1
$\begingroup$

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 $b\ne-1$.

$\endgroup$
5
  • $\begingroup$ 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

8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.