2016/045) Show that there exists infinitely many pairs of coprime integers (a,b) such that both $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer solution

from the above b can be expressed as product of 2 numbers in 2 different ways that is at least 3 integers say uvw.
now a = uv + w
and 2a = u + vw
so 2(uv+w) = u + vw
or u(2v-1) = w (v- 2)
the above holds if we choose (49) w = 2v- 1 and u = v-2.
that gives
$a = v(v-2) + 2v-1 = v^2 - 1 = (v-1)(v+1)$
$b= v(v-2)(2v-1)$
for a b to be coprime (v-1) should be co prime to v, (v-2) and (2v-1) which is true for any v
and (v+1) should be co prime to v, (v-2) and (2v-1) which is true unless v+1 is divisible by 3.
so we can choose v not divisible by 3 and get corresponding a,b for the same.
hence there are inifinite of them. for example v = 7 giving a = 48 and b= 13 * 35 = 455
$x^2+48x + 455 = 0$ gives solution x = -33,-35
$x^2+96x + 455 = 0$ gives solution -91,- 5