Waks. you say it's easy but I solved it using derivative, monotonic blah-blah. I'm too shy to post a solution. Can you give a sketch of your solution please?

X-man wrote: Solve in positive integers \[ x^y + y = y^x + x \] Obviously $x=y$ works... anybody with $x\neq y$ ???

$x=2, y=3$ for example.

WLOG consider $x\geq y$ then we make the substitution $x=y+k, k\geq 0$ So we have: $(y+k)^y+y=y^y^+^k+(y+k)$ Dividing by $y^y$ we get $(1+k/y)^y=y^k+k/(y^y)$ The left side is less than $e^k$ so we have $e^k>y^k+k/(y^y)$ If $k=0$ then $y=x$. If $k\geq 1$ then $y<e$ which implies that $y=1, 2$ If $y=1$ then $x=1$. If $y=2$ we have $2^x+x=x^2+2$ which gives us $x=2, 3$