I will check standard conventions. For example, in the past, for problems with multiple answers, they are often put in a single box separated by commas. For example, on AoPS or similar, so I think \boxed{-2, 1} is acceptable.

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

Wait, but the system might require a single box. Let me check the original problem's instructions again. The user says: "put your final answer within \boxed{}". So, maybe they want all answers in one box. For example, if the answer is multiple numbers, they are separated by commas inside the box. So, I think the correct format is \boxed{-2, 1}.

**Final Answer**
\boxed{-2}, \boxed{1}

Wait, but I'm confused. Let me check previous examples. For example, if the answer is multiple numbers, like in some problems, they use multiple boxes. However, since the user hasn't specified, but given that the original problem says "Enter all such integers, separated by commas," and the instruction says "put your final answer within \boxed{}", maybe they want all answers in one box. For example, if you look at problems on AoPS or other platforms, sometimes they use multiple boxes, but sometimes they use one. However, since the user hasn't specified, but given that the original problem says "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will go with \boxed{-2, 1}.

**Final Answer**
\boxed{-2, 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 presence of the 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:

- Left-hand side:
  $$
  f(x) + f(y) = a(x^2 + y^2) + b(x + y) + 2c
  $$

- Right-hand side:
  $$
  f(x + y) - xy - 1 = a(x + y)^2 + b(x + y) + c - xy - 1
  $$
  Expanding:
  $$
  a(x^2 + 2xy + y^2) + b(x + y) + c - xy - 1 = a(x^2 + y^2) + (2a - 1)xy + b(x + y) + (c - 1)
  $$

Equating both sides:

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

Subtracting the right-hand side from the left:

$$
(2a - 1)xy + (c - 1) - 2c = 0
\Rightarrow (2a - 1)xy - c - 1 = 0
$$

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, using the given condition $ f(1) = 1 $:

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

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:

$$
n^2 + 3n - 2 = 2n \Rightarrow n^2 + n - 2 = 0
$$

Factor the quadratic:

$$
(n + 2)(n - 1) = 0 \Rightarrow n = -2 \text{ or } n = 1
$$

---

### Step 3: Verify the Integer Solutions

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

Both values satisfy $ f(n) = n $, and no other integers satisfy the equation.

---

### Final Answer

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

[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's assume 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 we 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:

f(x + y) - xy - 1 = 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 combine like terms:

For x² and y² terms: a x² + a y².

For the xy terms: 2a xy - xy = (2a - 1)xy.

For the linear terms: b x + b y.

Constants: c - 1.

So, putting it all together, f(x + y) - xy - 1 equals:

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

Now, according to the functional equation, this should equal f(x) + f(y), which is:

a(x² + y²) + b(x + y) + 2c.

Therefore, setting them equal:

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

Subtracting the right-hand side from both sides:

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

Simplifying term by term:

a(x² + y²) - a(x² + y²) cancels out.

(2a - 1)xy remains.

b(x + y) - b(x + y) cancels out.

(c - 1) - 2c = -c - 1.

Therefore, the equation reduces to:

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

This 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 equation must hold for all x and y, the coefficient of xy must be zero, and the constant term must also be zero.

So, setting coefficients equal to zero:

For the xy term: 2a - 1 = 0 ⇒ 2a = 1 ⇒ a = 1/2.

For the constant term: -c - 1 = 0 ⇒ -c = 1 ⇒ c = -1.

So, we have found a = 1/2 and c = -1. The coefficient b hasn't been determined yet. Wait, but in the equation above, after subtracting, the only terms left were (2a - 1)xy - c - 1. There was no term involving b, because both sides had b(x + y). Therefore, the value of b is not determined by this equation? That seems odd. Maybe I need to check my steps again.

Wait, let me check again. The original equation after substitution was:

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

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

Therefore, equating them:

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

Subtracting the left side from both sides:

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

Which simplifies to:

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

So, yes, that's correct. Therefore, the equation reduces to (2a - 1)xy - c - 1 = 0 for all x, y. Therefore, since this must hold for all x and y, the coefficients of xy and the constant term must be zero. Therefore, 2a - 1 = 0 and -c - 1 = 0. Therefore, a = 1/2 and c = -1. The value of b is not determined here. Wait, but how?

But in the original functional equation, there is no term involving b? Wait, no. Wait, when we equated f(x) + f(y) and f(x + y) - xy - 1, the terms involving b(x + y) canceled out. Therefore, the value of b is arbitrary? But that can't be. Wait, but maybe there is another condition? Because we have f(1) = 1 given. So, maybe we can use that to find b?

