If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite. 
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.

$\alpha=\frac{-a\pm\sqrt{a^2-4(b+1)}}{2}$is an integer. But manipulating this expression is leading me to nowhere. Please help.

If $u=\alpha$ and $v$ are the roots
$$
\begin{align}
x^2+ax+b+1
&=(x-u)(x-v)\\
&=x^2-(u+v)x+uv
\end{align}
$$
Then $a=-(u+v)$ and $b=uv-1$. Since $u=\alpha\in\mathbb{Z}$, and $a\in\mathbb{Z}$, we know that $v\in\mathbb{Z}$. Therefore,
$$
\begin{align}
a^2+b^2
&=(u+v)^2+(uv-1)^2\\
&=u^2+v^2+u^2v^2+1\\
&=\left(u^2+1\right)\left(v^2+1\right)
\end{align}
$$
