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 $n = \boxed{1,-2}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
The function $f$ satisfies the functional equation
\[f(x) + f(y) = f(x + y) - xy - 1\]for all real numbers $x$ and $y.$  If $f(1) = 1,$ then find all integers $n$ such that $f(n) = n.$  Enter all such integers, separated by commas.
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
1,-2
The function $f$ satisfies the functional equation
\[f(x) + f(y) = f(x + y) - xy - 1\]for all real numbers $x$ and $y.$  If $f(1) = 1,$ then find all integers $n$ such that $f(n) = n.$  Enter all such integers, separated by commas.
test