Show that there are infinitely many pairs of nonzero coprime integers a,b such that both quadratics x^2+ax+b=0 and x^2+2ax+b=0 have integral roots.
Start with one such pair: (5,-24)
(x+8)(x-3)=x^2+5x-24
(x+12)(x-2)=x^2+10x-24
For any given pair we can get another pair by multiplying the roots by 2.
Ie given
(x+s)(x+t)=x^2+(s+t)x+st
(x+u)(x+v)=x^2+2(s+t)x+st
So that
[1] (u+v)=2(s+t) and
[2] uv=st
we have
(x+2s)(x+2t)=x^2+2(s+t)x+4st
(x+2u)(x+2v)=x^2+4(s+t)x+4st
But from [1] we have 2(u+v)=4(s+t)
and from [2] 4uv=4st
So there are an infinite number of solutions from just one.
(you can multiply by any number, I just chose 2)
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Edit. This is wrong. Multiplying a and b by 2 gives another pair, but 2a and 2b are not coprime.
Edited on April 2, 2022, 8:23 am
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Posted by Jer
on 2022-04-01 13:02:46 |
Different sort of solution
