Number Theory – Primes as Differences of Fourth Powers of Two Integers

Question

I was doing some basic Number Theory problems and came across this problem :

Find all primes that are the difference of the fourth powers of two integers

How can I go about it ?

Answer 1

$x^4-y^4=(x^2-y^2)(x^2+y^2)$

Note that $x^2+y^2\ge 2,\forall x,y\in\mathbb Z$, so $$x^4-y^4\in\mathbb P\implies x^2-y^2=1\iff x-y=x+y=\pm 1\iff x=\pm 1, y=0$$

But $(\pm 1)^4-0^4=1\not\in\mathbb P$, so there are no such primes.