Junior Balkan Mathematical Olympiad

First notice that if both primes $q$ and $r$ differ from $3$, then $q^2 \equiv r^2 \equiv 1 \pmod{3}$, hence the left hand side of the given equation is congruent to zero modulo $3$, which is impossible since $26$ is not divisible by $3$. Thus, $q=3$ or $r=3$. We consider two cases.

Case 1. $q=3$.
The equation reduces to $3p^4 - 4r^2 = 431$.
If $p \neq 5$, by Fermat's little theorem, $p^4 \equiv 1 \pmod{5}$, which yields $3 - 4r^2 \equiv 1 \pmod{5}$, or equivalently, $r^2 + 2 \equiv 0 \pmod{5}$. The last congruence is impossible in view of the fact that a residue of a square of a positive integer belongs to the set $\{0, 1, 4\}$. Therefore $p=5$ and $r=19$.

Case 2. $r=3$.
The equation becomes $3p^4 - 5q^4 = 62$.
Obviously $p \neq 5$. Hence, Fermat's little theorem gives $p^4 \equiv 1 \pmod{5}$. But then $5q^4 \equiv 1 \pmod{5}$, which is impossible.

Hence, the only solution of the given equation is $p=5$, $q=3$, $r=19$.