Yes! Since we have 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 we are told that f(1) = 1. Therefore:

b - 1/2 = 1 ⇒ b = 1 + 1/2 = 3/2.

Therefore, b = 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 side equals right side. Good.

Another check: Let me take x = 0 and y = 0. Then:

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

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

Therefore, 2f(0) = f(0) - 1 ⇒ f(0) = -1. Let's check with our function: f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. Correct.

Another check: Let me take x = 1, y = 0. Then:

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

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

Another check: Let me take x = 2, y = 3. Let's compute both sides.

First, compute f(2) = 4 + 3 - 1 = 6? Wait, earlier I had f(2) = 4? Wait, let me recalculate:

Wait, f(x) = (1/2)x² + (3/2)x - 1. So for x = 2:

(1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Yes, that's correct. Similarly, f(3) = (1/2)(9) + (3/2)(3) - 1 = 4.5 + 4.5 - 1 = 8. So f(3) = 8.

Left side: f(2) + f(3) = 4 + 8 = 12.

Right side: f(2 + 3) - (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. Correct. So that works.

Therefore, the function f(x) = (1/2)x² + (3/2)x - 1 seems to satisfy the functional equation and the given condition f(1) = 1. Great, so that's our function.

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 the left side:

n² + 3n - 2 - 2n = n² + n - 2 = 0.

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

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

n = [-1 ± √9]/2 = [-1 ± 3]/2.

Therefore, the solutions are:

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) = 1, which is given. Correct.

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

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Yes, that equals -2. So, f(-2) = -2. Correct.

Therefore, the integers are 1 and -2. Wait, but the problem says "all integers n", so maybe there are more? Wait, but the quadratic equation only has two roots. So, unless there are other solutions?

Wait, but the equation f(n) = n reduces to a quadratic equation, which can have at most two real roots. Since we found two integer roots, 1 and -2, those are the only solutions. Therefore, the answer should be n = -2 and 1.

But let me check if there are other integer solutions. Suppose someone might think maybe there are more? Let me check for n = 0: f(0) = -1 ≠ 0. n = 2: f(2) = 4 ≠ 2. n = -1: f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 0.5 - 1.5 - 1 = -2 ≠ -1. n = 3: f(3) = 8 ≠ 3. n = -3: f(-3) = (1/2)(9) + (3/2)(-3) - 1 = 4.5 - 4.5 - 1 = -1 ≠ -3. So, no other integers satisfy f(n) = n. Therefore, only -2 and 1.

But let me check again the quadratic equation. The equation was n² + n - 2 = 0. So, factors: (n + 2)(n - 1) = 0. Yes, so roots at n = -2 and 1. Therefore, only those two integers. Therefore, the answer is -2 and 1.

But let me just make sure that my function is correct. Let me check another integer. For example, n = 2: f(2) = 4, which is not equal to 2. n = -2: f(-2) = -2. Correct. So, seems okay.

Wait, but let me check if there's a possibility that the function isn't quadratic? Wait, the functional equation is similar to Cauchy's equation but with extra terms. However, since we were able to find a quadratic function that satisfies the equation and the given condition, and since the functional equation is of degree 2 (because of the xy term), it's likely that the function is indeed quadratic. Also, since we derived the function by assuming it's quadratic and found a solution that works, and the problem gives f(1) = 1, which allowed us to find the coefficients, so I think that's solid.

Alternatively, maybe there's another approach to solving the functional equation without assuming the form of f. Let me try that as a check.

Another approach: Let me try to find f(0). Let me set x = 0 and y = 0 in the functional equation:

f(0) + f(0) = f(0 + 0) - 0*0 - 1 ⇒ 2f(0) = f(0) - 1 ⇒ f(0) = -1. Which matches our previous result.

Now, let me try to find f(-x) in terms of f(x). Let me set y = -x in the functional equation:

f(x) + f(-x) = f(x + (-x)) - x*(-x) - 1 ⇒ f(x) + f(-x) = f(0) + x² - 1 ⇒ f(x) + f(-x) = -1 + x² - 1 = x² - 2.

Therefore, f(-x) = x² - 2 - f(x).

Hmm, interesting. Maybe we can use this relation later.

Alternatively, let me try to find a recursive formula. Let me set y = 1. Then, the functional 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.

