Junior Balkan Mathematical Olympiad

First notice that if both primes $q$ and $r$ differ from $3$, then $q^2\equiv r^2\equiv 1(\mathrm{mod}3)$, 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 $p\ne 5$, by Fermat's little theorem, $p^4\equiv 1(\mathrm{mod}5)$, which yields $3-4r^2\equiv 1(\mathrm{mod}5)$, or equivalently, $r^2+2\equiv 0(\mathrm{mod}5)$. 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 $p\ne 5$. Hence, Fermat's little theorem gives $p^4\equiv 1(\mathrm{mod}5)$. But then $5q^4\equiv 1(\mathrm{mod}5)$, which is impossible.

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