jmc

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, $$\begin{array}{rl}f(2)-f(1)& =1+2,\\ f(3)-f(2)& =2+2,\\ f(4)-f(3)& =3+2,\\ & \dots ,\\ f(n)-f(n-1)& =(n-1)+2.\end{array}$$Adding all the equations, we get $$f(n)-f(1)=1+2+3+\cdots +(n-1)+2(n-1)=\frac{(n-1)n}{2}+2n-2=\frac{n^2+3n-4}{2},$$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 $n=\boxed{1,-2}.$