interesting one

If p≠q, then we have (p-1)+(p-1)a=2q, for some integer a, that gives p-1=2 or p-1=q, excluding the case when p=2. That gives p>q, implying n≤2

ahh! at last i got it!

with n= 2k, k>1 it is

p(p^k+ 1)(p^k-1+ ... + p+ 1)= q(q+1)>p(p+1)>30

so q>2

gcd((p^k+ 1),P^k-1+ ... + 1) = gcd(p^k+1, p^k-1p-1)≤ gcd(p^k+1, p^k-1)≤2

easy to see that p|q+1 (*)

Case 1: q|p^k-1+ ...+ 1,(then q≤ p^k-1+ ...+ 1) ,so q doesn't divide p^k+ 1, i.e gcd(q,p^k+1)= 1

easy to see that p^k+1|q+1, implying p^k+1≤q+1

since gcd(p,p^k+1)= 1, p(p^k+1)|q+1 from (*)

so p(p^k+1)≤q+1≤p^k-1+ ...+ p+2

which is wrong since p^k+1 -1 + p> p^k-1> (p^k-1p-1)

Case 2: q| p^k+1,(-) so q does not divide p^k-1+ ... + p+ 1 i.e gcd(q,p^k-1+ ... + p+ 1)= 1

easy to see that p^k-1+ ... + p+ 1|q+1

agin gcd(p,p^k-1+ ... + p+ 1)= 1

so p(p^k-1+ ... + p+ 1)|q+1

it follows that p(p^k-1+ ... + p+ 1)≤q+1≤p^k+2, from (-)

which means p^k+ ...+ p≤p^k+ 2

i.e p(p^k-2+ ...+ 1)≤2

only possibility p=2,k= 2 giving q= 5

so the only solution is p=2,q=5,n= 4, and its easy to check that this indeed is a solution..

PS. @devil: i dont understand your hint, can you pls put details to it?

@h4hemag: hey is this a real olympiad question ?