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 $b\ne-1$.

  • $\begingroup$ What is $b=/=1$ ? $\endgroup$ Jan 14, 2019 at 9:33
  • $\begingroup$ I meant not equal to -1. I don't know how to type that symbol. $\endgroup$ Jan 14, 2019 at 9:34
  • 1
    $\begingroup$ Compute the discriminant. $\endgroup$
    – Wuestenfux
    Jan 14, 2019 at 9:35
  • 2
    $\begingroup$ 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 $r,s$. $\endgroup$ Jan 14, 2019 at 9:49
  • $\begingroup$ Thanks. I got it. $\endgroup$ Jan 14, 2019 at 9:54

1 Answer 1


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.


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