So, this gives a recursive formula for f(x + 1) in terms of f(x). Since we know f(1) = 1, maybe we can compute f(n) for integer n?

Let me try that. Let me compute f(0) = -1, as before.

Then, for x = 0: f(0 + 1) = f(0) + 0 + 2 ⇒ f(1) = -1 + 0 + 2 = 1. Which matches the given value.

For x = 1: f(1 + 1) = f(1) + 1 + 2 ⇒ f(2) = 1 + 1 + 2 = 4. Which matches our previous result.

For x = 2: f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which matches.

For x = 3: f(4) = 8 + 3 + 2 = 13.

Similarly, for x = -1: Let me see. Wait, the recursive formula is for f(x + 1) in terms of f(x). If I want to compute f(-1), maybe I need another approach. Let me set x = -1 and y = 1 in the original equation:

f(-1) + f(1) = f(0) - (-1)(1) - 1 ⇒ f(-1) + 1 = -1 + 1 - 1 ⇒ f(-1) + 1 = -1 ⇒ f(-1) = -2. Which matches our previous calculation for f(-2) when x = -2? Wait, no. Wait, f(-1) = -2? Wait, earlier when I calculated f(-1), I had:

f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 0.5 - 1.5 - 1 = -2. Yes, correct. So f(-1) = -2.

Alternatively, using the relation f(-x) = x² - 2 - f(x). Let me check for x = 1: f(-1) = (1)^2 - 2 - f(1) = 1 - 2 - 1 = -2. Correct. Similarly, for x = 2: f(-2) = (2)^2 - 2 - f(2) = 4 - 2 - 4 = -2. Which matches our previous result. So that works.

So, using the recursive formula f(x + 1) = f(x) + x + 2, we can compute f(n) for integer n. Let me try to find a general formula for f(n) where n is an integer.

Let me consider n as an integer. Let me compute f(n) using the recursive formula. Starting from f(0) = -1.

For n = 0: f(0) = -1.

For n = 1: f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1.

For n = 2: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4.

For n = 3: f(3) = f(2) + 2 + 2 = 4 + 4 = 8.

For n = 4: f(4) = f(3) + 3 + 2 = 8 + 5 = 13.

For n = 5: f(5) = 13 + 4 + 2 = 19.

Wait, but according to our quadratic function, f(n) = (1/2)n² + (3/2)n - 1. Let me check for n = 4: (1/2)(16) + (3/2)(4) - 1 = 8 + 6 - 1 = 13. Correct. For n = 5: (1/2)(25) + (3/2)(5) - 1 = 12.5 + 7.5 - 1 = 19. Correct. So, the recursive formula gives the same results as the quadratic function. Therefore, the general formula for integer n is f(n) = (1/2)n² + (3/2)n - 1.

Alternatively, we can derive this formula from the recursive relation. Let me try that.

The recursive formula is f(n + 1) - f(n) = n + 2. This is a recurrence relation. Let me solve it for integer n.

This is a linear recurrence. Let me write it as:

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

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

For integer n ≥ 0, starting from f(0) = -1.

Then, f(1) = f(0) + 0 + 2.

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

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

...

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

Therefore, summing from k = 0 to k = n - 1:

f(n) = f(0) + Σ_{k=0}^{n - 1} [k + 2].

Compute the sum:

Σ_{k=0}^{n - 1} [k + 2] = Σ_{k=0}^{n - 1} k + Σ_{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 quadratic function, f(n) = (1/2)n² + (3/2)n - 1, which is the same as (n² + 3n - 2)/2. Yes, exactly. Therefore, the formula is correct.

Therefore, for integer n, f(n) = (n² + 3n - 2)/2. Therefore, solving f(n) = n:

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

Therefore, the answer is \boxed{-2}, \boxed{1}.

But the problem says "Enter all such integers, separated by commas." So, probably they want both numbers in a box separated by commas. But the instructions say "put your final answer within \boxed{}". Maybe as multiple boxed numbers? Wait, the original problem says "find all integers n such that f(n) = n. Enter all such integers, separated by commas."

But the standard practice for such problems is to put all answers in a single box, separated by commas. For example, \boxed{-2}, \boxed{1} if multiple answers, but maybe here they want \boxed{-2}, \boxed{1} or \boxed{-2, 1}? Wait, the exact instruction says "put your final answer within \boxed{}". So maybe they want all answers inside a single box, separated by commas. Let me check previous problems. For example, if the answer is multiple numbers, they are separated by commas inside the box. For example, \boxed{-2, 1}.

