Explanation:
Okay, let me try to figure out this problem. So, Denali and Nate are dog walkers, right? Denali has 16 dogs, and Nate has 12. The company has a new policy where they can assign or unassign new dogs in groups of x dogs. The key part here is that the ratio of their pay would be the same in two different scenarios:

1. If Denali starts walking 4x more dogs and Nate stays at 12.
2. If x of Nate's dogs are reassigned to Denali.

And we need to find x, where x is not zero. Alright, let me break this down step by step.

First, let's recall that the ratio of their pay would be the same in both cases. Since they are paid per dog they walk, I assume their pay is directly proportional to the number of dogs they walk. So, the ratio of their pay is just the ratio of the number of dogs they each walk. That makes sense.

So initially, Denali has 16 dogs, Nate has 12. So the initial ratio of their pay is 16:12, which simplifies to 4:3. But maybe we need to keep it as fractions for the equations? Let me think.

But the problem says that under the new policy, they can assign or unassign new dogs in groups of x. Wait, does that mean that they can get new dogs assigned to them in groups of x? Or maybe they can reassign existing dogs? The wording says "assigned or unassigned new dogs in groups of x dogs." Hmm.

Wait, maybe the problem is about adjusting the number of dogs they have by adding or removing groups of x dogs? So for example, if they get more dogs, it's in groups of x, and if they lose dogs, it's also in groups of x. But then the problem mentions two different scenarios:

First scenario: Denali starts walking 4x more dogs and Nate stays at 12. So Denali's total dogs would be 16 + 4x? Wait, but the problem says "assigned or unassigned new dogs in groups of x dogs." So maybe adding or subtracting groups of x dogs. So if Denali is assigned 4x dogs? But 4x is not a multiple of x? Wait, 4x is 4 times x. Hmm, but if they are assigned in groups of x, then adding 4x would be adding 4 groups of x dogs. So maybe that's allowed. So maybe the first scenario is that Denali is assigned 4x dogs, so her total becomes 16 + 4x, and Nate remains at 12.

Second scenario: x of Nate's dogs are reassigned to Denali. So Nate loses x dogs, and Denali gains x dogs. Therefore, Denali's total dogs would be 16 + x, and Nate's total would be 12 - x.

In both scenarios, the ratio of Denali's pay to Nate's pay is the same. So the ratio in the first scenario (Denali has 16 + 4x, Nate has 12) is equal to the ratio in the second scenario (Denali has 16 + x, Nate has 12 - x).

Therefore, we can set up an equation:

(16 + 4x)/12 = (16 + x)/(12 - x)

Is that right? Let me check again.

Yes, the ratio of Denali's pay to Nate's pay in both cases should be equal. So first case: Denali's dogs are 16 + 4x, Nate's are still 12. Second case: Denali's dogs are 16 + x, Nate's are 12 - x. Therefore, the equation is:

(16 + 4x)/12 = (16 + x)/(12 - x)

Okay, so now we need to solve this equation for x. Let me write that down:

(16 + 4x)/12 = (16 + x)/(12 - x)

First, maybe cross-multiplying to eliminate the fractions. So multiply both sides by 12*(12 - x):

(16 + 4x)*(12 - x) = 12*(16 + x)

Let me compute both sides:

Left side: (16 + 4x)(12 - x)
Let me expand that:

First, distribute 16 over (12 - x): 16*12 - 16x
Then distribute 4x over (12 - x): 4x*12 - 4x*x
So total left side: 16*12 -16x + 48x -4x²
Simplify:

16*12 is 192
-16x + 48x is 32x
So left side is 192 + 32x -4x²

Right side: 12*(16 + x) = 12*16 + 12x = 192 + 12x

So now, set left side equal to right side:

192 + 32x -4x² = 192 + 12x

Subtract 192 from both sides:

32x -4x² = 12x

Subtract 12x from both sides:

20x -4x² = 0

Let me rearrange terms:

-4x² + 20x = 0

Multiply both sides by -1:

4x² - 20x = 0

Factor out 4x:

4x(x - 5) = 0

So, solutions are 4x = 0 => x = 0 or x - 5 = 0 => x = 5

But the problem says x ≠ 0, so x = 5.

Wait, so x is 5? Let me check this answer to make sure.

Let me verify if x = 5 satisfies the original equation.

First scenario: Denali has 16 + 4x = 16 + 20 = 36 dogs, Nate has 12 dogs. Ratio is 36:12 = 3:1.

Second scenario: Denali has 16 + x = 21 dogs, Nate has 12 - x = 7 dogs. Ratio is 21:7 = 3:1.

Yes, both ratios are 3:1. So that works. Therefore, x is indeed 5.

But let me double-check if there are any other constraints or if I misinterpreted the problem.

Wait, the problem says "assigned or unassigned new dogs in groups of x dogs." So when they say Denali started walking 4x more dogs, does that mean adding 4x dogs? But groups are of size x. So adding 4x would be 4 groups. That seems okay. Also, when x of Nate's dogs are reassigned, that would be removing x dogs from Nate and adding them to Denali. So that's a reassignment of x dogs. So that's allowed.

Is there a possibility that x has to be such that the number of dogs is positive? Let me check.

In the second scenario, Nate has 12 - x dogs. So 12 - x must be positive. Therefore, 12 - x > 0 => x < 12. Since x=5, that's okay. Also, in the first scenario, Denali's dogs are 16 + 4x, which is positive for any x. So x=5 is acceptable. Therefore, the answer is 5.

But let me check again if the original problem says anything else. Wait, the company's new policy is about assigning or unassigning new dogs in groups of x dogs. So maybe the first scenario is not adding 4x dogs, but something else? Let me read again:

"the ratio of Denali's pay to Nate's pay would be the same if Denali started walking 4x more dogs and Nate stayed at 12 dogs or if x of Nate's dogs were reassigned to Denali."

So, the first case: Denali started walking 4x more dogs. So that would mean Denali's total dogs is original 16 + 4x. Nate stays at 12. The second case: x of Nate's dogs are reassigned to Denali. So Nate's dogs become 12 - x, Denali's become 16 + x. So that's how I set up the equation.

Therefore, solving that gives x=5, and checking shows that it works. Therefore, the answer is 5. I think that is correct.

Wait, but just to make sure, let me think again about the ratio. Original ratio is 16:12, which reduces to 4:3. After the first scenario, Denali has 16 + 4x, Nate has 12. So the ratio is (16 + 4x)/12. In the second scenario, the ratio is (16 + x)/(12 - x). Setting them equal gives the equation.

But maybe I should check with x=5:

First case: (16 + 20)/12 = 36/12 = 3.

Second case: (16 + 5)/(12 - 5) = 21/7 = 3.

Yes, so both ratios are equal. Therefore, x=5 is correct.

Therefore, the answer is 5.

I think that's it. I don't see any mistakes in my reasoning. So I think x is 5.

**Final Answer**
The value of $ x $ is \boxed{5}.

To determine the value of $ x $, we analyze the given scenario involving Denali and Nate, who work for a dog-walking service. The key point is that the **ratio of their pay** remains the same in **two different assignments** of dogs:

1. **Case A:** Denali walks $ 16 + 4x $ dogs, while Nate walks 12 dogs.
2. **Case B:** Denali walks $ 16 + x $ dogs, while Nate walks $ 12 - x $ dogs.

Since their pay is directly proportional to the number of dogs they walk, the ratio of their pay is equivalent to the ratio of the number of dogs they walk. Thus, we equate the two ratios:

$$
\frac{16 + 4x}{12} = \frac{16 + x}{12 - x}
$$

---

### Step 1: Eliminate the Denominators

To simplify this equation, we cross-multiply:

$$
(16 + 4x)(12 - x) = 12(16 + x)
$$

---

### Step 2: Expand Both Sides

**Left-hand side:**

$$
(16 + 4x)(12 - x) = 16 \cdot 12 - 16x + 4x \cdot 12 - 4x^2 = 192 - 16x + 48x - 4x^2 = 192 + 32x - 4x^2
$$

**Right-hand side:**

$$
12(16 + x) = 192 + 12x
$$

Now we equate both expressions:

$$
192 + 32x - 4x^2 = 192 + 12x
$$

---

### Step 3: Simplify the Equation

Subtract 192 from both sides:

$$
32x - 4x^2 = 12x
$$

Subtract $ 12x $ from both sides:

$$
20x - 4x^2 = 0
$$

Factor the left-hand side:

$$
4x(5 - x) = 0
$$

Set each factor equal to zero:

$$
4x = 0 \quad \text{or} \quad 5 - x = 0
$$

$$
x = 0 \quad \text{or} \quad x = 5
$$

Since the problem specifies $ x \neq 0 $, the only valid solution is:

$$
x = 5
$$

---

### Step 4: Verify the Solution

**Case A:**

- Denali: $ 16 + 4x = 16 + 20 = 36 $
- Nate: $ 12 $
- Ratio = $ \frac{36}{12} = 3 $

**Case B:**

- Denali: $ 16 + x = 16 + 5 = 21 $
- Nate: $ 12 - x = 12 - 5 = 7 $
- Ratio = $ \frac{21}{7} = 3 $

Both cases yield the same ratio of 3:1, confirming that the solution is valid.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, Denali and Nate are dog walkers, right? Denali has 16 dogs, and Nate has 12. The company has a new policy where they can assign or unassign new dogs in groups of x dogs. The key part here is that the ratio of their pay would be the same in two different scenarios:

1. If Denali starts walking 4x more dogs and Nate stays at 12.
2. If x of Nate's dogs are reassigned to Denali.

And we need to find x, where x is not zero. Alright, let me break this down step by step.

First, let's recall that the ratio of their pay would be the same in both cases. Since they are paid per dog they walk, I assume their pay is directly proportional to the number of dogs they walk. So, the ratio of their pay is just the ratio of the number of dogs they each walk. That makes sense.

So initially, Denali has 16 dogs, Nate has 12. So the initial ratio of their pay is 16:12, which simplifies to 4:3. But maybe we need to keep it as fractions for the equations? Let me think.

But the problem says that under the new policy, they can assign or unassign new dogs in groups of x. Wait, does that mean that they can get new dogs assigned to them in groups of x? Or maybe they can reassign existing dogs? The wording says "assigned or unassigned new dogs in groups of x dogs." Hmm.

Wait, maybe the problem is about adjusting the number of dogs they have by adding or removing groups of x dogs? So for example, if they get more dogs, it's in groups of x, and if they lose dogs, it's also in groups of x. But then the problem mentions two different scenarios:

First scenario: Denali starts walking 4x more dogs and Nate stays at 12. So Denali's total dogs would be 16 + 4x? Wait, but the problem says "assigned or unassigned new dogs in groups of x dogs." So maybe adding or subtracting groups of x dogs. So if Denali is assigned 4x dogs? But 4x is not a multiple of x? Wait, 4x is 4 times x. Hmm, but if they are assigned in groups of x, then adding 4x would be adding 4 groups of x dogs. So maybe that's allowed. So maybe the first scenario is that Denali is assigned 4x dogs, so her total becomes 16 + 4x, and Nate remains at 12.

