Let's prepare for RMO #3

Nice one!

By Fermat's Little Theorem we get

$5^p=5 \mod p$ & $2^p=2 \mod p$

So Subtracting $5^p-3^p = 3 \mod p$

Which means $(p,q)=(3,3)$ is a solution.

Now Either Case I $(p=3 and q=q)$ or Case 2 $(q=3 and p=p)$

For case I, When we put p=3, We get q=13.

The opposite is the solution for q=3.

So $(3+3+13+3+3+13)=\boxed{38}$

Suppose for the moment that $p \ge q.$ Now either $p | (5^p-2^p)$ or $q|(5^p-2^p).$ If the former, then by Fermat's little theorem we get $0 \equiv 5^p - 2^p \equiv 5-2$ mod $p,$ so $p=3.$ Since $q\le p,$ the only solution in this case is $(3,3)$ by inspection.

If the latter, then get $(5/2)^p \equiv 1$ mod $q.$ (We know $q\ne 2$ so it's all right to divide through by $2^p.$ )

This means that the order of $5/2$ mod $q$ divides $p.$ Since $p$ is prime, the order is either $1$ or $p.$ But Fermat's little theorem says that the order is at most $q-1 < p,$ so it has to be $1.$ Hence $5/2 \equiv 1$ mod $q$ and $q=3.$

Now we get that $\frac{(5^p-2^p)117}{3p}$ is an integer, and if $p \ne 3$ then $p$ doesn't divide $5^p-2^p,$ so it must divide $117,$ and the only possibility is $p=13.$

Putting these all together and dropping the condition that $p \ge q,$ we get the three solutions $(3,3),(3,13),(13,3).$ The answer is $\fbox{38}.$

• Case 1 : $p \mid 5^{p} - 2^{p}$ and $q \mid 5^{q} - 2^{q}$ .

• Case 2 : $p \mid 5^{q} - 2^{q}$ and $q \mid 5^{p} - 2^{p}$ .

Case 1 ) - Suppose $p \neq 2/5$ .Then we have from Fermat 's Little Theorem that $5^{p} \equiv 5 \mod p$ and $2^{p} \equiv 2 \mod p$ .This implies $5^{p} - 2^{p} \equiv 3 \mod p$ .If $p \mid 5^{p} - 2^{p}$ then this forces $p = 3$ .Similarly we deduce $q =3$ .

Case 2) - Suppose $p \neq 5$ , $p > 3$ . Then $p \mid 5^{q} - 2^{q}$ . Or we can say $5^{q} \equiv 2^{q} \mod p$ .Since we have again from Fermat 's little theorem that $5^{p-1} \equiv 2^{p-1} \equiv 1 \mod p$ .These 2 deductions imply $5^{gcd(p-1,q)} \equiv 2^{gcd(p-1,q)} \mod p$ . But since $p,q$ are primes it means $gcd(p-1,q) = 1$ . Thus $5 \equiv 2 \mod p$ .This again forces $p = 3$ a contradiction. Hence one out of $p,q$ must be $3$ always.In case $1$ both are together $3$ however in case $2$ only one is $3$ .Putting $p = 3$ we get that $q \mid 5^{3} -2^{3} = 117$ and thus $q =13$ .Similarly $q = 3$ renders $p =13$ .

Thus we have 3 ordered solutions $(p,q) = [(3,3) ; (3,13) ; (13,3)]$

$Just~ working ~with~ the ~primes~ from~~ 2 ~to~ 41~~ on~ 50~ digit~ calculator~ noting~ that~~~p~\Longleftrightarrow~q.\\ (3,3)~ and ~(3,13) ~and~so~~(13,3) ~were~ the~ results. ~~So~2*(3+3+13)=38.\\ p=q=41, ~was~ a ~53~ digit~ round~ number. ~~~I~ am ~not~ much~ familiar~ with~ Theory~ of~ Numbers.$