Find all distinct prime numbers $p, q$ and $r$ such that

$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
First notice that if both primes $q$ and $r$ differ from 3, then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, 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 $3 p^{4}-4 r^{2}=431$

If $p \neq 5$, by Fermat's little theorem, $p^{4} \equiv 1(\bmod 5)$, which yields $3-4 r^{2} \equiv 1(\bmod 5)$, or equivalently, $r^{2}+2 \equiv 0(\bmod 5)$. 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 $3 p^{4}-5 q^{4}=62$

Obviously $p \neq 5$. Hence, Fermat's little theorem gives $p^{4} \equiv 1(\bmod 5)$. But then $5 q^{4} \equiv 1(\bmod 5)$, which is impossible.

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