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

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.

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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 ...

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Does this factorization imply that there can only be on solution? Since the product equals a prime, one of them must be $1$, and in $\Bbb N$ that easily shows the only solution could be $x=1$? – Clayton Jan 19 '13 at 20:59
Yes, that is how I would continue the argument. – Old John Jan 19 '13 at 21:01

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 $\{x^4+4:x \in \mathbb{Z}\}.$

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