I'm asked to solve this equation $$ x^y+y^x=499 $$ only positive integer solutions are permitted.
First I found the apparent solutions $x=498, y=1$ and $x=1, y=498$. I want to look for a way to solve for the other pairs that could be the answer. I noticed that $499$ is a prime but I don't know if it helps knowing this.
I'm trying this: let $f(y)=x^y+y^x$, then $f'(y)=x^y\ln x+xy^{x-1}>0$. So I'm considering cases like $22^2<499$, $7^3<499$, $4^4<499$, and $3^5<499$. But all the above cannot be the answer pair for $(x,y)$. While $23^2>499$, $7^4>499$ and so on so forth. If $x=2$, then $2^8+8^2\neq 499$ and $2^9>499$. Thus neither $x$ nor $y$ could be $2$. Namely the only solution is either $x$ or $y$ must be 1.
I am wondering if there is easier way to solve this problem? Like number theory method? Or any other approaches.
Thank you for any help!
WLOG x is at least y. Then if y is more than 3, the LHS is at least $2 \cdot 4^4 =512$ which fails. So we now have $y=1,2,3$ which can be easily checked individually; only $y=1$ yields a solution which you’ve already found. $$\\$$
(note: WLOG means “without loss of generality” and in this case it’s saying we can assume x is at least y since we can swap the labels of x and y and the equation is the same.)