Chinese Mathematical Olympiad

If $2\mid pq$, we suppose that $p=2$ without loss of generality, and then $q\mid 5^q+25$. By Fermat's theorem we have $q\mid 5^q-5$, so $q\mid 30$, where $(2,3)$ and $(2,5)$ are solutions [(2, 2) does not fit]. If $5\mid pq$, we suppose that $p=5$ without loss of generality and then $5q\mid 5^q+5^5$. By Fermat's theorem we have $q\mid 5^{5^q-1}$, so $q\mid 313$, where $(5,5)$ and $(5,313)$ are solutions. Otherwise, we have $pq\mid 5^{p-1}+5^{q-1}$, and so $$5^{p-1}+5^{q-1}\equiv 0(\mathrm{mod}p).$$ By Fermat's theorem, $5^{p-1}\equiv 1(\mathrm{mod}p)$, and because of the above, $5^{q-1}\equiv -1(\mathrm{mod}p)$. Denote by $p-1=2^k(2r-1)$, $q-1=2^l(2s-1)$, where $k,l,r,s$ are positive integers. If $k\le l$, because of the previous congruences we get $$\begin{array}{rl}1& =1^{2^{l-k}(2s-1)}\equiv (5^{p-1}{)}^{2^{l-k}(2s-1)}\\ & =5^{2^l(2r-1)(2s-1)}=(5^{q-1}{)}^{2r-1}=(-1{)}^{2r-1}\\ & =-1(\mathrm{mod}p),\end{array}$$ a contradiction of $p\ne 2$. So $k>l$. But we have $k<l$ by a similar argument — a contradiction. Therefore, all possible pairs of primes $(p,q)$ are $(2,3)$, $(3,2)$, $(2,5)$, $(5,2)$, $(5,5)$, $(5,313)$ and $(313,5)$.