# Irreducibility of a simple polynomial

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.

• 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. – 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:

$$x^4+a^2=(x^2-bx+a)(x^2+bx+a)=x^4+(2a-b^2)x^2+a^2$$

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:

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

Which is also known as Sophie Germain Identity.

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