Second scenario: x of Nate's dogs are reassigned to Denali. So Nate loses x dogs, and Denali gains x dogs. Therefore, Denali's total dogs would be 16 + x, and Nate's total would be 12 - x.

In both scenarios, the ratio of Denali's pay to Nate's pay is the same. So the ratio in the first scenario (Denali has 16 + 4x, Nate has 12) is equal to the ratio in the second scenario (Denali has 16 + x, Nate has 12 - x).

Therefore, we can set up an equation:

(16 + 4x)/12 = (16 + x)/(12 - x)

Is that right? Let me check again.

Yes, the ratio of Denali's pay to Nate's pay in both cases should be equal. So first case: Denali's dogs are 16 + 4x, Nate's are still 12. Second case: Denali's dogs are 16 + x, Nate's are 12 - x. Therefore, the equation is:

(16 + 4x)/12 = (16 + x)/(12 - x)

Okay, so now we need to solve this equation for x. Let me write that down:

(16 + 4x)/12 = (16 + x)/(12 - x)

First, maybe cross-multiplying to eliminate the fractions. So multiply both sides by 12*(12 - x):

(16 + 4x)*(12 - x) = 12*(16 + x)

Let me compute both sides:

Left side: (16 + 4x)(12 - x)
Let me expand that:

First, distribute 16 over (12 - x): 16*12 - 16x
Then distribute 4x over (12 - x): 4x*12 - 4x*x
So total left side: 16*12 -16x + 48x -4x²
Simplify:

16*12 is 192
-16x + 48x is 32x
So left side is 192 + 32x -4x²

Right side: 12*(16 + x) = 12*16 + 12x = 192 + 12x

So now, set left side equal to right side:

192 + 32x -4x² = 192 + 12x

Subtract 192 from both sides:

32x -4x² = 12x

Subtract 12x from both sides:

20x -4x² = 0

Let me rearrange terms:

-4x² + 20x = 0

Multiply both sides by -1:

4x² - 20x = 0

Factor out 4x:

4x(x - 5) = 0

So, solutions are 4x = 0 => x = 0 or x - 5 = 0 => x = 5

But the problem says x ≠ 0, so x = 5.

Wait, so x is 5? Let me check this answer to make sure.

Let me verify if x = 5 satisfies the original equation.

First scenario: Denali has 16 + 4x = 16 + 20 = 36 dogs, Nate has 12 dogs. Ratio is 36:12 = 3:1.

Second scenario: Denali has 16 + x = 21 dogs, Nate has 12 - x = 7 dogs. Ratio is 21:7 = 3:1.

Yes, both ratios are 3:1. So that works. Therefore, x is indeed 5.

But let me double-check if there are any other constraints or if I misinterpreted the problem.

Wait, the problem says "assigned or unassigned new dogs in groups of x dogs." So when they say Denali started walking 4x more dogs, does that mean adding 4x dogs? But groups are of size x. So adding 4x would be 4 groups. That seems okay. Also, when x of Nate's dogs are reassigned, that would be removing x dogs from Nate and adding them to Denali. So that's a reassignment of x dogs. So that's allowed.

Is there a possibility that x has to be such that the number of dogs is positive? Let me check.

In the second scenario, Nate has 12 - x dogs. So 12 - x must be positive. Therefore, 12 - x > 0 => x < 12. Since x=5, that's okay. Also, in the first scenario, Denali's dogs are 16 + 4x, which is positive for any x. So x=5 is acceptable. Therefore, the answer is 5.

But let me check again if the original problem says anything else. Wait, the company's new policy is about assigning or unassigning new dogs in groups of x dogs. So maybe the first scenario is not adding 4x dogs, but something else? Let me read again:

"the ratio of Denali's pay to Nate's pay would be the same if Denali started walking 4x more dogs and Nate stayed at 12 dogs or if x of Nate's dogs were reassigned to Denali."

So, the first case: Denali started walking 4x more dogs. So that would mean Denali's total dogs is original 16 + 4x. Nate stays at 12. The second case: x of Nate's dogs are reassigned to Denali. So Nate's dogs become 12 - x, Denali's become 16 + x. So that's how I set up the equation.

Therefore, solving that gives x=5, and checking shows that it works. Therefore, the answer is 5. I think that is correct.

Wait, but just to make sure, let me think again about the ratio. Original ratio is 16:12, which reduces to 4:3. After the first scenario, Denali has 16 + 4x, Nate has 12. So the ratio is (16 + 4x)/12. In the second scenario, the ratio is (16 + x)/(12 - x). Setting them equal gives the equation.

But maybe I should check with x=5:

First case: (16 + 20)/12 = 36/12 = 3.

Second case: (16 + 5)/(12 - 5) = 21/7 = 3.

Yes, so both ratios are equal. Therefore, x=5 is correct.

Therefore, the answer is 5.

I think that's it. I don't see any mistakes in my reasoning. So I think x is 5.

**Final Answer**
The value of $ x $ is \boxed{5}.

To determine the value of $ x $, we analyze the given scenario involving Denali and Nate, who work for a dog-walking service. The key point is that the **ratio of their pay** remains the same in **two different assignments** of dogs:

1. **Case A:** Denali walks $ 16 + 4x $ dogs, while Nate walks 12 dogs.
2. **Case B:** Denali walks $ 16 + x $ dogs, while Nate walks $ 12 - x $ dogs.

Since their pay is directly proportional to the number of dogs they walk, the ratio of their pay is equivalent to the ratio of the number of dogs they walk. Thus, we equate the two ratios:

$$
\frac{16 + 4x}{12} = \frac{16 + x}{12 - x}
$$

---

### Step 1: Eliminate the Denominators

To simplify this equation, we cross-multiply:

$$
(16 + 4x)(12 - x) = 12(16 + x)
$$

---

### Step 2: Expand Both Sides

**Left-hand side:**

$$
(16 + 4x)(12 - x) = 16 \cdot 12 - 16x + 4x \cdot 12 - 4x^2 = 192 - 16x + 48x - 4x^2 = 192 + 32x - 4x^2
$$

**Right-hand side:**

$$
12(16 + x) = 192 + 12x
$$

Now we equate both expressions:

$$
192 + 32x - 4x^2 = 192 + 12x
$$

---

### Step 3: Simplify the Equation

Subtract 192 from both sides:

$$
32x - 4x^2 = 12x
$$

Subtract $ 12x $ from both sides:

$$
20x - 4x^2 = 0
$$

Factor the left-hand side:

$$
4x(5 - x) = 0
$$

Set each factor equal to zero:

$$
4x = 0 \quad \text{or} \quad 5 - x = 0
$$

$$
x = 0 \quad \text{or} \quad x = 5
$$

Since the problem specifies $ x \neq 0 $, the only valid solution is:

$$
x = 5
$$

---

### Step 4: Verify the Solution

**Case A:**

- Denali: $ 16 + 4x = 16 + 20 = 36 $
- Nate: $ 12 $
- Ratio = $ \frac{36}{12} = 3 $

**Case B:**

- Denali: $ 16 + x = 16 + 5 = 21 $
- Nate: $ 12 - x = 12 - 5 = 7 $
- Ratio = $ \frac{21}{7} = 3 $

Both cases yield the same ratio of 3:1, confirming that the solution is valid.

---

### Final Answer

$$
\boxed{5}
$$

Solution:
Rewriting the sentence "the ratio of Denali's pay to Nate's pay would be the same if Denali started walking $4x$ more dogs and Nate stayed at $12$ dogs or if $x$ of Nate's dogs were reassigned to Denali" as an equation, we have \[\frac{16+4x}{12}=\frac{16+x}{12-x}.\]Clearing denominators, \begin{align*}
(16+4x)(12-x)&=(16+x)(12)\quad \Rightarrow\\
192-16x+48x-4x^2&=192+12x\quad \Rightarrow\\
32x-4x^2&=12x\quad \Rightarrow\\
0&=4x^2-20x\quad \Rightarrow\\
0&=4x(x-5).
\end{align*}Because $x$ cannot be $0$, $x=\boxed{5}$.


Explanation:
Okay, so I need to solve the equation $ x = \sqrt{11 - 2x} + 4 $. Hmm, let me think. First, 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 before I do that, I should note any restrictions on the variable because square roots can't be negative.

So, the expression under the square root is $ 11 - 2x $. That must be greater than or equal to zero. So:

$ 11 - 2x \geq 0 $

Let me solve that inequality. Subtract 11 from both sides:

$ -2x \geq -11 $

Divide both sides by -2. Wait, dividing by a negative number reverses the inequality sign. So:

$ x \leq \frac{11}{2} $

Which is $ x \leq 5.5 $. Okay, so whatever solution I get, it must be less than or equal to 5.5. Got it.

Now, back to the original equation: $ x = \sqrt{11 - 2x} + 4 $. Let me try to isolate the square root. If I subtract 4 from both sides:

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

Okay, so now the square root is by itself on the right side. Now, I can square both sides 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: $ 11 - 2x $

So now the equation becomes:

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

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

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

Simplify the terms:

Combine the $ -8x + 2x $: that's $ -6x $

Combine the constants: 16 - 11 = 5

So the equation is:

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

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

$ (x - 1)(x - 5) = x^2 - 6x + 5 $. Yes, that's correct.

So, the equation factors as:

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

Therefore, the solutions are:

$ x = 1 $ or $ x = 5 $

But wait, since we squared both sides of the equation earlier, there might be extraneous solutions. So I need to check both solutions in the original equation to make sure they work.

First, let's 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. So, $ x = 1 $ is an extraneous solution. It doesn't work.

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 the valid solution.

Wait, but hold on. Let me double-check my steps to make sure I didn't make any mistakes. Let me go through again.

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

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

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

Which expands to $ x^2 - 8x + 16 = 11 - 2x $

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

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

Factor: $ (x - 1)(x - 5) = 0 $. That's correct.

So solutions x=1 and x=5. Check x=1: 1 vs sqrt(9) +4 = 3+4=7. Not equal. So extraneous. x=5: sqrt(11 - 10) +4 = sqrt(1) +4=5. That works.

So only x=5 is the solution. So the answer is x=5.

But just to make sure, let me think if there's another way to approach this problem or if I missed something. For example, maybe when I squared both sides, I introduced an extraneous solution, but is there a possibility of another solution?

Alternatively, maybe I could have moved terms differently. Let me try solving the original equation again step by step.

Original equation:

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

Let me subtract 4 from both sides again:

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

Then square both sides:

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

Right side: $ 11 - 2x $

Set equal:

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

Bring all terms to left:

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

Which is:

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

Same as before. So same solutions. So checking again gives only x=5 as valid. Therefore, I think that's the only solution.

But just to be thorough, let me check if there could be another solution if I had manipulated the equation differently. For example, suppose I tried to isolate the square root another way, but I don't think that would change anything. Let me see.

