Find all groups of positive integers $ (a,x,y,n,m)$ that satisfy $ a(x^n - x^m) = (ax^m - 4) y^2$ and $ m \equiv n \pmod{2}$ and $ ax$ is odd.
Problem
Source: China TST 2002 Quiz
Tags: modular arithmetic, induction, number theory unsolved, number theory
16.06.2012 10:33
Sorry to revive an old topic. But the problem is hard. Can someone solve it?
17.06.2012 06:07
HINT : $ax^m-4$ divides $x^{n-m}-1$ by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$ so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x^m)$ hence $ax^m-4$ divides $ax^m+4$ thus we get $x\leq 4$ checking all case we get $x=1$ is only solution.
17.06.2012 06:54
subham1729 wrote: by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$ Why is this?
17.06.2012 07:11
dinoboy wrote: subham1729 wrote: by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$ Why is this? $ax^m-4$ divides $a.x^{n-m}-4x^{n-2m}$ and also $ax^{n-m}-a$ substracting we get it for $k=2$ similarly going on we get for all k such that $n\geq mk$
17.06.2012 07:32
Sorry, I quoted the wrong line. What I meant is how did you get this: subham1729 wrote: so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x)$
17.06.2012 07:39
dinoboy wrote: Sorry, I quoted the wrong line. What I meant is how did you get this: subham1729 wrote: so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x)$ Put $(k-1)$ in place of $k$ and substract both and multyply by $4$
17.06.2012 07:45
That'll get you a $4 - x^m$ term or something, not $4-x$...
17.06.2012 07:54
dinoboy wrote: That'll get you a $4 - x^m$ term or something, not $4-x$... oops, now edited
17.06.2012 08:00
Your solution is still completely wrong, if you expand properly you simply get $ax^m - 4$ divides $ax^m - 4$ which is useless... To actually solve this problem the fact that $y^2$ is a perfect square is crucial.
18.06.2012 11:26
@dinoboy Can you post your solution?
24.06.2012 11:27
I am very interested in a solution to this difficult problem, so come on please post some solution!