<s>[INST] 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.
 [/INST] 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.
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