Find all $p, q$ primes such that $p q$ divides $2^{p}+2^{q}$.
First, note that if $p=2, 2q$ divides $4+2^q$ if and only if either $q=2$ or $2q$ divides 6, since for any odd $q$, $q$ divides $2^{q-1}-1$, hence $2q$ divides $2^q-2$. Thus, the solutions are: $(p, q)=(2,2),(2,3)$ or $(3,2)$. We will now assume $p$ and $q$ are odd. Let $\omega_p$ and $\omega_q$ be the orders of 2 modulo $p$ and $q$ respectively. If $p$ divides $2^p+2^q$, then $2^{p-1}+2^{q-1}$, and since $p$ divides $2^{p-1}-1$, $p$ divides $2^{q-1}+1$, hence $2^{2(q-1)}-1$. Therefore, $\omega_p$ divides $p-1$ and $2(q-1)$ but does not divide $q-1$. If the highest power of 2 dividing $\omega_p$ (resp. $\omega_q$) is $2^{v_p}$ (resp. $2^{v_q}$), the fact that $\omega_p$ divides $2(q-1)$ and not $q-1$ implies that $v_p > v_q$, because $q-1$ is divisible by $\omega_q$ and thus by $2^{v_q}$ but not by $2^{v_p}$. The same reasoning, by swapping $p$ and $q$, leads to $v_q > v_p$, which is clearly incompatible. Therefore, there are no pairs of odd prime numbers that satisfy this condition.