Show that there are infinitely many pairs (a,b) such that both $x^2+ax+b$ and $x^2+2ax+b$ have integer roots The question is :
Show that there are infinitely many pairs (a,b) of relatively prime integers (not necessarily positive ) such that both quadratic equations $x^2+ax+b$ =0
and $x^2+2ax+b$ =0 have integer roots.
This is the problem .I tried it to do in many way.
At first we let that there are finely many pairs then we shall show it is a contradiction. It is an INMO  problem. Somebody please help me.
Thank you 

Wanting $a^2-b=c^2$ and $a^2-4b=d^2$, we can see that we want $4c^2-d^2=3a^2$. If $a$ is odd, we can solve $2c-d=3$, $2c+d=a^2$ getting:
$$c=\frac{a^2+3}{4}, d=\frac{a^2-3}{2}$$
Then use $3b=c^2-d^2$ to determine $b$ in terms of $a$, and then do a lot of arithmetic to get an explicit infinite family. There is another (simple) condition required on $a$ to make $a,b$ relatively prime.
You get, after some work, that if $a=2k-3$ then $b=-k(k-1)(k-2)(k-3)$, then  $$x^2+ax+b=0$$ has two integer solutions, $k(k-2)$ and $-(k-1)(k-3)$, and $$x^2+2ax+b=0$$ has two integer solutions $k^2-k$ and $-(k-2)(k-3)$. 
