For an integer $a$, I'm trying to find a criterion to tell me if $x^4+a^2$ is irreducible over $\mathbb{Q}$.

What I've done so far is shown that if $a$ is odd, then $a^2$ is congruent to $1 \mod 4$, and so the Eisenstein criterion with $p = 2$ tells me that $(x+1)^4 + a^2$ is irreducible and hence $x^4+a^2$ is irreducible.

For $a$ even I have had no such luck. I know that the polynomial is not irreducible for all such $a$ because when $a = 2$, for example, we have the factorization $x^4+4 = (x^2-2x+2)(x^2+2x+2)$. It's easy to show that this polynomial has no linear factors, but I'm at a loss trying to decide when there are a pair of irreducible quadratic factors.

A thought I had was to try and consider when $\mathbb{Q}(\sqrt{ai})$ is a degree $4$ extension, but this didn't seem to help.

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    $\begingroup$ If $a=2n^2$, then $x^4+a^2=x^4+(2n^2)^2=(x^2-2nx+2n^2)(x^2+2nx+2n^2)$. In other cases it seems to be irreducible. $\endgroup$ – Sil Mar 26 at 21:08

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as $$ x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(\sqrt{2a}x)^2= (x^2-\sqrt{2a}x+a)(x^2+\sqrt{2a}x+a) $$ and it's obvious that the two polynomials are irreducible over $\mathbb{R}$. By uniqueness of factorization, this is a factorization in $\mathbb{Q}[x]$ if and only if $2a$ is a square in $\mathbb{Q}$.


Notice that we are trying to reduce that polynomial by this way:


We need:

$$2a-b^2=0$$ $$b=\sqrt{2a}$$

But since we are working on integers then $$a=2k^2$$ .So your polynomial is reducible if and only if it can be written i nthis form:


Which is also known as Sophie Germain Identity.

  • $\begingroup$ I think you should justify why $(x^2-bx+a)(x^2+bx+a)$ is the only possible form, and not generic $(x^2+mx+n)(x^2+px+q)$ $\endgroup$ – Sil Mar 26 at 21:27
  • $\begingroup$ @Sil It's pretty easy with complex factorization. And it can also be done with a system of equations. $\endgroup$ – Eureka Mar 26 at 21:30
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    $\begingroup$ @Sil Try expanding out $(x^2+mx+n)(x^2+px+q)$ and look at the $x^3$ term to why $m=-p$. Then expand $(x^2-bx+n)(x^2+bx+q)$ and look at the $x$ term to see why $n=q$. $\endgroup$ – Ethan MacBrough Mar 26 at 21:40
  • $\begingroup$ @EthanMacBrough I understand, my point though was that things like that should be in answer itself. $\endgroup$ – Sil Mar 26 at 21:42

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