Alternatively, perhaps I can consider substituting variables? Let me think. Let me let y = sqrt(11 - 2x). Then the original equation is x = y + 4. Also, since y = sqrt(11 - 2x), then squaring both sides gives y² = 11 - 2x. So now, we have a system of equations:

1. x = y + 4

2. y² = 11 - 2x

Substitute equation 1 into equation 2:

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. Therefore, only y = 1 is valid.

Then from equation 1: x = y + 4 = 1 + 4 = 5. So same result. Therefore, x=5 is the only solution.

This confirms the earlier result. So even when approaching it through substitution, same answer. Therefore, seems like x=5 is the only valid solution.

Another check: Let me plug x=5 into original equation. Left side is 5. Right side is sqrt(11 - 2*5) + 4 = sqrt(11 -10) +4 = sqrt(1) +4 =1 +4=5. Perfect.

And for x=1, which was extraneous, let me check the step before squaring. When I had x - 4 = sqrt(11 - 2x). For x=1, left side is 1 - 4 = -3. Right side is sqrt(11 - 2*1)=sqrt(9)=3. So -3 vs 3. They are not equal. Which is why squaring both sides introduced that extraneous solution. Because when you square both sides, you can get solutions where the original sides are negatives of each other. So that's why checking is important.

Therefore, the only solution is x=5.

I think that's thorough. I considered different methods, checked for extraneous solutions, verified the answer. Seems solid.

**Final Answer**
The solution is \boxed{5}.

To solve the equation:

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

we begin by identifying any domain restrictions. Since the square root is defined only when its argument is non-negative:

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

So, any valid solution must satisfy $ x \leq 5.5 $.

---

### Step 1: Isolate the square root

We isolate the square root by subtracting 4 from both sides:

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

---

### Step 2: Square both sides

To eliminate the square root, we square both sides:

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

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

---

### Step 3: Simplify and solve the quadratic

Bring all terms to one side:

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

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

Factor the quadratic:

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

So, the potential solutions are:

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

---

### Step 4: Check for extraneous solutions

We must verify both solutions against the original equation to ensure they are valid.

- **For $ x = 1 $**:

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

So, $ 1 \ne 7 $, making $ x = 1 $ an **extraneous solution**.

- **For $ x = 5 $**:

$$
\text{LHS: } x = 5 \\
\text{RHS: } \sqrt{11 - 2(5)} + 4 = \sqrt{1} + 4 = 1 + 4 = 5
$$

So, $ 5 = 5 $, and $ x = 5 $ is a valid solution.

---

### Step 5: Final verification

Let’s verify that the solution satisfies the domain condition:

$$
x = 5 \leq \frac{11}{2} = 5.5
$$

Yes, it does.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $ x = \sqrt{11 - 2x} + 4 $. Hmm, let me think. First, 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 before I do that, I should note any restrictions on the variable because square roots can't be negative.

So, the expression under the square root is $ 11 - 2x $. That must be greater than or equal to zero. So:

$ 11 - 2x \geq 0 $

Let me solve that inequality. Subtract 11 from both sides:

$ -2x \geq -11 $

Divide both sides by -2. Wait, dividing by a negative number reverses the inequality sign. So:

$ x \leq \frac{11}{2} $

Which is $ x \leq 5.5 $. Okay, so whatever solution I get, it must be less than or equal to 5.5. Got it.

Now, back to the original equation: $ x = \sqrt{11 - 2x} + 4 $. Let me try to isolate the square root. If I subtract 4 from both sides:

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

Okay, so now the square root is by itself on the right side. Now, I can square both sides 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: $ 11 - 2x $

So now the equation becomes:

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

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

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

Simplify the terms:

Combine the $ -8x + 2x $: that's $ -6x $

Combine the constants: 16 - 11 = 5

So the equation is:

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

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

$ (x - 1)(x - 5) = x^2 - 6x + 5 $. Yes, that's correct.

So, the equation factors as:

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

Therefore, the solutions are:

$ x = 1 $ or $ x = 5 $

But wait, since we squared both sides of the equation earlier, there might be extraneous solutions. So I need to check both solutions in the original equation to make sure they work.

First, let's 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. So, $ x = 1 $ is an extraneous solution. It doesn't work.

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 the valid solution.

Wait, but hold on. Let me double-check my steps to make sure I didn't make any mistakes. Let me go through again.

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

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

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

Which expands to $ x^2 - 8x + 16 = 11 - 2x $

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

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

Factor: $ (x - 1)(x - 5) = 0 $. That's correct.

So solutions x=1 and x=5. Check x=1: 1 vs sqrt(9) +4 = 3+4=7. Not equal. So extraneous. x=5: sqrt(11 - 10) +4 = sqrt(1) +4=5. That works.

So only x=5 is the solution. So the answer is x=5.

But just to make sure, let me think if there's another way to approach this problem or if I missed something. For example, maybe when I squared both sides, I introduced an extraneous solution, but is there a possibility of another solution?

Alternatively, maybe I could have moved terms differently. Let me try solving the original equation again step by step.

Original equation:

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

Let me subtract 4 from both sides again:

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

Then square both sides:

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

Right side: $ 11 - 2x $

Set equal:

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

Bring all terms to left:

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

Which is:

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

Same as before. So same solutions. So checking again gives only x=5 as valid. Therefore, I think that's the only solution.

But just to be thorough, let me check if there could be another solution if I had manipulated the equation differently. For example, suppose I tried to isolate the square root another way, but I don't think that would change anything. Let me see.

Alternatively, perhaps I can consider substituting variables? Let me think. Let me let y = sqrt(11 - 2x). Then the original equation is x = y + 4. Also, since y = sqrt(11 - 2x), then squaring both sides gives y² = 11 - 2x. So now, we have a system of equations:

1. x = y + 4

2. y² = 11 - 2x

Substitute equation 1 into equation 2:

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. Therefore, only y = 1 is valid.

Then from equation 1: x = y + 4 = 1 + 4 = 5. So same result. Therefore, x=5 is the only solution.

This confirms the earlier result. So even when approaching it through substitution, same answer. Therefore, seems like x=5 is the only valid solution.

Another check: Let me plug x=5 into original equation. Left side is 5. Right side is sqrt(11 - 2*5) + 4 = sqrt(11 -10) +4 = sqrt(1) +4 =1 +4=5. Perfect.

And for x=1, which was extraneous, let me check the step before squaring. When I had x - 4 = sqrt(11 - 2x). For x=1, left side is 1 - 4 = -3. Right side is sqrt(11 - 2*1)=sqrt(9)=3. So -3 vs 3. They are not equal. Which is why squaring both sides introduced that extraneous solution. Because when you square both sides, you can get solutions where the original sides are negatives of each other. So that's why checking is important.

Therefore, the only solution is x=5.

I think that's thorough. I considered different methods, checked for extraneous solutions, verified the answer. Seems solid.

**Final Answer**
The solution is \boxed{5}.

To solve the equation:

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

we begin by identifying any domain restrictions. Since the square root is defined only when its argument is non-negative:

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

So, any valid solution must satisfy $ x \leq 5.5 $.

---

### Step 1: Isolate the square root

We isolate the square root by subtracting 4 from both sides:

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

---

### Step 2: Square both sides

To eliminate the square root, we square both sides:

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

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

---

### Step 3: Simplify and solve the quadratic

Bring all terms to one side:

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

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

Factor the quadratic:

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

So, the potential solutions are:

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

---

### Step 4: Check for extraneous solutions

We must verify both solutions against the original equation to ensure they are valid.

- **For $ x = 1 $**:

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

So, $ 1 \ne 7 $, making $ x = 1 $ an **extraneous solution**.

- **For $ x = 5 $**:

$$
\text{LHS: } x = 5 \\
\text{RHS: } \sqrt{11 - 2(5)} + 4 = \sqrt{1} + 4 = 1 + 4 = 5
$$

So, $ 5 = 5 $, and $ x = 5 $ is a valid solution.

---

### Step 5: Final verification

Let’s verify that the solution satisfies the domain condition:

$$
x = 5 \leq \frac{11}{2} = 5.5
$$

Yes, it does.

---

### 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 he 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. The question is asking for the minimal compound interest rate the account must provide. Alright, let me try to figure this out step by step.

First, let me recall how compound interest works. If you deposit money into an account, the interest is calculated on the current amount in the account, including previous interest. The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, including interest. P is the principal amount, r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the time the money is invested for in years. But in this case, the worker is making annual deposits, so it's not a single lump sum deposit but a series of deposits. Therefore, this seems like an annuity problem.

Since the deposits are made at the end of each year, it's an ordinary annuity. The formula for the future value of an ordinary annuity is:

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

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

In this problem, the worker deposits $20,000 each year for 3 years, so PMT = $20,000, n = 3. The future value after 3 years should be at least $66,200. Therefore, we need to solve for r in the equation:

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

Hmm, okay. Let me write that equation down more clearly:

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

So we need to solve for r here. Since this is a nonlinear equation, it might not be straightforward to solve algebraically. Maybe I can rearrange it or use some trial and error? Or perhaps there's a way to simplify it?

Let me first divide both sides by 20,000 to make the equation simpler.

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

Calculating the right-hand side: 66200 divided by 20000. Let me compute that. 20000 goes into 66200 three times (20000 * 3 = 60000), then 66200 - 60000 = 6200. So that's 3.31. Wait, 66200 / 20000 = 3.31? Let me check with decimals. 66200 divided by 20000. Dividing numerator and denominator by 100: 662 / 200 = 3.31. Yes, correct. So:

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

So now, we need to find r such that ((1 + r)^3 - 1)/r = 3.31.

This seems a bit tricky. Maybe I can try plugging in some interest rates and see which one gives me approximately 3.31?

Alternatively, perhaps I can expand the left-hand side?

Let me think. Let me denote x = 1 + r. Then the equation becomes:

(x^3 - 1)/ (x - 1) ) = 3.31

Wait, because if x = 1 + r, then (x^3 - 1)/r = (x^3 - 1)/(x - 1). Since r = x - 1. Therefore, the equation simplifies to:

(x^3 - 1)/(x - 1) = 3.31

But x^3 - 1 factors as (x - 1)(x^2 + x + 1). Therefore, (x^3 - 1)/(x - 1) is equal to x^2 + x + 1. So the equation simplifies to:

x^2 + x + 1 = 3.31

Which is:

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

Wait, hold on! That seems too easy. Let me check:

Original equation: [(1 + r)^3 - 1]/r = 3.31. Then, substituting x = 1 + r, so r = x - 1. So [(x^3 - 1)/ (x - 1)] = 3.31. Then, since x^3 - 1 = (x - 1)(x^2 + x + 1), dividing both sides by (x - 1) gives x^2 + x + 1 = 3.31. So indeed, the equation simplifies to x^2 + x + 1 = 3.31, which is x^2 + x - 2.31 = 0. Then, solving this quadratic equation for x:

x = [-1 ± sqrt(1 + 4 * 2.31)] / 2

Compute discriminant:

sqrt(1 + 9.24) = sqrt(10.24). Wait, sqrt(10.24) is 3.2? Because 3.2^2 = 10.24. Yes, so:

x = [ -1 ± 3.2 ] / 2

Since x = 1 + r must be positive, we take the positive root:

x = (-1 + 3.2)/2 = 2.2 / 2 = 1.1

Therefore, x = 1.1, so 1 + r = 1.1, so r = 0.1, or 10%.

