find all primes $p$ and $q$ such that $p \cdot q | 2^p + 2^q$ 
I have to find all prime numbers $p,q$ such that $p\cdot q | 2^p + 2^q$. 

I don't know from what I have to start. 

If $p=q$, then, $p^2|2^{p+1}$. So, $p=q=2$.
Now,suppose $p\ne q$.
If $p=2$, we have $q|2+2^{q-1}$. By Fermat's Little Theorem(FLT), $q=3$. Similarly, $q=2,p=3$ is another solution.
Now, suppose $p$ and $q$ are odd primes. Let $p-1=2^ms$ and $q-1=2^nr$, where $r$ and $s$ odd numbers.
$2^q\equiv -2^p\equiv -2\pmod p$. So, $2^{(q-1)}\equiv -1\pmod p$ and $2^{2(q-1)}\equiv 1\pmod p$. Let $k$ be the order of $2$ in $\bmod p$.  Then, $k|2(q-1)=2^{n+1}r$. So, $k=2^{n_0}r_0$ for some $n_0\le n+1$ and $r_0|r$. On the other hand, $k\not\lvert (q-1)$, thus, $n_0> n$.
So, $k=2^{n+1}r_0$. However, by FLT, $k|p-1$, too. So, $m\ge n+1$.
However, $2^p\equiv -2^q\equiv -2\pmod q$. So, $2^{(p-1)}\equiv -1\pmod q$ and $2^{2(p-1)}\equiv 1\pmod q$. Let $l$ be the order of $2$ in $\bmod q$.  Then, $l|2(p-1)=2^{m+1}s$. So, $l=2^{m_0}s_0$ for some $m_0\le m+1$ and $s_0|s$. On the other hand, $l\not\lvert (p-1)$, thus, $m_0> m$.
So, $l=2^{m+1}s_0$. However, by FLT, $l|q-1$, too. So, $n\ge m+1$.
However, we get 
$$n\ge m+1\ge n+2$$
Contradiction.Thus, $(2,2),(2,3),(3,2)$ are the only solutions.