But to confirm, let me check the problem statement again: "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So, the answer is two integers: -2 and 1. Therefore, the final answer should be \boxed{-2}, \boxed{1}? Or \boxed{-2, 1}? Depending on the convention. In many Olympiad problems, multiple answers are put in a single box separated by commas. For example, on AoPS or similar, they might expect \boxed{-2}, \boxed{1} if separate, but since the user instruction says "put your final answer within \boxed{}", maybe they want all in one box. Let me check the original problem's exact wording: "Enter all such integers, separated by commas." So, likely, they want something like \boxed{-2}, \boxed{1} if multiple boxes, but since the user says "put your final answer within \boxed{}", maybe they want \boxed{-2, 1}.

But to be safe, maybe check if the system accepts multiple boxes. However, since the user hasn't specified, but given that in many cases for multiple answers, they are put in one box separated by commas. For example, previous problems on this platform often have multiple answers in one box. So, I think \boxed{-2, 1} is acceptable.

But let me check once again. If I write \boxed{-2}, \boxed{1}, that would be two boxes. However, the instruction says "put your final answer within \boxed{}", which suggests a single box. Therefore, I should write \boxed{-2, 1}.

But to make sure, let me check the problem again. The original problem says "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So, the answer is two integers, so they should be separated by commas. Therefore, the final answer is \boxed{-2}, \boxed{1} if multiple boxes are allowed, but if only one box is allowed, then \boxed{-2, 1}.

But since the user hasn't specified, but given that in the problem statement they say "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will go with \boxed{-2, 1}.

But let me check once again if there are other integer solutions. Suppose n is an integer, and f(n) = n. We have the equation n² + n - 2 = 0, which factors as (n + 2)(n - 1) = 0. Therefore, only solutions are n = -2 and n = 1. Therefore, no other integers. Therefore, the answer is definitely -2 and 1.

Therefore, I think the answer is \boxed{-2} and \boxed{1}, but since the user might expect comma-separated in one box, I will check standard conventions. For example, in the past, for problems with multiple answers, they are often put in a single box separated by commas. For example, on AoPS or similar, so I think \boxed{-2, 1} is acceptable.

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

Wait, but the system might require a single box. Let me check the original problem's instructions again. The user says: "put your final answer within \boxed{}". So, maybe they want all answers in one box. For example, if the answer is multiple numbers, they are separated by commas inside the box. So, I think the correct format is \boxed{-2, 1}.

**Final Answer**
\boxed{-2}, \boxed{1}

Wait, but I'm confused. Let me check previous examples. For example, if the answer is multiple numbers, like in some problems, they use multiple boxes. However, since the user hasn't specified, but given that the original problem says "Enter all such integers, separated by commas," and the instruction says "put your final answer within \boxed{}", maybe they want all answers in one box. For example, if you look at problems on AoPS or other platforms, sometimes they use multiple boxes, but sometimes they use one. However, since the user hasn't specified, but given that the original problem says "put your final answer within \boxed{}", I think they expect a single box. Therefore, I will go with \boxed{-2, 1}.

**Final Answer**
\boxed{-2, 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 presence of the 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:

- Left-hand side:
  $$
  f(x) + f(y) = a(x^2 + y^2) + b(x + y) + 2c
  $$

- Right-hand side:
  $$
  f(x + y) - xy - 1 = a(x + y)^2 + b(x + y) + c - xy - 1
  $$
  Expanding:
  $$
  a(x^2 + 2xy + y^2) + b(x + y) + c - xy - 1 = a(x^2 + y^2) + (2a - 1)xy + b(x + y) + (c - 1)
  $$

Equating both sides:

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

Subtracting the right-hand side from the left:

$$
(2a - 1)xy + (c - 1) - 2c = 0
\Rightarrow (2a - 1)xy - c - 1 = 0
$$

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, using the given condition $ f(1) = 1 $:

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

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:

$$
n^2 + 3n - 2 = 2n \Rightarrow n^2 + n - 2 = 0
$$

Factor the quadratic:

$$
(n + 2)(n - 1) = 0 \Rightarrow n = -2 \text{ or } n = 1
$$

---

### Step 3: Verify the Integer Solutions

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

Both values satisfy $ f(n) = n $, and no other integers satisfy the equation.

---

### Final Answer

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

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 $n = \boxed{1,-2}.$