Wait, so the interest rate is 10%? Let me check if this is correct. Let me compute the future value with 10% interest rate.

First, let's compute the future value of each deposit:

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

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

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

So total future value:

FV = 20000*(1 + 0.1)^2 + 20000*(1 + 0.1)^1 + 20000*(1 + 0.1)^0

Compute each term:

First term: 20000*(1.21) = 24,200

Second term: 20000*(1.1) = 22,000

Third term: 20000*1 = 20,000

Total FV: 24,200 + 22,000 + 20,000 = 66,200

Oh wow! That's exactly the amount needed. So with a 10% interest rate, the future value is exactly $66,200. So then the minimal compound interest rate is 10%. Therefore, the answer is 10.

Wait, but let me check if maybe there's a lower rate? Because the problem says "minimal compound interest rate", so maybe there's a rate lower than 10% that also gives at least $66,200? Let me check with 9%.

Let me compute FV with 9% interest.

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: 23,762 + 21,800 + 20,000 = 65,562

Which is less than 66,200. So 9% is not enough. How about 10% gives exactly 66,200, so that's the minimal rate. Because if you try 9.5%, let's check.

Compute 9.5% interest.

First term: 20000*(1.095)^2. Let me calculate 1.095 squared. 1.095^2 = ?

Well, 1.095 * 1.095. Let me compute:

First, 1 * 1.095 = 1.095

Then 0.095 * 1.095: Let's compute 1.095 * 0.095.

Wait, maybe better to compute (1 + 0.095)^2 = 1 + 2*0.095 + 0.095^2 = 1 + 0.19 + 0.009025 = 1.199025

Therefore, first term is 20000 * 1.199025 = 23,980.5

Second term: 20000 * 1.095 = 21,900

Third term: 20,000

Total FV: 23,980.5 + 21,900 + 20,000 = 65,880.5

Still less than 66,200. So even at 9.5%, it's only about 65,880.5. Still less. What about 10% gives exactly 66,200. Therefore, 10% is the minimal rate.

Alternatively, maybe there's a rate between 9.5% and 10% that gives exactly 66,200? Wait, but according to the algebra earlier, the equation simplified to a quadratic which gave exactly r = 10%. Therefore, in the equation [(1 + r)^3 - 1]/r = 3.31, solving gives r = 10%. Therefore, that is the exact rate. Therefore, there's no lower rate. So 10% is the minimal rate.

Wait, but let me think again. How did we arrive at the quadratic equation? Because the original problem was with annual deposits, and when we did the substitution, we assumed that the deposits are made at the end of each year, and the interest is compounded annually? Wait, but is that how it is? The problem says "compound interest rate". Typically, compound interest can be compounded multiple times per year, but since the deposits are made annually, perhaps the interest is compounded annually. The problem doesn't specify the compounding frequency. Hmm. Wait, the problem says "compound interest rate", and since the deposits are made annually, it's common to assume annual compounding unless stated otherwise. However, if it was compounded more frequently, the calculation might be different. However, the original equation we used assumes annual compounding, since the formula for the future value of an ordinary annuity with annual compounding is indeed FV = PMT * [ ((1 + r)^n - 1)/r ].

Therefore, if the problem is using annual compounding, then our calculation is correct, and 10% is the answer. However, if the interest is compounded more frequently, then the effective rate might be different. But since the problem doesn't specify compounding frequency, and in most cases, when talking about an annual interest rate for such problems, it's safe to assume annual compounding. Moreover, since the equation simplified nicely to 10% with annual compounding, and the answer checks out exactly, that suggests that this is the correct answer.

Therefore, the minimal compound interest rate is 10%.

But let me just confirm once again. Let me think if there's another approach. For example, using the formula for future value of annuity:

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

We have:

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

Divide both sides by 20000:

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

Let me check with r = 0.1, as before:

((1.1)^3 - 1)/0.1 = (1.331 - 1)/0.1 = 0.331 / 0.1 = 3.31, which matches. Therefore, this is correct.

So, solving this equation gives exactly r = 10%. Hence, the minimal rate is 10%.

Alternatively, if someone tried to use the formula for compound interest with different compounding periods, but since the deposits are annual, and the problem says "compound interest rate", I think the answer is 10%.

Therefore, the answer is 10.

**Final Answer**
\boxed{10}

To determine the minimal compound interest rate required for the worker to accumulate at least \$66,200 in his savings account by the end of the third year, we treat the situation as an **ordinary annuity** (deposits at the end of each year). The formula for the future value (FV) of an ordinary annuity with regular annual deposits is:

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

Where:
- $ PMT = 20,000 $ (annual deposit),
- $ r $ is the annual compound interest rate (in decimal form),
- $ n = 3 $ (number of years),
- $ FV = 66,200 $ (desired amount after 3 years).

---

### Step 1: Set up the equation

Substituting the known values:

$$
66,200 = 20,000 \times \left( \frac{(1 + r)^3 - 1}{r} \right)
$$

Divide both sides by 20,000:

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

---

### Step 2: Simplify the equation

Let $ x = 1 + r $. Then the equation becomes:

$$
\frac{x^3 - 1}{x - 1} = 3.31
$$

Using polynomial factorization:

$$
x^3 - 1 = (x - 1)(x^2 + x + 1)
$$

So:

$$
\frac{(x - 1)(x^2 + x + 1)}{x - 1} = x^2 + x + 1 = 3.31
$$

Now solve the quadratic:

$$
x^2 + x + 1 = 3.31 \Rightarrow x^2 + x - 2.31 = 0
$$

Use the quadratic formula:

$$
x = \frac{-1 \pm \sqrt{1 + 4 \cdot 2.31}}{2} = \frac{-1 \pm \sqrt{10.24}}{2}
$$

$$
x = \frac{-1 \pm 3.2}{2}
$$

Since $ x = 1 + r $ must be positive, we take the positive root:

$$
x = \frac{-1 + 3.2}{2} = \frac{2.2}{2} = 1.1
$$

So:

$$
1 + r = 1.1 \Rightarrow r = 0.1
$$

---

### Step 3: Verify the solution

Let’s double-check with $ r = 0.1 $, or 10%, to confirm that the future value is exactly \$66,200:

$$
FV = 20000 \times \left( \frac{(1.1)^3 - 1}{0.1} \right)
$$

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

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

Yes, the calculation matches exactly.

---

### Step 4: Consider lower rates

Check with $ r = 0.09 $ (9%) to ensure 10% is the **minimal** rate:

$$
(1.09)^3 ≈ 1.295029 \Rightarrow 1.295029 - 1 = 0.295029
$$

$$
\frac{0.295029}{0.09} ≈ 3.2781 \Rightarrow 20000 \times 3.2781 ≈ 65562
$$

Which is less than 66,200. So, **9% is insufficient** to reach the target.

Thus, **10% is the minimal** compound interest rate that meets the requirement.

---

### Final Answer

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so there's this worker who gets $20,000 every year, and he 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. The question is asking for the minimal compound interest rate the account must provide. Alright, let me try to figure this out step by step.

First, let me recall how compound interest works. If you deposit money into an account, the interest is calculated on the current amount in the account, including previous interest. The formula for compound interest is A = P(1 + r/n)^(nt), where A is the amount of money accumulated after n years, including interest. P is the principal amount, r is the annual interest rate, n is the number of times that interest is compounded per year, and t is the time the money is invested for in years. But in this case, the worker is making annual deposits, so it's not a single lump sum deposit but a series of deposits. Therefore, this seems like an annuity problem.

Since the deposits are made at the end of each year, it's an ordinary annuity. The formula for the future value of an ordinary annuity is:

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

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

In this problem, the worker deposits $20,000 each year for 3 years, so PMT = $20,000, n = 3. The future value after 3 years should be at least $66,200. Therefore, we need to solve for r in the equation:

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

Hmm, okay. Let me write that equation down more clearly:

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

So we need to solve for r here. Since this is a nonlinear equation, it might not be straightforward to solve algebraically. Maybe I can rearrange it or use some trial and error? Or perhaps there's a way to simplify it?

Let me first divide both sides by 20,000 to make the equation simpler.

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

Calculating the right-hand side: 66200 divided by 20000. Let me compute that. 20000 goes into 66200 three times (20000 * 3 = 60000), then 66200 - 60000 = 6200. So that's 3.31. Wait, 66200 / 20000 = 3.31? Let me check with decimals. 66200 divided by 20000. Dividing numerator and denominator by 100: 662 / 200 = 3.31. Yes, correct. So:

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

So now, we need to find r such that ((1 + r)^3 - 1)/r = 3.31.

This seems a bit tricky. Maybe I can try plugging in some interest rates and see which one gives me approximately 3.31?

Alternatively, perhaps I can expand the left-hand side?

Let me think. Let me denote x = 1 + r. Then the equation becomes:

(x^3 - 1)/ (x - 1) ) = 3.31

Wait, because if x = 1 + r, then (x^3 - 1)/r = (x^3 - 1)/(x - 1). Since r = x - 1. Therefore, the equation simplifies to:

(x^3 - 1)/(x - 1) = 3.31

But x^3 - 1 factors as (x - 1)(x^2 + x + 1). Therefore, (x^3 - 1)/(x - 1) is equal to x^2 + x + 1. So the equation simplifies to:

x^2 + x + 1 = 3.31

Which is:

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

Wait, hold on! That seems too easy. Let me check:

Original equation: [(1 + r)^3 - 1]/r = 3.31. Then, substituting x = 1 + r, so r = x - 1. So [(x^3 - 1)/ (x - 1)] = 3.31. Then, since x^3 - 1 = (x - 1)(x^2 + x + 1), dividing both sides by (x - 1) gives x^2 + x + 1 = 3.31. So indeed, the equation simplifies to x^2 + x + 1 = 3.31, which is x^2 + x - 2.31 = 0. Then, solving this quadratic equation for x:

x = [-1 ± sqrt(1 + 4 * 2.31)] / 2

Compute discriminant:

sqrt(1 + 9.24) = sqrt(10.24). Wait, sqrt(10.24) is 3.2? Because 3.2^2 = 10.24. Yes, so:

x = [ -1 ± 3.2 ] / 2

Since x = 1 + r must be positive, we take the positive root:

x = (-1 + 3.2)/2 = 2.2 / 2 = 1.1

Therefore, x = 1.1, so 1 + r = 1.1, so r = 0.1, or 10%.

Wait, so the interest rate is 10%? Let me check if this is correct. Let me compute the future value with 10% interest rate.

