# Solving $y^{ax}=x^2b$ over integers

Let $x,y,a,b \in \mathbb{Z}>1$and $\gcd(x,b)=1,$ $y^{ax}=x^2b$, I cannot find any integral solution.

What I have done so far: I assume there must be 2 coprime integers $c, d>1$ such that $$y=c^2d$$ $$x=c^{ax}$$ $$b=d^{ax}$$ And conclude that there is no integer $x>1$ such that $x=c^{ax}$. Am I correct?Please any insight will be greatly appreciated.

• Are you allowed to use Lambert's W-function? – Aaron Maroja Jan 13 '15 at 12:40
• Anything goes, I guess. – user97615 Jan 13 '15 at 12:43

By prime factorisation, $y=cd$ and $c^{ax}=x^2$, $d^{ax}=b$.
Then $x/\log x=2/(a\log c)$. But $x/\log x>2>2/(a\log c)$