(Find the triplets of prime numbers $(p, q, r)$ such that $3 p^{4}-5 q^{4}-4 r^{2}=26$.

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We look modulo 5, knowing that the only squares are $1, -1$, and 0. If $p \neq 5$, it's impossible, so $p=5$. Then we look modulo 3, and similarly we get $q=3$. The calculation then yields $r=19$. The solution is $(5,3,19)$.