$x^{5x}=y^y$, $x, y \in \mathbb{Z}^+$, find largest value of $x$.

Question

Let $x$ and $y$ be positive integers satisfying $x^{5x} = y^y$. What is the largest possible value for $x$?

I'm stuck on this question in an Olympiad past paper. Anyone have any ideas about this one?

Answer 1

First, we have: $$x^{5x}=y^y<y^{5y} \implies x<y$$ As $x^{5x}=y^y$ and $x<y$, it follows that $x \mid y$. Substituting $y=kx$ where $k \in \mathbb{N}$ : $$x^{5x}=(kx)^{kx} \implies x^5=(kx)^k \implies k^k=x^{5-k}$$ Clearly. $k<5$. Substituting $k=1,2,3,4$ : $$k=1 \implies 1=x^4 \implies x=1 \implies (x,y)=(1,1)$$ $$k=2 \implies 4=x^3 \text{ (impossible)}$$ $$k=3 \implies 27=x^2 \text{ (impossible)}$$ $$k=4 \implies 256=x \implies (x,y)=(256,1024)$$

Thus, the solutions are $(x,y)=(1,1),(256,1024)$.