Problem: Prove that there exist infinitely many pairs of
relatively prime positive integers $a,b > 1$
for which $a+b$ divides $a^b+b^a$.

Solution: One construction: let $d \equiv 1 \pmod 4$, $d > 1$.
Let $x = \frac{d^d+2^d}{d+2}$. Then set
\[ a = \frac{x+d}{2}, \qquad
  b = \frac{x-d}{2}. \]
To see this works, first check that $b$ is odd and $a$ is even.
Let $d = a-b$ be odd.
Then:
\begin{align*}
  a+b \mid a^b+b^a &\iff
  (-b)^b + b^a \equiv 0 \pmod{a+b} \\
  &\iff b^{a-b} \equiv 1 \pmod{a+b} \\
  &\iff b^d \equiv 1 \pmod{d+2b} \\
  &\iff (-2)^d \equiv d^d \pmod{d+2b} \\
  &\iff d+2b \mid d^d + 2^d.
\end{align*}
So it would be enough that
\[ d+2b = \frac{d^d+2^d}{d+2}
  \implies b = \half \left( \frac{d^d+2^d}{d+2} - d \right) \]
which is what we constructed.
Also, since $\gcd(x,d) = 1$ it follows $\gcd(a,b) = \gcd(d,b) = 1$.

\begin{remark*}
  Ryan Kim points out that in fact,
  $(a,b) = (2n-1,2n+1)$ is always a solution.
\end{remark*}