First, let's compute the future value of each deposit:

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

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

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

So total future value:

FV = 20000*(1 + 0.1)^2 + 20000*(1 + 0.1)^1 + 20000*(1 + 0.1)^0

Compute each term:

First term: 20000*(1.21) = 24,200

Second term: 20000*(1.1) = 22,000

Third term: 20000*1 = 20,000

Total FV: 24,200 + 22,000 + 20,000 = 66,200

Oh wow! That's exactly the amount needed. So with a 10% interest rate, the future value is exactly $66,200. So then the minimal compound interest rate is 10%. Therefore, the answer is 10.

Wait, but let me check if maybe there's a lower rate? Because the problem says "minimal compound interest rate", so maybe there's a rate lower than 10% that also gives at least $66,200? Let me check with 9%.

Let me compute FV with 9% interest.

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: 23,762 + 21,800 + 20,000 = 65,562

Which is less than 66,200. So 9% is not enough. How about 10% gives exactly 66,200, so that's the minimal rate. Because if you try 9.5%, let's check.

Compute 9.5% interest.

First term: 20000*(1.095)^2. Let me calculate 1.095 squared. 1.095^2 = ?

Well, 1.095 * 1.095. Let me compute:

First, 1 * 1.095 = 1.095

Then 0.095 * 1.095: Let's compute 1.095 * 0.095.

Wait, maybe better to compute (1 + 0.095)^2 = 1 + 2*0.095 + 0.095^2 = 1 + 0.19 + 0.009025 = 1.199025

Therefore, first term is 20000 * 1.199025 = 23,980.5

Second term: 20000 * 1.095 = 21,900

Third term: 20,000

Total FV: 23,980.5 + 21,900 + 20,000 = 65,880.5

Still less than 66,200. So even at 9.5%, it's only about 65,880.5. Still less. What about 10% gives exactly 66,200. Therefore, 10% is the minimal rate.

Alternatively, maybe there's a rate between 9.5% and 10% that gives exactly 66,200? Wait, but according to the algebra earlier, the equation simplified to a quadratic which gave exactly r = 10%. Therefore, in the equation [(1 + r)^3 - 1]/r = 3.31, solving gives r = 10%. Therefore, that is the exact rate. Therefore, there's no lower rate. So 10% is the minimal rate.

Wait, but let me think again. How did we arrive at the quadratic equation? Because the original problem was with annual deposits, and when we did the substitution, we assumed that the deposits are made at the end of each year, and the interest is compounded annually? Wait, but is that how it is? The problem says "compound interest rate". Typically, compound interest can be compounded multiple times per year, but since the deposits are made annually, perhaps the interest is compounded annually. The problem doesn't specify the compounding frequency. Hmm. Wait, the problem says "compound interest rate", and since the deposits are made annually, it's common to assume annual compounding unless stated otherwise. However, if it was compounded more frequently, the calculation might be different. However, the original equation we used assumes annual compounding, since the formula for the future value of an ordinary annuity with annual compounding is indeed FV = PMT * [ ((1 + r)^n - 1)/r ].

Therefore, if the problem is using annual compounding, then our calculation is correct, and 10% is the answer. However, if the interest is compounded more frequently, then the effective rate might be different. But since the problem doesn't specify compounding frequency, and in most cases, when talking about an annual interest rate for such problems, it's safe to assume annual compounding. Moreover, since the equation simplified nicely to 10% with annual compounding, and the answer checks out exactly, that suggests that this is the correct answer.

Therefore, the minimal compound interest rate is 10%.

But let me just confirm once again. Let me think if there's another approach. For example, using the formula for future value of annuity:

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

We have:

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

Divide both sides by 20000:

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

Let me check with r = 0.1, as before:

((1.1)^3 - 1)/0.1 = (1.331 - 1)/0.1 = 0.331 / 0.1 = 3.31, which matches. Therefore, this is correct.

So, solving this equation gives exactly r = 10%. Hence, the minimal rate is 10%.

Alternatively, if someone tried to use the formula for compound interest with different compounding periods, but since the deposits are annual, and the problem says "compound interest rate", I think the answer is 10%.

Therefore, the answer is 10.

**Final Answer**
\boxed{10}

To determine the minimal compound interest rate required for the worker to accumulate at least \$66,200 in his savings account by the end of the third year, we treat the situation as an **ordinary annuity** (deposits at the end of each year). The formula for the future value (FV) of an ordinary annuity with regular annual deposits is:

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

Where:
- $ PMT = 20,000 $ (annual deposit),
- $ r $ is the annual compound interest rate (in decimal form),
- $ n = 3 $ (number of years),
- $ FV = 66,200 $ (desired amount after 3 years).

---

### Step 1: Set up the equation

Substituting the known values:

$$
66,200 = 20,000 \times \left( \frac{(1 + r)^3 - 1}{r} \right)
$$

Divide both sides by 20,000:

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

---

### Step 2: Simplify the equation

Let $ x = 1 + r $. Then the equation becomes:

$$
\frac{x^3 - 1}{x - 1} = 3.31
$$

Using polynomial factorization:

$$
x^3 - 1 = (x - 1)(x^2 + x + 1)
$$

So:

$$
\frac{(x - 1)(x^2 + x + 1)}{x - 1} = x^2 + x + 1 = 3.31
$$

Now solve the quadratic:

$$
x^2 + x + 1 = 3.31 \Rightarrow x^2 + x - 2.31 = 0
$$

Use the quadratic formula:

$$
x = \frac{-1 \pm \sqrt{1 + 4 \cdot 2.31}}{2} = \frac{-1 \pm \sqrt{10.24}}{2}
$$

$$
x = \frac{-1 \pm 3.2}{2}
$$

Since $ x = 1 + r $ must be positive, we take the positive root:

$$
x = \frac{-1 + 3.2}{2} = \frac{2.2}{2} = 1.1
$$

So:

$$
1 + r = 1.1 \Rightarrow r = 0.1
$$

---

### Step 3: Verify the solution

Let’s double-check with $ r = 0.1 $, or 10%, to confirm that the future value is exactly \$66,200:

$$
FV = 20000 \times \left( \frac{(1.1)^3 - 1}{0.1} \right)
$$

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

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

Yes, the calculation matches exactly.

---

### Step 4: Consider lower rates

Check with $ r = 0.09 $ (9%) to ensure 10% is the **minimal** rate:

$$
(1.09)^3 ≈ 1.295029 \Rightarrow 1.295029 - 1 = 0.295029
$$

$$
\frac{0.295029}{0.09} ≈ 3.2781 \Rightarrow 20000 \times 3.2781 ≈ 65562
$$

Which is less than 66,200. So, **9% is insufficient** to reach the target.

Thus, **10% is the minimal** compound interest rate that meets the requirement.

---

### Final Answer

$$
\boxed{10}
$$

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 know that f(1) = 1. The task is to find all integers n such that f(n) = n. Alright, let me try to work through this step by step.

First, functional equations often require finding the general form of the function or figuring out specific values. Since we have a condition on f(1), maybe we can find a general formula for f(x) and then check when f(n) = n for integer n.

Let me recall that functional equations sometimes can be turned into differential equations or related to known functions. But since this is a Cauchy-like equation, perhaps we can find a quadratic function? Let me think.

The given equation is f(x) + f(y) = f(x + y) - xy - 1. Let me rearrange this equation:

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

This looks similar to the Cauchy equation f(x + y) = f(x) + f(y), but with an extra term xy + 1. So maybe f(x) is a quadratic function? Let me assume that f(x) is a quadratic function. Let me suppose that f(x) is a quadratic function of the form f(x) = ax² + bx + c, where a, b, c are constants. Then I can substitute this into the equation and see if I can find a, b, c.

Let me try that. So substitute f(x) = ax² + bx + c into the equation f(x + y) = f(x) + f(y) + xy + 1.

First compute f(x + y):

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

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

[ax² + bx + c] + [ay² + by + c] + xy + 1 = ax² + ay² + bx + by + 2c + xy + 1.

So set these equal:

Left side: a x² + 2a xy + a y² + b x + b y + c

Right side: ax² + ay² + bx + by + 2c + xy + 1

Subtract the right side from the left side:

Left - Right = [a x² + 2a xy + a y² + b x + b y + c] - [ax² + ay² + bx + by + 2c + xy + 1] =

Let me compute term by term:

a x² - a x² = 0

a y² - a y² = 0

2a xy - xy = (2a - 1)xy

b x - b x = 0

b y - b y = 0

c - 2c - 1 = -c - 1.

Therefore, Left - Right = (2a - 1)xy - c - 1.

But since the original equation is f(x + y) = f(x) + f(y) + xy + 1, the difference should be zero for all x, y. Therefore, the coefficients of the terms must be zero. So:

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

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

So we have a = 1/2, c = -1. Then the coefficient b remains? Wait, but in the equation above, after subtracting, there were no terms with b, so the coefficients for bx and by canceled out. Therefore, there is no restriction on b from this? Wait, but the equation is supposed to hold for all x and y, so unless the coefficients of all terms are zero. However, we only accounted for the xy term and the constant term. The terms with x and y? Wait, no, when we subtracted, they canceled out. Therefore, there's no constraints on b? But then how do we find b?

Wait, but maybe there is another condition. Remember, we also have the value f(1) = 1. Let me check that. So once we have a = 1/2, c = -1, then f(x) = (1/2)x² + bx - 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 this.

So let me compute f(x) + f(y):

f(x) + f(y) = [ (1/2)x² + (3/2)x - 1 ] + [ (1/2)y² + (3/2)y - 1 ] = (1/2)(x² + y²) + (3/2)(x + y) - 2.

Now compute f(x + y) - xy - 1:

First, compute f(x + y):

f(x + y) = (1/2)(x + y)^2 + (3/2)(x + y) - 1 = (1/2)(x² + 2xy + y²) + (3/2)x + (3/2)y - 1 = (1/2)x² + xy + (1/2)y² + (3/2)x + (3/2)y - 1.

Then subtract xy and 1:

f(x + y) - xy - 1 = [ (1/2)x² + xy + (1/2)y² + (3/2)x + (3/2)y - 1 ] - xy - 1 = (1/2)x² + (1/2)y² + (3/2)x + (3/2)y - 2.

Compare with f(x) + f(y) which was (1/2)(x² + y²) + (3/2)(x + y) - 2. These are the same! So yes, the function satisfies the equation. Great. Therefore, our 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 let's set f(n) = n and solve for n.

So:

(1/2)n² + (3/2)n - 1 = n.

Multiply both sides by 2 to eliminate denominators:

n² + 3n - 2 = 2n.

Subtract 2n from both sides:

n² + n - 2 = 0.

Now solve the quadratic equation n² + n - 2 = 0.

Using quadratic formula:

n = [ -1 ± sqrt(1 + 8) ] / 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.

Wait, let me check. Let me verify these values.

First, check n = 1:

f(1) = (1/2)(1)^2 + (3/2)(1) - 1 = 1/2 + 3/2 - 1 = (1 + 3)/2 - 1 = 2 - 1 = 1. Correct, since f(1) = 1.

