Three. (Full marks 20 points) 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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Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$. Also, let $x=1$, $y=-1$ to get $f(1)=1$. Then let $x=1$, we have
$$
f(y+1)=f(y)+y+2 \text {. }
$$

Therefore, $f(y+1)-f(y)=y+2$, 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 $y, f(y)>0$. Thus, when $y \in \mathbf{N}$, $f(y+1)=f(y)+y+2>y+1$, meaning for all positive numbers $t$ greater than 1, $f(t)>t$.
Also from equation $(1), f(-3)=-1, f(-4)=1$.
Next, we prove that when the integer $t \leqslant-4$, $f(t)>0$. Since $t \leqslant$
-4, hence $-(t+2)0 \text {, }
$$

which means $\square$
$$
\begin{array}{l}
f(-5)-f(-4)>0, \\
f(-6)-f(-5)>0, \cdots, \\
f(t+1)-f(t+2)>0, \\
f(t)-f(t+1)>0 .
\end{array}
$$

Adding them up, we get $f(t)-f(-4)>0$,
which means $f(t)>f(-4)=1>0$.
Since $t \leqslant-4$, hence $f(t)>t$.
In summary, the integers that satisfy $f(t)=t$ are only $t=1,-2$.