Find the integer solutions of $y^x=x^{50}$

Question

I can't solve this olympiad problem, I tried with simple cases: $4^2=2^k$

And I think maybe that $y=50^k$. But I can't keep going?

Answer 1

I assume $x,y$ are positive, otherwise there is an issue with non-integer powers of negative numbers.

You have: $y=x^{50/x}$.

Case 1. $x$ divides $50$. For each divisor $x|50$ you find $y$, these $(x,y)$ are integer solutions. For example $x=50, y=50$ or $x=25$, $y=625$.

Case 2. $x$ does not divide $50$. Let $50/x=p/q$ be in lowest terms. Then $x$ should have an integer root $s$ of degree $q$: $x=s^q$ otherwise $x^{p/q}$ is irrational. Hence $50/s^q=p/q$, $50q=s^qp$. So $p$ divides $50$ because $p,q$ are coprime. This limits the choices of $p=1,2,5,10, 25, 50$, and for each of them you can find $q$ such that $s^q/q=50/p$ for some $s$. For example if $p=50$, we need $s^q/q=1$, so $s^q=q$ which can only happen if $q=1, s=1$. This gives $x=1$ and we are in Case 1. If $p=25$, we have $s^q=2q$, so $s=q=2$, $x=4$, $y=2^{25}$ which is a solution.

Answer 2

I would start by letting $x=2^a5^b$ because $2,5$ are the factors of $50$ and see where that leads. Then $$y^x=y^{2^a5^b}=x^{50}=(2^a5^b)^{50}=2^{50a}5^{50b}$$ Now $y=2^c5^d$ and we get $$2^{(2^ac)}5^{(5^bd)}=2^{50a}5^{50b}\\ 2^ac=50a\\5^bd=50b$$ $c$ must be a multiple of $25$ and have all the factors of $a$ except $2$.
$d$ must be a multiple of $2$ and have all the factors of $b$ except $5$.
Some obvious solutions are $(a=1,c=25),(a=2,c=25)$ Any higher $a$ has too many factors of $2$ on the left. $(b=1,d=10), (b=2,d=4)$ and again if $b$ gets any higher there are too many $5$s on the left. We can mix and match these to get the following solutions $$\begin {array} {r |r}x&y\\ \hline 10&2^{25}5^{10}\\ 20& 2^{25}5^{10}\\50&2^{25}5^4\\100&2^{25}5^4 \end {array}$$ I don't guarantee that there are no more, but it seems unlikely. You could let $x=2^a5^bp^c$ for some other prime $p$. I think the factors will not work out.

Answer 3

I will assume $x$ and $y$ are positive. Since some power of $x$ equals some power of $y$, both $x$ and $y$ must be powers of a common integer; say $x = z^a$ and $y = z^b$, where $z$, $a$, and $b$ are positive integers. Rearranging $y^x = x^{50}$ gives $$z^{b z^a} = z^{50a}.$$ If $z = 1$ then $x = y = 1$; otherwise this equation is equivalent to $b z^a = 50a$. Note that if $a \geq 9$, then $z^a > 50a$. So it suffices to check $a = 1, \dots, 8$, and in each case find all possible factorizations of $50a$ as a positive integer times an $a$-th power. You can check that all $a > 2$ give rise to only the trivial solution $z = 1$, whereas $a = 1$ and $a = 2$ give several other solutions. The full list is: $$ (x, y) = (1, 1), (2, 2^{25}), (4, 2^{25}), (5, 5^{10}), (10, 10^5), (25, 625), (50, 50), (100, 10). $$