Okay, so I need to find all integers n where f(n) equals n, given that the function f satisfies the equation f(x) + f(y) = f(x + y) - xy - 1 for all real numbers x and y, and f(1) = 1. Hmm, functional equations can be tricky, but let's take it step by step.

First, let's recall that functional equations often require us to find the general form of the function f. Once we have the general form, we can plug in specific values like n and set f(n) = n to solve for n. So maybe the first step is to figure out what f(x) looks like.

The given equation is f(x) + f(y) = f(x + y) - xy - 1. This reminds me of Cauchy's functional equation, which is f(x + y) = f(x) + f(y). But here, there's an extra term involving xy and a constant. So maybe f(x) is similar to a linear function plus some other terms. Let's try to manipulate the equation to find a form for f(x).

Let me try setting y = 0. Then the equation becomes f(x) + f(0) = f(x + 0) - x*0 - 1. Simplifying, f(x) + f(0) = f(x) - 0 - 1. Subtracting f(x) from both sides gives f(0) = -1. So f(0) is -1. That's a start.

Next, maybe set x = 1 and y = 1. Then the equation becomes f(1) + f(1) = f(2) - (1)(1) - 1. Since f(1) = 1, this is 1 + 1 = f(2) - 1 - 1. So 2 = f(2) - 2. Therefore, f(2) = 4. Hmm, interesting. So f(2) is 4. Let's see if there's a pattern here.

Let me compute f(3). Let's set x = 2 and y = 1. Then f(2) + f(1) = f(3) - (2)(1) - 1. So 4 + 1 = f(3) - 2 - 1. That gives 5 = f(3) - 3. Therefore, f(3) = 8. Wait, f(1) = 1, f(2) = 4, f(3) = 8. That seems like f(n) = n^2 - 1? Let's check: 1^2 -1 = 0, which is not 1. Hmm, maybe not. Wait, 1, 4, 8. Maybe f(n) is quadratic? Let's test another value.

Compute f(4). Let x = 3 and y = 1. Then f(3) + f(1) = f(4) - 3*1 -1. So 8 + 1 = f(4) - 3 -1. So 9 = f(4) -4. Therefore, f(4) = 13. So 1, 4, 8, 13. Hmm, that's 0, 3, 7, 12... Wait, maybe the differences are 3, 4, 5? Let's see: 1 to 4 is +3, 4 to 8 is +4, 8 to 13 is +5. So the differences are increasing by 1 each time. If that's the case, the next difference would be +6, so f(5) would be 13 + 6 = 19. Let's check if that's correct.

Set x = 4, y = 1: f(4) + f(1) = f(5) -4*1 -1. So 13 + 1 = f(5) -4 -1. 14 = f(5) -5. Then f(5) = 19. Yes, that works. So the sequence is 1,4,8,13,19,... So each time, the increment is increasing by 1. So the general term might be a quadratic function. Let's suppose f(n) = an^2 + bn + c. Then, since we know f(0) = -1, so c = -1. Then f(1) = a + b -1 = 1, so a + b = 2. f(2) = 4a + 2b -1 = 4. So 4a + 2b = 5. Since a + b = 2, then 2a + 2b = 4. Subtracting that from 4a + 2b =5 gives 2a =1, so a=1/2. Then b=2 -1/2=3/2. So f(n) = (1/2)n^2 + (3/2)n -1. Let's test this with n=3: (1/2)*9 + (3/2)*3 -1 = 4.5 + 4.5 -1=8. Correct. n=4: (1/2)*16 + (3/2)*4 -1=8 +6 -1=13. Correct. So seems like f(n) is (n² + 3n -2)/2. Let's check n=5: (25 +15 -2)/2=38/2=19. Correct. So f(n) = (n² + 3n -2)/2.

But wait, let's verify if this function satisfies the original functional equation. Let's take arbitrary x and y. Let's compute f(x) + f(y) and see if it equals f(x + y) - xy -1.

f(x) + f(y) = [(x² +3x -2)/2 + (y² +3y -2)/2] = [x² + y² +3x +3y -4]/2.

On the other hand, f(x + y) - xy -1 = [( (x + y)² +3(x + y) -2)/2] - xy -1.

Compute (x + y)² =x² + 2xy + y². So numerator becomes x² +2xy + y² +3x +3y -2. Divide by 2: [x² +2xy + y² +3x +3y -2]/2.

Subtract xy and 1: [x² +2xy + y² +3x +3y -2]/2 - xy -1 = [x² +2xy + y² +3x +3y -2 -2xy -2]/2 = [x² + y² +3x +3y -4]/2. Which matches f(x) + f(y). So yes, the function satisfies the equation. Therefore, f(n) = (n² +3n -2)/2.

Now, we need to find all integers n such that f(n) = n. So set (n² +3n -2)/2 = n. Multiply both sides by 2: n² +3n -2 = 2n. Bring all terms to left: n² +3n -2 -2n = n² +n -2 =0. Solve quadratic equation: n² +n -2=0. Using quadratic formula: n = [-1 ±√(1 +8)]/2 = [-1 ±3]/2. So solutions are n=(-1 +3)/2=1 and n=(-1 -3)/2=-2. So integers n=1 and n=-2. Let's check if these satisfy f(n)=n.

For n=1: f(1)=(1 +3 -2)/2=(2)/2=1. Correct.

For n=-2: f(-2)=((-2)^2 +3*(-2) -2)/2=(4 -6 -2)/2=(-4)/2=-2. Correct.

Therefore, the integers are 1 and -2. So the answer is \boxed{-2}, \boxed{1}. Wait, but the question says "enter all such integers, separated by commas." So I should write them together. So the final answer is -2, 1. But in the box, we need to put them in one box. Wait, the instruction says "Enter all such integers, separated by commas." So maybe inside the box, separated by commas. So like \boxed{-2, 1}.

Wait, but the original problem statement says "find all integers n such that f(n) = n". So according to our calculation, the solutions are n=1 and n=-2. Therefore, the integers are -2 and 1. So the answer is \boxed{-2}, \boxed{1}? But the instructions say "Enter all such integers, separated by commas." So perhaps they should be in one box, like \boxed{-2, 1}. Let me check.

The final answer should be within a single box. So yeah, the integers are -2 and 1. So the final answer is \boxed{-2, 1}.

But just to make sure, let me verify again. Our function f(n) is (n² +3n -2)/2. Let's check n=-2: (4 -6 -2)/2=(-4)/2=-2. Correct. n=1: (1 +3 -2)/2=2/2=1. Correct. So yes, they both satisfy f(n)=n. Therefore, the answer is -2 and 1.

**Final Answer**
The integers are \boxed{-2} and \boxed{1}.

Wait, but the problem says "Enter all such integers, separated by commas." So maybe the answer should be written as \boxed{-2, 1} in a single box. Let me check the instructions again. The user says: "Enter all such integers, separated by commas." So in the final answer, I should present them in one box, separated by commas. So the final answer is \boxed{-2, 1}.

Yes, that makes sense. So the final answer is \boxed{-2, 1}.
Therefore, the integers \( n \) such that \( f(n) = n \) are \(\boxed{-2}\) and \(\boxed{1}\).

However, following the user's instruction to present all integers separated by commas within a single box, the final answer should be:

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

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