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 some roots e.g.
x = 1/2(sqrt(a^2-4b)-a) and x = sqrt(a^2-b)-a
These will give integer x if (a^2-4b) and (a^2-b) are both square: say, (a^2-4b) = p^2, (a^2-b) = q^2,
Add (a^2-4b) = p^2, 4(a^2-b) = 4q^2,
a^2 = p^2+4b, 4q^2=4a^2-4b
a^2+4q^2 = p^2+4a^2
p^2+3a^2=4q^2
say p=1 then 1+3a^2=4q^2, 0<a<10000 gives solutions (a,q) (1,1)(15,13)(209,181)(2911,2521)... a very familiar sequence, which is infinite, see
A001570,
A001353.
The corresponding b values are 0,56,10290,2118480,..
A059989.e.g
x^2+15x+56=0: x=-7,-8: x^2+30x+56=0; x=-28,x=-2, x^2+2911x+2118480=0: x=-1456,x=-1455, x^2+2*2911x+2118480=0;x=-5432,x=-390, etc.
Trivially, from 1+3a^2=4q^2, a and q have no common factors (indeed a/q converges to 2/sqrt(3)), so from (a^2-b) = q^2, neither do a and b.
Edited on April 1, 2022, 11:41 pm
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Posted by broll
on 2022-04-01 10:29:37 |
Possible Solution
