How do I find all 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} $, with $ax$ odd?
$(x^n-x^m)a=(ax^m-4)y^2$ in positive integers
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0$m$ and $n$ are of same parity, $a$ and $x$ both are odd, $y$ is even. – 2012-07-12
2 Answers
Some partial results. Case 1) $n
Then $ax^m<4$.
The solutions of this inequality are: $ a=1, m=1, x=1,2,3,\\ m\geq2, x=1, $ and $ a=2,3, x=1, m\geq 1. $
From these we obtain two candidates (i) $x=1$ and (ii) $x=3$.
(i) Substituting into the original equation we get $ 0=(a-4)y^2. $ Since $y>0$ and $ax^m=a<4$ there is no solution.
(ii) Substituting into the original equation we get $ 3^n-3=(3-4)y^2. $ Since $n
Case 2) $n=m$.
Substituting into the original equation we have $ (ax^m-4)y^2=0. $ Since $y>0$ we get $ ax^m=4. $ Since $ax$ is odd there is no solution.
Thus we have justified that $n>m$.
So we can write $ x^m(x^{n-m}-1)a=(ax^m-4)y^2. $ Since $a$ is odd and $(a,ax^m-4)=1$, therefore $a|y^2$, and similarly $x^m|y^2$. Furthermore $n=m+2k$ where $k$ is positive integer. Thus we obtain $ ax^m(x^{2k}-1)=(ax^m-4)y^2. $ Introducing the new variables $u:=ax^m$, $z:=x^k$ we obtain $ u(z^2-1)=(u-4)y^2, $ where $u,z$ are odd and $z>1$. Obviously $8|z^2-1$, thus $4|y$. Since $ u(z^2-1)=(u-4)y^2
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0@PeterPhipps it was assumed in the question that $m\equiv n \pmod{2}$. – 2012-07-17
Brute force finds three solutions so far:
a=3, x=3, y=12, n=5, m=1
a=1, x=3, y=12, n=6, m=2
a=1, x=9, y=12, n=3, m=1
which are just different ways of saying that $729 - 9 = (9-4)\times12^2.$