Problem #674: Are there any integer solutions to $x^xy^y=z^z$ with...

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Progress

Ko [Ko40] proved there are none if $(x,y)=1$, but there are in fact infinitely many solutions in general - for example,\[x=2^{12}3^6, y = 2^83^8,\textrm{ and } z = 2^{11}3^7.\]More generally, writing $a=2^{n+1}$ and $b=2^n-1$,\[x = 2^{a(b-n)}b^{2b}\cdot 2^{2n},\]\[y = 2^{a(b-n)}b^{2b}\cdot b^2,\]and\[z = 2^{a(b-n)}b^{2b}\cdot 2^{n+1}b.\]In [Er79] Erdős asked if the infinite families found by Ko are the only solutions.

Mills [Mi] proved that there are no non-trivial solutions if $4xy>z^2$, and that the only non-trivial solutions when $4xy=z^2$ are those given by Ko above. Mills found no non-trivial solutions when $4xy<z^2$ if $(x,y)< 6^{150}$.

Schinzel [Sc58] proved that in any non-trivial solution which is not in Ko's family either every prime factor of $x$ divides $y$ or vice versa, and conjectured that in any non-trivial solution all of $x,y,z$ have the same prime factors. This conjecture was proved by Dem'janenko [De75b].

Uchiyama [Uc84] proved that, for any fixed rational $Q\in (0,1/4)$, there are only finitely many non-trivial solutions with $Q=xy/z^2$, and proved that there are no non-trivial solutions for various families of $Q$, for example $Q=(1-k^2)/4$ for rational $k\in (0,1)$ or $Q=a/b<1/4$ with $(a,b)=1$ and $1\leq a\leq 5$.