# Show that $x^4+1$ is reducible in p-adic numbers $\mathbb{Q}_p$ for p>2 prime.

This is a homework problem for algebraic number theory but I'm having trouble getting started. Do I use induction in general, or show this holds for $p \equiv 1,3$ (mod 4)?

Any help would be appreciated!

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Hint: do you have a square root of 2 available? [Clearly you are done if you have a fourth root of -1 to hand] –  Mark Bennet Nov 12 '12 at 21:57
Mark's hint is on exactly the right path, but it's worth asking: what would make you even think to use induction on this problem? What would you induct over? –  Steven Stadnicki Nov 12 '12 at 22:00
@StevenStadnicki I think he wants to use Newton approximation / Hensel lemma, i.e. stepping from $\mod p^n$ to $\mod p^{n+1}$. –  Hagen von Eitzen Nov 12 '12 at 22:06
@HagenvonEitzen Ahhh, okay, that makes a lot more sense. I'd been thinking about stepping from one prime to the next, which seemed crazy on the face of it. –  Steven Stadnicki Nov 12 '12 at 22:12
(1) Use the fact that $$X^4+1 = (X^2+\sqrt{-1})(X^2-\sqrt{-1}) = (X^2+\sqrt{2}X+1)(X^2-\sqrt{2}X+1) = (X^2+\sqrt{-2}X-1)(X^2-\sqrt{-2}X-1),$$ to show that $X^4+1$ is reducible in $\mathbb{F}_p[X]$ (even for $p=2$). You may need the law of quadratic reciprocity.