11. For every pair of real numbers $x, y$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)=-2$, find the number of integer solutions $a$ that satisfy $f(a)=a$.

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Note: The last part is a note and not part of the problem statement, so it is not translated.
11. Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, and $f(-2)=-2$, we get $f(-1)=-2$; let $x=1, y=-1$, we get $f(1)=1$; let $x=1$, we get $f(y+1)=f(y)+y+2$, so $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$. From $f(1)=1$, we know that for all positive integers $y, f(y)>0$. Therefore, when $y \in \mathbf{N}_{+}$, $f(y+1)=f(y)+y+2>y+1$, i.e., when $t>1$, $f(t)>t$. We also get $f(-3)=-1, f(-4)=1$. When $t \leqslant-4$, $-(t+2)>0$, then $f(t)-f(t+1)=-(t+2)>0$, i.e., $f(-5)-f(-4)>0, f(-6)-f(-5)>0, \cdots, f(t+1)-f(t+2)>0, f(t)-f(t+1)>0$. Adding them up, we get $f(t)-f(-4)>0$, i.e., $f(t)>f(-4)=1>0$, which also means $f(t)>t$. In summary, the integer $t$ for which $f(t)=t$ is $t=1$ or -2.