Now check n = -2:

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Which equals -2. Correct.

So both n = 1 and n = -2 satisfy f(n) = n.

But wait, the problem says "find all integers n such that f(n) = n." So are there any other integer solutions? The quadratic equation only has two roots, so unless there are other solutions, but since we solved the equation f(n) = n, which is a quadratic equation, so only two solutions. So the integers are -2 and 1. So the answer should be -2 and 1.

Wait, but let me check again if there's a possibility that maybe my assumption that f is quadratic is wrong? Because sometimes functional equations can have multiple solutions, but given that we have the equation f(x + y) = f(x) + f(y) + xy + 1, and assuming f is quadratic seems reasonable. But let me check if the function is indeed uniquely determined by the given conditions.

In functional equations, sometimes you have to check if the solution is unique. Let me think. Since the equation is for all real numbers x and y, and we found a quadratic function that satisfies it, and given that the equation is similar to Cauchy's equation but with an extra term, which is quadratic. Also, since the equation is linear in f(x + y) and the rest, but with a quadratic term in x and y, so the solution is likely unique. Therefore, our function f(x) = (1/2)x² + (3/2)x - 1 is the unique solution. Therefore, the equation f(n) = n only has integer solutions at n = -2 and n = 1. Hence, those are the answers.

But just to make sure, let me check another integer value to see if maybe the function has more fixed points? For example, let me compute f(0):

f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. So f(0) = -1 ≠ 0. So 0 is not a solution.

How about n = 2?

f(2) = (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Which is not equal to 2. So 2 is not a solution.

n = -1:

f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 1/2 - 3/2 - 1 = (-1) - 1 = -2 ≠ -1. Not a solution.

n = 3:

f(3) = (1/2)(9) + (3/2)(3) - 1 = 9/2 + 9/2 - 1 = 9 - 1 = 8 ≠ 3.

n = -3:

f(-3) = (1/2)(9) + (3/2)(-3) -1 = 9/2 - 9/2 -1 = -1 ≠ -3.

So seems like only -2 and 1 are solutions. Therefore, the integers are -2 and 1.

But wait, the problem says "all integers n such that f(n) = n". So the answer is -2 and 1. Therefore, the answer should be boxed as per instructions. Wait, the problem says "Enter all such integers, separated by commas." So maybe \boxed{-2}, \boxed{1}? But the instructions say "put your final answer within \boxed{}". Maybe as a list inside a single box? Let me check the original problem statement again.

"Enter all such integers, separated by commas."

But the user instruction says "put your final answer within \boxed{}". Maybe they expect multiple boxed numbers? But in the problem statement, the user says "Enter all such integers, separated by commas." So probably the answer is -2 and 1, so written as -2, 1 inside a box. But standard practice in math competitions is to have multiple boxed answers if needed, but since the user might expect comma-separated inside one box. Let me check previous problems. For example, if the answer is multiple numbers, they might expect \boxed{-2}, \boxed{1}, but maybe here they want \boxed{-2}, \boxed{1} or \boxed{-2, 1}? The original problem says "put your final answer within \boxed{}". So maybe they want both numbers in one box, separated by commas. For example: \boxed{-2}, \boxed{1} or \boxed{-2, 1}? Wait, in some problems, multiple answers are boxed separately. However, since the user hasn't specified, but given that the initial problem says "Enter all such integers, separated by commas," then I think it's safer to put them in one box as \boxed{-2}, \boxed{1} or maybe \boxed{-2, 1}? But in standard math notation, if multiple answers, perhaps inside a single box separated by commas. Let me check previous problems from the same user. Wait, I can't check previous problems, but in general, for problems asking for multiple numbers, sometimes they are listed in a single box separated by commas. For example, if you look at problems from AoPS or similar, sometimes they expect multiple answers in one box. However, since the user mentioned "put your final answer within \boxed{}", which might indicate a single box. To be safe, perhaps write \boxed{-2} and \boxed{1}, but maybe the system expects comma-separated. Since the original problem says "Enter all such integers, separated by commas," then the answer is -2, 1. So in the required format, maybe \boxed{-2}, \boxed{1}? But the instruction says "put your final answer within \boxed{}", which is singular. Maybe the user expects multiple boxed answers? However, the original problem is from an Olympiad style, where sometimes they have multiple answers. However, since I can't be sure, but given that the initial problem says "put your final answer within \boxed{}", maybe they want all answers in one box. So I will write \boxed{-2}, \boxed{1} but if that's not allowed, maybe \boxed{-2, 1}. However, standard practice on some platforms is to put multiple answers in separate boxes. But since the user hasn't specified, but their instruction says "put your final answer within \boxed{}", I think they want a single box. Maybe they expect comma-separated values inside a single box. For example, \boxed{-2, 1}. Let me check similar problems. For example, if you look at problems on AoPS or Art of Problem Solving, sometimes they have multiple answers in a single box separated by commas. So I think that is acceptable. So I will go with \boxed{-2}, \boxed{1} if multiple boxes are allowed, but if only one box is allowed, then \boxed{-2, 1}.

But since the original instruction says "put your final answer within \boxed{}", maybe it's expecting a single box. Therefore, I think \boxed{-2} and \boxed{1} are the answers. However, since the user might have intended comma-separated in one box, but given the ambiguity, maybe safest to check. Wait, in the problem statement, it says "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So they probably expect the answer as -2, 1. So if the user wants the answer in a single box, like \boxed{-2}, \boxed{1}, but since the system might require one box, then maybe \boxed{-2, 1}. However, standard practice on some platforms (like Brilliant or AoPS) is to have comma-separated inside a single box. Since the original problem says "put your final answer within \boxed{}", maybe they want all answers inside one box. Therefore, I think I should go with \boxed{-2}, \boxed{1} if multiple boxes are allowed, but since the user might have a system that requires one box, maybe \boxed{-2, 1} is better. However, to be safe, I can check the quadratic equation solution again. The equation n² + n - 2 = 0, so roots are 1 and -2, so integers are 1 and -2. Therefore, I think the answer is -2 and 1, so box them as \boxed{-2} and \boxed{1}.

But since the original problem is for a specific platform, maybe the answer should be written as \boxed{-2} and \boxed{1} separated by commas. But since I can't be sure, maybe check for possible mistakes.

Wait, is there a possibility of another solution? For example, could there be another function f satisfying the functional equation? Suppose that the function isn't quadratic? Wait, the functional equation is f(x + y) = f(x) + f(y) + xy + 1. Let me think of another approach. Maybe find f(0).

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

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

Which matches with our previous result where f(0) = -1. So that's consistent.

Now, let me check another value. Let me set y = 1, and see if I can find a recurrence.

Let me denote f(x + 1) = f(x) + f(1) + x*1 + 1. Since the original equation is f(x) + f(y) = f(x + y) - xy - 1. If I set y = 1, then:

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

Since f(1) = 1, this gives:

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

Therefore, f(x + 1) = f(x) + x + 2.

This is a recurrence relation. So if I know f(x), I can find f(x + 1). Let me try to compute f(n) for integer n. Let me suppose that n is integer, and try to find a formula.

Let me compute f(0) = -1.

Then f(1) = 1 (given).

Compute f(2) = f(1 + 1) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Wait, according to our earlier quadratic formula, f(2) should be 8? Wait no, earlier when I calculated f(2) with the quadratic formula, I had f(2) = 8? Wait no, let me check:

Wait, no, earlier I said f(2) = 8? Wait let me recheck.

Wait, f(x) = (1/2)x² + (3/2)x - 1. Therefore, f(2) = (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. So f(2) = 4. Which matches the recurrence: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Correct. Then f(3) = f(2) + 2 + 2? Wait, no. Wait the recurrence is f(x + 1) = f(x) + x + 2. So if x = 2, then f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which matches the quadratic formula: f(3) = (1/2)(9) + (3/2)(3) -1 = 4.5 + 4.5 -1 = 8. Correct. Similarly, f(4) = f(3) + 3 + 2 = 8 + 5 = 13. Let me check with the quadratic formula: f(4) = (1/2)(16) + (3/2)(4) -1 = 8 + 6 -1 = 13. Correct. So the recurrence works.

But if I want to find f(n) for integer n, maybe we can derive a formula. Let me try to find f(n) for integer n. Let me consider that for integer n, f(n) can be computed via recurrence:

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

Wait, hold on, the recurrence is f(x + 1) = f(x) + x + 2. So if I let x = n - 1, then f(n) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n + 1. Wait, but that seems inconsistent with previous calculation? Wait for example, when x = 1: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Which is correct. For x = 2, f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which is correct. So the general recurrence is f(n) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n + 1? Wait, no:

Wait the recurrence is f(x + 1) = f(x) + x + 2. Therefore, if we have f(n) = f((n - 1) + 1) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n - 1 + 2 = f(n - 1) + n + 1. Wait, that's correct. So f(n) = f(n - 1) + n + 1. So we can write this as a recurrence relation. Let's try to solve it.

Let me compute f(n) for integer n:

We have f(0) = -1.

f(1) = f(0) + 0 + 2? Wait no, the recurrence is f(x + 1) = f(x) + x + 2. So for x = 0, f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1, which is correct. Then f(2) = f(1) + 1 + 2 = 1 + 3 = 4, as before. Similarly, f(3) = f(2) + 2 + 2 = 4 + 4 = 8, and so on.

So the recurrence can be written as:

f(n) = f(n - 1) + n + 1 for n ≥ 1, with f(0) = -1.

Therefore, to find f(n), we can expand the recurrence:

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

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

Wait, more systematically:

This is a linear recurrence. Let me write out the terms:

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

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

Substitute into f(n):

f(n) = [f(n - 2) + (n - 1) + 1] + n + 1 = f(n - 2) + (n - 1 + 1 + n + 1) = f(n - 2) + 2n

Wait, let me check again:

Wait f(n) = f(n - 1) + n + 1

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

Therefore:

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

Wait, no, wait. Let me do it step by step:

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

= [f(n - 2) + (n - 1) + 1] + n + 1

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

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

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

Wait, that's not quite. Let me compute the constants:

(n - 1) + 1 = n. Then adding n + 1:

So total is n + n + 1 = 2n + 1. Therefore, f(n) = f(n - 2) + 2n + 1.

But this seems to complicate. Maybe another approach: since it's a linear recurrence, we can solve it by summing the terms.

Let me consider expanding f(n):

f(n) = f(0) + sum_{k=1}^n [f(k) - f(k - 1)] ?

Wait, no, the recurrence is f(k) = f(k - 1) + k + 1 for k ≥ 1, with f(0) = -1.

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

Therefore:

f(n) = -1 + sum_{k=1}^n (k + 1) = -1 + sum_{k=1}^n k + sum_{k=1}^n 1.

Which is:

-1 + [n(n + 1)/2] + n = -1 + [n(n + 1)/2 + n] = -1 + [n(n + 1 + 2)/2] = -1 + [n(n + 3)/2].

