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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.

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  • $\begingroup$ Are you allowed to use Lambert's W-function? $\endgroup$ – Aaron Maroja Jan 13 '15 at 12:40
  • $\begingroup$ Anything goes, I guess. $\endgroup$ – user97615 Jan 13 '15 at 12:43
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By prime factorisation, $y=cd$ and $c^{ax}=x^2$, $d^{ax}=b$.
Then $x/\log x=2/(a\log c)$. But $x/\log x>2>2/(a\log c)$

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