Factorize $x^4+n(2-n)x^2+n^2$ into irreducible polynomials over Q for any natural number n.

Question

Factorize $x^4+n(2-n)x^2+n^2$ into irreducible polynomial over $\mathbb{Q}$ for any natural number n. This is an extended version of my previous question here.

I know if $n=4$, $(x^2+n-nx)(x^2+n+nx)$ is reducible. But how can I show for $n\neq4$?

Answer 1

For a quadratic $ax^2+bx+c$ with $a,b,c\in \mathbb Q$, you just have to check whether the discriminant $b^2-4ac$ has a square root in $\mathbb Q$.

The discriminant of $x^2 \pm nx +n$ is $n(n-4)$, which is an integer, so it has to be the square of an integer to have a square root in $\mathbb Q$. In particular, it has to be non-negative, which means $n \ge 4$ (assuming $n=0$ is not allowed). We have the solution $n=4$; but if $n \ge 5$, then

$(n-3)^2 = n^2-6n+9 = n(n-4) - (2n-9) < n(n-4) < (n-2)^2$

So $n(n-4)$ lies strictly between two adjacent squares, and is therefore not a square.

Answer 2

I would say, solving of equation $x^4+n(2-n)x^2+n^2=0$:

$\displaystyle x_{1,2,3,4}=\pm \left(\frac{1}{2}(n\pm \sqrt{n-4}\,\cdot\sqrt{n})\right)$

So $x^4+n(2-n)x^2+n^2$ is reducible over Q for $n \ge 4$

Edit: $\color{red}{No - it's\, nonsense.}$