Therefore:

f(n) = [n(n + 3)/2] - 1.

Simplify:

Multiply out:

n(n + 3)/2 - 1 = (n² + 3n)/2 - 1 = (n² + 3n - 2)/2.

Wait, which is the same as our quadratic function:

Earlier, f(x) = (1/2)x² + (3/2)x - 1 = (x² + 3x - 2)/2. Yes, exactly. Therefore, 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 equation as before, with roots 1 and -2. So that's consistent. Therefore, there are only two integer solutions: 1 and -2. Therefore, the answer is those two integers.

Thus, after verifying through multiple approaches—assuming a quadratic function, solving the recurrence, and checking the functional equation—it seems that the only integer solutions are n = -2 and n = 1. Therefore, the answer should be these two values.

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

To solve the given functional equation:

$$
f(x) + f(y) = f(x + y) - xy - 1 \quad \text{for all real } x, y
$$

and the condition $ f(1) = 1 $, we aim to find all integers $ n $ such that $ f(n) = n $.

---

### Step 1: Functional Equation Analysis

Rewriting the functional equation:

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

This form is similar to a **Cauchy-type equation** but with an added quadratic term. This suggests a **quadratic function** might be the solution.

Let’s assume:

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

Now substitute into the functional equation.

---

### Step 2: Substitution

Compute both sides:

**Left-hand side (LHS):**
$$
f(x + y) = a(x + y)^2 + b(x + y) + c = ax^2 + 2axy + ay^2 + bx + by + c
$$

**Right-hand side (RHS):**
$$
f(x) + f(y) + xy + 1 = ax^2 + bx + c + ay^2 + by + c + xy + 1 = ax^2 + ay^2 + bx + by + 2c + xy + 1
$$

Now equate the two expressions:

$$
ax^2 + 2axy + ay^2 + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy + 1
$$

Subtracting RHS from LHS:

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

Impose that this must be true for all $ x, y $, so the coefficients of all terms must vanish:

- Coefficient of $ xy $: $ 2a - 1 = 0 \Rightarrow a = \frac{1}{2} $
- Constant term: $ c - 2c - 1 = -c - 1 = 0 \Rightarrow c = -1 $

So now we have:

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

Use 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 3: Solve for Integer Inputs Where $ f(n) = n $

Set up the equation:

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

Multiply both sides by 2 to eliminate fractions:

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

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 roots are:

$$
n = \frac{2}{2} = 1 \quad \text{and} \quad n = \frac{-4}{2} = -2
$$

---

### Step 4: Verification

Check that $ f(1) = 1 $:

$$
f(1) = \frac{1}{2}(1)^2 + \frac{3}{2}(1) - 1 = \frac{1}{2} + \frac{3}{2} - 1 = 2 - 1 = 1 \quad \text{✓}
$$

Check that $ f(-2) = -2 $:

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

Both values satisfy $ f(n) = n $, and hence are solutions.

---

### Final Answer

The integers $ n $ such that $ f(n) = n $ are:

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

[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 know that f(1) = 1. The task is to find all integers n such that f(n) = n. Alright, let me try to work through this step by step.

First, functional equations often require finding the general form of the function or figuring out specific values. Since we have a condition on f(1), maybe we can find a general formula for f(x) and then check when f(n) = n for integer n.

Let me recall that functional equations sometimes can be turned into differential equations or related to known functions. But since this is a Cauchy-like equation, perhaps we can find a quadratic function? Let me think.

The given equation is f(x) + f(y) = f(x + y) - xy - 1. Let me rearrange this equation:

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

This looks similar to the Cauchy equation f(x + y) = f(x) + f(y), but with an extra term xy + 1. So maybe f(x) is a quadratic function? Let me assume that f(x) is a quadratic function. Let me suppose that f(x) is a quadratic function of the form f(x) = ax² + bx + c, where a, b, c are constants. Then I can substitute this into the equation and see if I can find a, b, c.

Let me try that. So substitute f(x) = ax² + bx + c into the equation f(x + y) = f(x) + f(y) + xy + 1.

First compute f(x + y):

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

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

[ax² + bx + c] + [ay² + by + c] + xy + 1 = ax² + ay² + bx + by + 2c + xy + 1.

So set these equal:

Left side: a x² + 2a xy + a y² + b x + b y + c

Right side: ax² + ay² + bx + by + 2c + xy + 1

Subtract the right side from the left side:

Left - Right = [a x² + 2a xy + a y² + b x + b y + c] - [ax² + ay² + bx + by + 2c + xy + 1] =

Let me compute term by term:

a x² - a x² = 0

a y² - a y² = 0

2a xy - xy = (2a - 1)xy

b x - b x = 0

b y - b y = 0

c - 2c - 1 = -c - 1.

Therefore, Left - Right = (2a - 1)xy - c - 1.

But since the original equation is f(x + y) = f(x) + f(y) + xy + 1, the difference should be zero for all x, y. Therefore, the coefficients of the terms must be zero. So:

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

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

So we have a = 1/2, c = -1. Then the coefficient b remains? Wait, but in the equation above, after subtracting, there were no terms with b, so the coefficients for bx and by canceled out. Therefore, there is no restriction on b from this? Wait, but the equation is supposed to hold for all x and y, so unless the coefficients of all terms are zero. However, we only accounted for the xy term and the constant term. The terms with x and y? Wait, no, when we subtracted, they canceled out. Therefore, there's no constraints on b? But then how do we find b?

Wait, but maybe there is another condition. Remember, we also have the value f(1) = 1. Let me check that. So once we have a = 1/2, c = -1, then f(x) = (1/2)x² + bx - 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 this.

So let me compute f(x) + f(y):

f(x) + f(y) = [ (1/2)x² + (3/2)x - 1 ] + [ (1/2)y² + (3/2)y - 1 ] = (1/2)(x² + y²) + (3/2)(x + y) - 2.

Now compute f(x + y) - xy - 1:

First, compute f(x + y):

f(x + y) = (1/2)(x + y)^2 + (3/2)(x + y) - 1 = (1/2)(x² + 2xy + y²) + (3/2)x + (3/2)y - 1 = (1/2)x² + xy + (1/2)y² + (3/2)x + (3/2)y - 1.

Then subtract xy and 1:

f(x + y) - xy - 1 = [ (1/2)x² + xy + (1/2)y² + (3/2)x + (3/2)y - 1 ] - xy - 1 = (1/2)x² + (1/2)y² + (3/2)x + (3/2)y - 2.

Compare with f(x) + f(y) which was (1/2)(x² + y²) + (3/2)(x + y) - 2. These are the same! So yes, the function satisfies the equation. Great. Therefore, our 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 let's set f(n) = n and solve for n.

So:

(1/2)n² + (3/2)n - 1 = n.

Multiply both sides by 2 to eliminate denominators:

n² + 3n - 2 = 2n.

Subtract 2n from both sides:

n² + n - 2 = 0.

Now solve the quadratic equation n² + n - 2 = 0.

Using quadratic formula:

n = [ -1 ± sqrt(1 + 8) ] / 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.

Wait, let me check. Let me verify these values.

First, check n = 1:

f(1) = (1/2)(1)^2 + (3/2)(1) - 1 = 1/2 + 3/2 - 1 = (1 + 3)/2 - 1 = 2 - 1 = 1. Correct, since f(1) = 1.

Now check n = -2:

f(-2) = (1/2)(4) + (3/2)(-2) - 1 = 2 - 3 - 1 = -2. Which equals -2. Correct.

So both n = 1 and n = -2 satisfy f(n) = n.

But wait, the problem says "find all integers n such that f(n) = n." So are there any other integer solutions? The quadratic equation only has two roots, so unless there are other solutions, but since we solved the equation f(n) = n, which is a quadratic equation, so only two solutions. So the integers are -2 and 1. So the answer should be -2 and 1.

Wait, but let me check again if there's a possibility that maybe my assumption that f is quadratic is wrong? Because sometimes functional equations can have multiple solutions, but given that we have the equation f(x + y) = f(x) + f(y) + xy + 1, and assuming f is quadratic seems reasonable. But let me check if the function is indeed uniquely determined by the given conditions.

In functional equations, sometimes you have to check if the solution is unique. Let me think. Since the equation is for all real numbers x and y, and we found a quadratic function that satisfies it, and given that the equation is similar to Cauchy's equation but with an extra term, which is quadratic. Also, since the equation is linear in f(x + y) and the rest, but with a quadratic term in x and y, so the solution is likely unique. Therefore, our function f(x) = (1/2)x² + (3/2)x - 1 is the unique solution. Therefore, the equation f(n) = n only has integer solutions at n = -2 and n = 1. Hence, those are the answers.

But just to make sure, let me check another integer value to see if maybe the function has more fixed points? For example, let me compute f(0):

f(0) = (1/2)(0)^2 + (3/2)(0) - 1 = -1. So f(0) = -1 ≠ 0. So 0 is not a solution.

How about n = 2?

f(2) = (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. Which is not equal to 2. So 2 is not a solution.

n = -1:

f(-1) = (1/2)(1) + (3/2)(-1) - 1 = 1/2 - 3/2 - 1 = (-1) - 1 = -2 ≠ -1. Not a solution.

n = 3:

f(3) = (1/2)(9) + (3/2)(3) - 1 = 9/2 + 9/2 - 1 = 9 - 1 = 8 ≠ 3.

n = -3:

f(-3) = (1/2)(9) + (3/2)(-3) -1 = 9/2 - 9/2 -1 = -1 ≠ -3.

So seems like only -2 and 1 are solutions. Therefore, the integers are -2 and 1.

But wait, the problem says "all integers n such that f(n) = n". So the answer is -2 and 1. Therefore, the answer should be boxed as per instructions. Wait, the problem says "Enter all such integers, separated by commas." So maybe \boxed{-2}, \boxed{1}? But the instructions say "put your final answer within \boxed{}". Maybe as a list inside a single box? Let me check the original problem statement again.

"Enter all such integers, separated by commas."

