<s>[INST] Distinct prime number solution of an equation I am trying to solve the following question:
$$\textrm{Find all distinct prime numbers } p, q \textrm{ and } r \textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 \equiv p^2 \equiv 1(p \geq 5), 3(p = 3), 4(p = 2) \mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct. 
$$3\times(1, 3)-5\times(1, 3)-4\times(1, 3, 4) \equiv 2 \mod{6}$$
Possible solutions would be
$$\begin{align}
3\times1-5\times3-4\times1&\equiv\\
3\times1-5\times3-4\times4&\equiv\\
3\times3-5\times3-4\times1&\equiv\\
3\times3-5\times3-4\times4&\equiv2\mod{6}
\end{align}$$
So $q \equiv 3 \mod{6} \Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
 [/INST] $$3p^4=5q^4+4r^2+26$$
If $s\ne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and  $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5\mid 3p^4\implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3\cdot 5^4 -5\cdot 2^4 -26= 4r^2\implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 \implies q=3 \implies p=...$
If $r=5$ we have $3p^4=5q^4+126 \implies q=3 \implies p=...$
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