LHS can be either $1,2$ mod $4$, and RHS can be either $0,2$ mod $4$. Thus, we want $q^2+r^2\equiv 1$ mod $2$, which means one of them has to be $2$. WLOG $q=2$. Then LHS is either $0,1,2$ mod $9$ and RHS is either $1,6,8$ mod $9$. Thus, $p=3$, but then the LHS is obviously less than the RHS, so there are no solutions.

kootrapali wrote: . Then LHS is either $0,1,2$ mod $9$ Incorrect. WLOG $q=2$. Then $p^2+1\equiv 8+2r^2 (\bmod\; 9)$ LHS can be 1,2,5,8 mod 9, rhs can be 1,4,7,8 mod 9. Case 1: LHS is 1 mod 9, p=3, fail Case 2: RHS is 8 mod 9, r=3, then RHS=$74*13=962$, and $p=31$. All solutions: $(31,2,3)$ and $(31,3,2)$

After you find $q=2$, you get $p^2=74r^2+295$. Clearly $p>3$. If $r\neq 3$, interpreting this equation mod 3 gives $1\equiv 74+295$ which is a clear contradiction. This gives $r=3$, whence we get the two solutions as described above.

You mean 296 right? It's not a big deal after all. The first solution probably earns 5 points since it has the big idea. Though it may earn 2 because it's such an easy problem for an Olympiad.

$\pmod {6}$ also works.

Okay me taking mod 9 is dumb.

Easy problem. Forgive my bad style; I’m on mobile. We know that p^2 mod 3 is 1 for any prime. Taking mod 3 on both sides, we get that 0 mod 3 is 1 mod 3. Invalid. No solutions. Easy to see p=3,r=3, or q=3 doesn’t work.

AopsUser101 wrote: Easy problem. Forgive my bad style; I’m on mobile. We know that p^2 mod 3 is 1 for any prime. Taking mod 3 on both sides, we get that 0 mod 3 is 1 mod 3. Invalid. No solutions. Wrong For $n=3$, $p^2\equiv 0 \pmod{3}$ So you will have to see some cases. However your $\pmod{3}$ idea is much better than my $\pmod 6$ approach. Btw this problem solved reminds of an old JBMO problem in which the official solution used FLT(Little). So I think FLT can kill this problem too.

AopsUser101 wrote: Easy problem. Forgive my bad style; I’m on mobile. We know that p^2 mod 3 is 1 for any prime. Taking mod 3 on both sides, we get that 0 mod 3 is 1 mod 3. Invalid. No solutions. You're almost right After proving $q=2$, by mod 3 we get contradiction iff $p,r>3$. So, one of $p,r= 3$ $p= 3$ doesn't make sense. $r=3$ gives $p=31$, and we are done