But the user instruction says "put your final answer within \boxed{}". Maybe they expect multiple boxed numbers? But in the problem statement, the user says "Enter all such integers, separated by commas." So probably the answer is -2 and 1, so written as -2, 1 inside a box. But standard practice in math competitions is to have multiple boxed answers if needed, but since the user might expect comma-separated inside one box. Let me check previous problems. For example, if the answer is multiple numbers, they might expect \boxed{-2}, \boxed{1}, but maybe here they want \boxed{-2}, \boxed{1} or \boxed{-2, 1}? The original problem says "put your final answer within \boxed{}". So maybe they want both numbers in one box, separated by commas. For example: \boxed{-2}, \boxed{1} or \boxed{-2, 1}? Wait, in some problems, multiple answers are boxed separately. However, since the user hasn't specified, but given that the initial problem says "Enter all such integers, separated by commas," then I think it's safer to put them in one box as \boxed{-2}, \boxed{1} or maybe \boxed{-2, 1}? But in standard math notation, if multiple answers, perhaps inside a single box separated by commas. Let me check previous problems from the same user. Wait, I can't check previous problems, but in general, for problems asking for multiple numbers, sometimes they are listed in a single box separated by commas. For example, if you look at problems from AoPS or similar, sometimes they expect multiple answers in one box. However, since the user mentioned "put your final answer within \boxed{}", which might indicate a single box. To be safe, perhaps write \boxed{-2} and \boxed{1}, but maybe the system expects comma-separated. Since the original problem says "Enter all such integers, separated by commas," then the answer is -2, 1. So in the required format, maybe \boxed{-2}, \boxed{1}? But the instruction says "put your final answer within \boxed{}", which is singular. Maybe the user expects multiple boxed answers? However, the original problem is from an Olympiad style, where sometimes they have multiple answers. However, since I can't be sure, but given that the initial problem says "put your final answer within \boxed{}", maybe they want all answers in one box. So I will write \boxed{-2}, \boxed{1} but if that's not allowed, maybe \boxed{-2, 1}. However, standard practice on some platforms is to put multiple answers in separate boxes. But since the user hasn't specified, but their instruction says "put your final answer within \boxed{}", I think they want a single box. Maybe they expect comma-separated values inside a single box. For example, \boxed{-2, 1}. Let me check similar problems. For example, if you look at problems on AoPS or Art of Problem Solving, sometimes they have multiple answers in a single box separated by commas. So I think that is acceptable. So I will go with \boxed{-2}, \boxed{1} if multiple boxes are allowed, but if only one box is allowed, then \boxed{-2, 1}.

But since the original instruction says "put your final answer within \boxed{}", maybe it's expecting a single box. Therefore, I think \boxed{-2} and \boxed{1} are the answers. However, since the user might have intended comma-separated in one box, but given the ambiguity, maybe safest to check. Wait, in the problem statement, it says "find all integers n such that f(n) = n. Enter all such integers, separated by commas." So they probably expect the answer as -2, 1. So if the user wants the answer in a single box, like \boxed{-2}, \boxed{1}, but since the system might require one box, then maybe \boxed{-2, 1}. However, standard practice on some platforms (like Brilliant or AoPS) is to have comma-separated inside a single box. Since the original problem says "put your final answer within \boxed{}", maybe they want all answers inside one box. Therefore, I think I should go with \boxed{-2}, \boxed{1} if multiple boxes are allowed, but since the user might have a system that requires one box, maybe \boxed{-2, 1} is better. However, to be safe, I can check the quadratic equation solution again. The equation n² + n - 2 = 0, so roots are 1 and -2, so integers are 1 and -2. Therefore, I think the answer is -2 and 1, so box them as \boxed{-2} and \boxed{1}.

But since the original problem is for a specific platform, maybe the answer should be written as \boxed{-2} and \boxed{1} separated by commas. But since I can't be sure, maybe check for possible mistakes.

Wait, is there a possibility of another solution? For example, could there be another function f satisfying the functional equation? Suppose that the function isn't quadratic? Wait, the functional equation is f(x + y) = f(x) + f(y) + xy + 1. Let me think of another approach. Maybe find f(0).

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

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

Which matches with our previous result where f(0) = -1. So that's consistent.

Now, let me check another value. Let me set y = 1, and see if I can find a recurrence.

Let me denote f(x + 1) = f(x) + f(1) + x*1 + 1. Since the original equation is f(x) + f(y) = f(x + y) - xy - 1. If I set y = 1, then:

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

Since f(1) = 1, this gives:

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

Therefore, f(x + 1) = f(x) + x + 2.

This is a recurrence relation. So if I know f(x), I can find f(x + 1). Let me try to compute f(n) for integer n. Let me suppose that n is integer, and try to find a formula.

Let me compute f(0) = -1.

Then f(1) = 1 (given).

Compute f(2) = f(1 + 1) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Wait, according to our earlier quadratic formula, f(2) should be 8? Wait no, earlier when I calculated f(2) with the quadratic formula, I had f(2) = 8? Wait no, let me check:

Wait, no, earlier I said f(2) = 8? Wait let me recheck.

Wait, f(x) = (1/2)x² + (3/2)x - 1. Therefore, f(2) = (1/2)(4) + (3/2)(2) - 1 = 2 + 3 - 1 = 4. So f(2) = 4. Which matches the recurrence: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Correct. Then f(3) = f(2) + 2 + 2? Wait, no. Wait the recurrence is f(x + 1) = f(x) + x + 2. So if x = 2, then f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which matches the quadratic formula: f(3) = (1/2)(9) + (3/2)(3) -1 = 4.5 + 4.5 -1 = 8. Correct. Similarly, f(4) = f(3) + 3 + 2 = 8 + 5 = 13. Let me check with the quadratic formula: f(4) = (1/2)(16) + (3/2)(4) -1 = 8 + 6 -1 = 13. Correct. So the recurrence works.

But if I want to find f(n) for integer n, maybe we can derive a formula. Let me try to find f(n) for integer n. Let me consider that for integer n, f(n) can be computed via recurrence:

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

Wait, hold on, the recurrence is f(x + 1) = f(x) + x + 2. So if I let x = n - 1, then f(n) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n + 1. Wait, but that seems inconsistent with previous calculation? Wait for example, when x = 1: f(2) = f(1) + 1 + 2 = 1 + 1 + 2 = 4. Which is correct. For x = 2, f(3) = f(2) + 2 + 2 = 4 + 4 = 8. Which is correct. So the general recurrence is f(n) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n + 1? Wait, no:

Wait the recurrence is f(x + 1) = f(x) + x + 2. Therefore, if we have f(n) = f((n - 1) + 1) = f(n - 1) + (n - 1) + 2 = f(n - 1) + n - 1 + 2 = f(n - 1) + n + 1. Wait, that's correct. So f(n) = f(n - 1) + n + 1. So we can write this as a recurrence relation. Let's try to solve it.

Let me compute f(n) for integer n:

We have f(0) = -1.

f(1) = f(0) + 0 + 2? Wait no, the recurrence is f(x + 1) = f(x) + x + 2. So for x = 0, f(1) = f(0) + 0 + 2 = -1 + 0 + 2 = 1, which is correct. Then f(2) = f(1) + 1 + 2 = 1 + 3 = 4, as before. Similarly, f(3) = f(2) + 2 + 2 = 4 + 4 = 8, and so on.

So the recurrence can be written as:

f(n) = f(n - 1) + n + 1 for n ≥ 1, with f(0) = -1.

Therefore, to find f(n), we can expand the recurrence:

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

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

Wait, more systematically:

This is a linear recurrence. Let me write out the terms:

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

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

Substitute into f(n):

f(n) = [f(n - 2) + (n - 1) + 1] + n + 1 = f(n - 2) + (n - 1 + 1 + n + 1) = f(n - 2) + 2n

Wait, let me check again:

Wait f(n) = f(n - 1) + n + 1

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

Therefore:

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

Wait, no, wait. Let me do it step by step:

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

= [f(n - 2) + (n - 1) + 1] + n + 1

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

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

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

Wait, that's not quite. Let me compute the constants:

(n - 1) + 1 = n. Then adding n + 1:

So total is n + n + 1 = 2n + 1. Therefore, f(n) = f(n - 2) + 2n + 1.

But this seems to complicate. Maybe another approach: since it's a linear recurrence, we can solve it by summing the terms.

Let me consider expanding f(n):

f(n) = f(0) + sum_{k=1}^n [f(k) - f(k - 1)] ?

Wait, no, the recurrence is f(k) = f(k - 1) + k + 1 for k ≥ 1, with f(0) = -1.

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

Therefore:

f(n) = -1 + sum_{k=1}^n (k + 1) = -1 + sum_{k=1}^n k + sum_{k=1}^n 1.

Which is:

-1 + [n(n + 1)/2] + n = -1 + [n(n + 1)/2 + n] = -1 + [n(n + 1 + 2)/2] = -1 + [n(n + 3)/2].

Therefore:

f(n) = [n(n + 3)/2] - 1.

Simplify:

Multiply out:

n(n + 3)/2 - 1 = (n² + 3n)/2 - 1 = (n² + 3n - 2)/2.

Wait, which is the same as our quadratic function:

Earlier, f(x) = (1/2)x² + (3/2)x - 1 = (x² + 3x - 2)/2. Yes, exactly. Therefore, 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 equation as before, with roots 1 and -2. So that's consistent. Therefore, there are only two integer solutions: 1 and -2. Therefore, the answer is those two integers.

Thus, after verifying through multiple approaches—assuming a quadratic function, solving the recurrence, and checking the functional equation—it seems that the only integer solutions are n = -2 and n = 1. Therefore, the answer should be these two values.

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

To solve the given functional equation:

$$
f(x) + f(y) = f(x + y) - xy - 1 \quad \text{for all real } x, y
$$

and the condition $ f(1) = 1 $, we aim to find all integers $ n $ such that $ f(n) = n $.

---

### Step 1: Functional Equation Analysis

Rewriting the functional equation:

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

This form is similar to a **Cauchy-type equation** but with an added quadratic term. This suggests a **quadratic function** might be the solution.

Let’s assume:

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

Now substitute into the functional equation.

---

### Step 2: Substitution

Compute both sides:

**Left-hand side (LHS):**
$$
f(x + y) = a(x + y)^2 + b(x + y) + c = ax^2 + 2axy + ay^2 + bx + by + c
$$

**Right-hand side (RHS):**
$$
f(x) + f(y) + xy + 1 = ax^2 + bx + c + ay^2 + by + c + xy + 1 = ax^2 + ay^2 + bx + by + 2c + xy + 1
$$

Now equate the two expressions:

$$
ax^2 + 2axy + ay^2 + bx + by + c = ax^2 + ay^2 + bx + by + 2c + xy + 1
$$

Subtracting RHS from LHS:

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

Impose that this must be true for all $ x, y $, so the coefficients of all terms must vanish:

- Coefficient of $ xy $: $ 2a - 1 = 0 \Rightarrow a = \frac{1}{2} $
- Constant term: $ c - 2c - 1 = -c - 1 = 0 \Rightarrow c = -1 $

So now we have:

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

Use 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 3: Solve for Integer Inputs Where $ f(n) = n $

Set up the equation:

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

Multiply both sides by 2 to eliminate fractions:

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

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 roots are:

$$
n = \frac{2}{2} = 1 \quad \text{and} \quad n = \frac{-4}{2} = -2
$$

---

### Step 4: Verification

Check that $ f(1) = 1 $:

$$
f(1) = \frac{1}{2}(1)^2 + \frac{3}{2}(1) - 1 = \frac{1}{2} + \frac{3}{2} - 1 = 2 - 1 = 1 \quad \text{✓}
$$

Check that $ f(-2) = -2 $:

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

Both values satisfy $ f(n) = n $, and hence are solutions.

---

### Final Answer

The integers $ n $ such that $ f(n) = n $ are:

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

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