For which $p\in\mathbb{P}$ does $x^4+4=p$ have solution in $\mathbb{N}$?

Question

Find all primes $p$ for which equation $$x^4+4=p$$ has solutions in set of natural numbers.

I already have my own idea in mind, though the proof seems kinda weak to me at some points. But I'm more interested in different approaches to this problem by the community here, so I will leave my idea aside (since I don't want to form any directions to your thinking), and post it as an answer after a while.

Thanks.

Answer 1

My approach would be to notice that

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

and then look at the two factors ...

Answer 2

Fortunately, we can factorise $x^4+4$ as follows:

$$x^4 + 4 \equiv (x^2-2x+2)(x^2+2x+2).$$

If you want $x^4+4$ to be prime then its only factors can be one and itself. The first case: $$x^2-2x+2=1 \iff x^2-2x+1=0 \iff (x-1)^2=0 \iff x=1 \, . $$

If $x=1$ then $x^4+4=5$, which gives $p=5$ as one possibility. The second case:

$$x^2+2x+2 = 1 \iff x^2+2x+1=0 \iff (x+1)^2=0 \iff x=-1 \, . $$

If $x=-1$ then $x^4+4=5$, which again gives $p=5$ as a possibility.

It follows that $p=5$ is the only possible prime belonging to the set $\{x^4+4:x \in \mathbb{N}\}.$ Indeed, $p=5$ is the only possible prime belonging to the set $\{x^4+4:x \in \mathbb{Z}\}.$