2 quadratic equations

[photon]

Show that there are infinitely many pairs $(a,b)$ of relatively prime integers (not necessarily positive) such that both the equations $x^2 + ax + b=0 , x^2+2ax+b=0$ have integer roots.

[ Source-INMO-1995 ]

[Nayeemul Islam Swad]

Check my solution. I'm not sure if it's correct or not. Let $a$ be even$($I would prove there still exists infinite $(a,b))$. Then, let $a = 2p$. And let the solutions of the first equation be $p + q$ and $p - q$. So, $b = p^2 - q^2$. Again let the solutions of the second equation be $2p + r$ and $2p - r$. So here $b = 4p^2 - r^2$. Here observe that if we can prove that there are infinitely many triplets of integers $(p,q,r)$ satisfying $p^2 - q^2 = 4p^2 - r^2$ then it would also imply that there are infinitely many pairs $(a,b)$. Now $p^2 - q^2 = 4p^2 - r^2 \leftrightarrow 3p^2 = r^2 - q^2 \leftrightarrow 3p^2 = (r + q)(r - q)$. So, we just need to prove that there exists infinitely many $p$ such that $3p^2$ has two factors whose difference is even. But it's trivial. Just let $p = 2x$ and the factors $2, 6x^2$.

[photon]

I couldn't solve it yet , good approach ... you see $(x-m)(x-n)=0 \Rightarrow x^2-(m+n)x+mn=0$ - for $m,n$ as solution the coefficient of $x$ is $-(m+n)$ . Hence for $a=2p$ , the solutions should be $-p-q,-p+q$ of the first equation. Though it didn't influence on $b$ 's magnitude . For even number $x$ , $(p,q)$ and $(p,r)$ will co-prime , so will be $(a,b)$ . So the solution is correct .