# If $x^2+ax+b+1=0$ ($a,b\in\mathbb{Z}$) has integral roots, prove that $a^2+b^2$ is composite.

If $$x^2+ax+b+1=0$$ ($$a,b\in\mathbb{Z}$$) has integral roots, prove that $$a^2+b^2$$ is composite.

Would someone please help me to solve the above question? I'm not able to understand how I should proceed. Take $$b\ne-1$$.

• What is $b=/=1$ ? Jan 14, 2019 at 9:33
• I meant not equal to -1. I don't know how to type that symbol. Jan 14, 2019 at 9:34
• Compute the discriminant. Jan 14, 2019 at 9:35
• Let $$x^2+ax+b+1=(x-r)(x-s)$$ so that $r,s$ are the two roots, then try to compute $a^2+b^2$ in terms of $r,s$. Jan 14, 2019 at 9:49
• Thanks. I got it. Jan 14, 2019 at 9:54

From Vieta's $$x_1 + x_2 = -a$$ $$x_1 \cdot x_2 = b+1$$ or $$a^2+b^2=\left(x_1+x_2\right)^2+\left(x_1\cdot x_2-1\right)^2=\\ x_1^2+x_2^2+2x_1x_2+x_1^2x_2^2-2x_1x_2+1=\\ x_1^2+x_2^2+x_1^2x_2^2+1=\\ \left(x_1^2+1\right)\left(x_2^2+1\right)$$ Since $$b\ne-1$$, then none of $$x_1,x_2$$ is $$0$$ and the result follows.