Congruences and prime powers I have just a small question that probably is not hard to answer, but I could not find and elegant solution to this question.
Let $p$ and $q$ be prime numbers.
$$5^q\equiv 2^q \pmod p$$
$$5^p\equiv 2^p \pmod q$$
Find all solutions to this modular equation.
I know solutions are $p=q=3$ and $p=3; q=13$ (and the other way around)
But I do not find a way to prove these are the only ones (or find other solutions). 
 A: If $p=q$, we have $5^p \equiv 2^p \pmod{p}$ so by Fermat's little theorem, $5 \equiv 2 \pmod{p}$ so $p=3$. This gives the solution $(p, q)=(3, 3)$.
Otherwise $p \not =q$, so we may WLOG assume $p>q$. 
Clearly $q \not =2, 5$, so $(2, q)=(5, q)=1$. Let $d$ be the order of $5 \cdot 2^{-1} \pmod{q}$. Then $(5 \cdot 2^{-1})^p \equiv 1 \pmod{q}$ implies $d \mid p$, and $d \mid q-1$ by Fermat's little theorem. Since $p>q$, we have $(p, q-1)=1$, so $d=1$. Thus $5 \cdot 2^{-1} \equiv 1 \pmod{q}$, so $q=3$. 
Now $5^3 \equiv 2^3 \pmod{p}$ and $p>q=3$ so $p=13$. This gives $(p, q)=(13, 3)$.
In conclusion, the solutions are $(p, q)=(3, 3), (3, 13), (13, 3)$.

Edit: Here I have added some explanation about order.
The order of $x \pmod{q}$ is the smallest positive integer $d$ such that $x^d \equiv 1 \pmod{q}$. The following useful result is not too hard to prove:
Result: If $d$ is the order of $x \pmod{q}$ and $x^n \equiv 1 \pmod{q}$ then $d \mid n$.
Proof: Write $n=ad+b$, where $a, b \in \mathbb{Z}$ and $0 \leq b<d$. Then $$1 \equiv x^n \equiv (x^d)^ax^b \equiv x^b \pmod{q}$$ (Note $x^d \equiv 1 \pmod{q}$ by definition)
If $b>0$, then $b<d$ is a positive integer and $x^b \equiv 1 \pmod{q}$, contradicting minimality of $d$. Thus $b=0$ so $d \mid n$. 
Note: In my solution of the question above, I have used $x \equiv 5 \cdot 2^{-1} \pmod{q}$ and $n=p, q-1$ to conclude using this result that $d \mid p, d\mid q-1$.
