Olympiad prime

When p=2, $q^n$ =13 which is impossible.Other wise p is odd and 5 divides $2^p+3^p$ .Since n>1 we must have 25 dividing $2^p+3^p$ .....interesting??!!...so why not write 3 as (5-2)... $2^p+3^p$ = $2^p+(5-2)^p$ is congruent to $5*p*2^ (p-1)$ mod 25(This I expect u can show)...So 5 must divide p(WHY?)....Thus p=5 and $q^n$ =275...which has got ZERO SOLUTION :) ...This is an ITALIAN MATHEMATICAL OLYMPIAD QUESTION....Hope u liked it...:)