$x^a+y^b$ is divisible by $x^b+y^a$ for infinitely many $(x,y)$ Let $a\geq b>0$ be integers. For which $(a,b)$ do there exist infinitely many positive integers $x\neq y$ such that $x^a+y^b$ is divisible by $x^b+y^a$?
If $a=b$, we certainly have $x^a+y^a$ divisible by itself.
For $a>b$ maybe we can choose some form of $(x,y)$ in terms of $(a,b)$?

If $a$ is odd and $b=1$ and any positive $x>1$ with $y=1$ you have the integer $(x^a+1)/(x+1).$ There are other families of pairs $(a,b)$ with some good choices of $x$ or $y$. For example replacing $x$ above by $x^t$ gives the pair $(at,t)$ of exponents which, on choosing $y=1$, gives the integer $(x^{at}+1)/(x^t+1).$
