Irreducibility of a simple polynomial

Question

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.

Answer 1

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 $\mathbb{Q}$.

Answer 2

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.