Find all integers such that polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$ In my studies of abstract algebra and polynomials, I have the following question:

We are asked to find all integers $n$ such that the polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$.

The only thing I can think of is Eisenstein's criterion but that just tells me for $n$ that have a prime factor of power 1, it is irreducible. How would I handle the rest? I thank all helpers.
 A: Let's assume the polynomial is reducible. Then we can see that it must have a quadratic factor - for if it had a linear factor, it must be in form $x-p$, but then $p$ is a root and so is the $-p$ (since the polynomial has only even exponents), and so $(x-p)(x+p)=x^2-p^2$ is a quadratic factor.
So assume a generic factorization $x^4+n=(x^2+ax+b)(x^2+cx+d)$. Comparing the coefficients, we have $a+c=0$, $ac+b+d=0$, $ad+bc=0$ and $bd=n$. Hence $c=-a$, and so $0=ad+bc=ad-ab=a(d-b)$. So either $a=0$, or $d=b$.
If $a=0$, then $c=0$ and $b+d=0$. Thus we have $n=-b^2$ and $x^4+n=(x^2-b)(x^2+b)$.
If $d=b$, then $-a^2+2b=0$, and so $a=2k$ is even. But then $-4k^2+2b=0$, so $-2k^2+b=0$, and so $b=2l$ is even. So, $-k^2+l=0$, thus $a=2k$, $b=2k^2=d$. Thus we have $n=bd=4k^4$ and $x^4+n=(x^2+2kx+2k^2)(x^2-2kx+2k^2)$.
So we have found that if $x^4+n$ is reducible, then necessarily $n=-k^2$ or $n=4k^4$, and in both cases we have found these factorizations, so it is also sufficient.
A: Assume $m=-n > 0$.
If $m$ isn’t a fourth power, and $X^4+n$ is reducible, it means that there is a quadratic algebraic integer $\alpha$ with $\alpha^4=m$.
$\alpha$ is a quadratic algebraic integer so it can be written as $a+b\sqrt{D}$ for integers $a,b$ and square-free $D \neq 1$, or $\frac{a+b\sqrt{D}}{2}$ with furthermore $4|D-1$ and $2|a-b$.
Then $\alpha^2$ is real, so if $D < 0$, $\alpha$ is pure imaginary and $\alpha^2$ must be a real element of an imaginary quadratic field over $\mathbb{Q}$, so is rational, and $m$ is a perfect square – we are in the case $X^4-p^2$.
If $D > 0$, then $\alpha$ is real so $\alpha=\pm m^{1/4}$, so, $\alpha^2$ is $\sqrt{m}$ and is a rational linear combination of $1$ and $\sqrt{D}$, which implies $m=c^2D$ for some integer $c$.
Thus $\alpha=\pm \sqrt{c\sqrt{D}}$ is a quadratic integer over $\mathbb{Q}$. But $\alpha=a+b\sqrt{D}$ where $2a,2b$ are integers. Taking squares, we find $a^2+b^2D=0, 2ab=c$, which implies (as $D$ is square free) a contradiction.
So if $n < 0$, $X^4+n$ is reducible iff $-n$ is a perfect square ($X^4-m^2=(X^2-m)(X^2+m)$).
Assume $n > 0$ and $X^4+n$ reducible. Then there is a quadratic algebraic integer $\alpha$ such that $\alpha^4=-n$, ie $\alpha^2=\pm i\sqrt{n}$.
Let $d$ be the product of prime factors with odd valuation of $n$, (so that $\sqrt{n} \in \mathbb{N}\sqrt{d}$) it follows that $\alpha=a+ib\sqrt{d}$ where $2a,2b$ are integers. As $\alpha^2 \in i\mathbb{N}\sqrt{d}$, it follows $a^2=db^2$, which forces $d=1$ ie $n$ is a square.
As $n=c^2$, $c>0$, then $\alpha^2=\pm ic$, so that $\alpha=(\pm 1 \pm i)(c/2)^{1/2}$.
So (with complex conjugation), the question is as follows: when is $u=(1+i)\sqrt{c/2}$ a quadratic integer?
Assume $u$ quadratic integer. Then $K=\mathbb{Q}(u)$ is stable under conjugation and $\overline{u} \in K$. But $\frac{u}{\overline{u}}=(1+i)(1-i)^{-1}=(1+i)^2/2=i$, so $i \in K$ and $K=\mathbb{Q}(i)$. Thus the real part of $u$ is rational, ie $c/2$ is a square.
Finally, it follows that the only cases where $X^4+n$ is reducible are the “trivial ones”:

*

*if $-n$ is a square (difference of squares)


*if $n=4a^4$ (Sophie Germain)
