<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] If $p$ is not $5$, then reducing the equation modulo $5$ gives (using Fermat's little theorem)
$$3+r^2 \equiv 1 \pmod{5}$$
or $$r^2 \equiv -2 \pmod{5}$$
which has no solution.  So $p=5,$ and size considerations force $q<5$, so $q=2$ or $3$.  If $q=2$ then $3p^4$ is forced to be even, which it isn't, so $q=3$.  Now solve 
$$3\cdot 5^4-5\cdot 3^4 -4r^2 = 26$$ to get $r=19.$
</s>