Prove that$a^2+b^2$ is composite from the information provided. Suppose $\alpha$,a,b are integers and $b\neq-1$. Show that if $\alpha$ satisfies the equation $x^2+ax+b+1=0$,then prove $a^2+b^2$ is composite.
I am starting with this study course of polynomials and finding it very difficult to understand. Please help me with the question. Thanks in advance ! 
 A: $\alpha \in \mathbb{Z}$ and $a,b \in \mathbb{Z}$ together with Vieta's formula for roots of quadratic equation: $$\{\alpha,\beta\} \textrm{ are roots of }x^2+ax+b+1 = 0 \implies \begin{cases}\alpha+\beta\ = -a \\ \alpha\beta\ = b+1 \end{cases}$$
Implies $\beta \in \mathbb{Z}$.
So, $$a^2+b^2 = (\alpha+\beta)^2+(\alpha\beta - 1)^2 =\alpha^2\beta^2+ \alpha^2+\beta^2+1 = (\alpha^2+1)(\beta^2+1)$$
Hence, $a^2+b^2$ is composite. Note that $b \neq -1$ ensures $\alpha\beta \neq 0$, i.e., $(\alpha^2+1),(\beta^2+1)$ are factors not equal to $1$.
A: Hint $\ $ It's the norm of a $\rm\color{#c00}{composite\  Gaussian\ integer}$ since, by Vieta, if the roots are $\,\alpha,\beta$
$$\begin{eqnarray} -a &=& \alpha+\beta\\ b+1 &=& \alpha\beta\\ \Rightarrow\ \ b^2+a^2 &= &\ \ \ (\alpha\beta\!-\!1)^2\!+(\alpha\!+\!\beta)^2\\ &=& N(\color{#c00}{\alpha\beta\!-\!1 +\  (\alpha\!+\!\beta)\, i})\\ &=& N(\color{#c00}{(\alpha+i)\,(\beta+i)})\\ &=& N(\alpha+i)N(\beta+i)\\ &=& \ \,(\alpha^2+1)\,(\beta^2+1)\end{eqnarray}\qquad\qquad\qquad\qquad $$
