<s>[INST] 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.
 [/INST] 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.
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