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.
 A: 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,
$$
\begin{align}
a^2+b^2
&=(u+v)^2+(uv-1)^2\\
&=u^2+v^2+u^2v^2+1\\
&=\left(u^2+1\right)\left(v^2+1\right)
\end{align}
$$
A: Hint: $a^2-4(b+1)=a^2+b^2-(b^2+4b+4)=c^2$ for some integer $c$.
Do you know any facts about expressing primes as the sum of two squares?
A: Let the roots be $r,s$.

By hypothesis, at least one of $r,s$ is an integer.

By Vieta's formulas
\begin{align*}
r + s &= -a\\[4pt]
rs &= b + 1\\[4pt]
\end{align*}
From the first of the above equations, since one of $r,s$ is an integer, and $a$ is an integer, $r,s$ must both be integers.

From the second of the above equations, since $b \ne -1$, it follows that $r,s$ are both nonzero.
\begin{align*}
\text{Then}\;
a^2+b^2&=(r+s)^2 + (rs-1)^2\\[4pt]
&=(r^2 + 2rs + s^2) + (r^2s^2 - 2rs + 1)\\[4pt]
&=r^2s^2 + r^2 + s^2 + 1\\[4pt]
&=(r^2+1)(s^2+1)\\[4pt]
\end{align*}
Since $r,s$ are nonzero integers, it follows that $a^2+b^2$ is composite.
A: We have 
\begin{eqnarray*}
\alpha^2 + a \alpha +b+1=0.
\end{eqnarray*}
Multiply by $4$ and add $a^2$
\begin{eqnarray*}
(2 \alpha + a)^2 =a^2 -4(b+1).
\end{eqnarray*}
Now add & subtract $b^2$ & rearrange to
\begin{eqnarray*}
a^2+b^2=(2 \alpha + a+b+2)(2 \alpha + a-b-2).
\end{eqnarray*}
