Mrthdrb

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

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 $mathbbQ(sqrtai)$ is a degree $4$ extension, but this didn't seem to help.

Assuming, without loss of generality, $a>0$, the polynomial can be rewritten as
$$
x^4+2ax^2+a^2-2ax^2=(x^2+a)^2-(sqrt2ax)^2=
(x^2-sqrt2ax+a)(x^2+sqrt2ax+a)
$$
and it's obvious that the two polynomials are irreducible over $mathbbR$. By uniqueness of factorization, this is a factorization in $mathbbQ[x]$ if and only if $2a$ is a square in $mathbbQ$.

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:

$$2a-b^2=0$$
$$b=sqrt2a$$

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.