Sorry to revive an old topic. But the problem is hard. Can someone solve it?

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.

subham1729 wrote: by induction $ax^m-4$ divides $4^{k-1}x^{n-mk}-a^{k-1}$ Why is this?

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 $n\geq mk$

Sorry, I quoted the wrong line. What I meant is how did you get this: subham1729 wrote: so $ax^m-4$ divides $4a^{k-2}(a-1)-4^{k-1}x^{n-mk}(4-x)$

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 $4$

That'll get you a $4 - x^m$ term or something, not $4-x$...

dinoboy wrote: That'll get you a $4 - x^m$ term or something, not $4-x$... oops, now edited

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.

@dinoboy Can you post your solution?

I am very interested in a solution to this difficult problem, so come on please post some solution!