Distinct prime number solution of an equation

Question

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

Answer 1

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 $r=19.$

Answer 2

$$3p^4=5q^4+4r^2+26$$

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 $3p^4=5q^4+126 \implies q=3 \implies p=...$