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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?

Fawkes4494d3
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Maths explorer
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1 Answers1

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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 $(x,y)=(1,1),(256,1024)$.

Haran
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  • Why is x a factor of y? – Maths explorer Jul 05 '20 at 08:20
  • Because the powers of the primes in the factorization of $x$ and $y$ are in proportion. This means that all the powers are greater (or equal) in one compared to the other, and thus, the smaller number, which is $x$, is a factor of the larger number, which is $y$. – Haran Jul 05 '20 at 08:28