How to show the following polynomial is reducible

Question

Prove that $f(x)=x^4-10x^2+1$ is reducible modulo $p$ for every prime $p$

In order to show this I will have to show that $f(x)$ can be factored into two lower degree polynomials of degree $\le 4$.

But I am not getting how should I find the factorization.Please give some hints to proceed.

@CarryonSmiling yes it is ..... you have mistaken ...$2x^2$ instead of $4x^2$ — Paladin, Mar 29 '16 at 17:17

score 3 · Answer 1 · answered Mar 29 '16 at 17:40

3

$x^4 - 10x^2 + 1$ has the following three factorisations: \begin{align*} x^4 - 2x^2 + 1 - 8x^2 & = (x^2 - 1)^2 - 2(2x)^2 \\ x^4 + 2x^2 + 1 - 12x^2 & = (x^2 + 1)^2 - 3(2x)^2 \\ x^4 - 10x^2 + 25 - 24 & = (x^2 - 5)^2 - 6(2^2) \end{align*} If $2$ is a quadratic residue modulo $p$, then the first is a difference of two squares, hence reducible.

If $3$ is a quadratic residue modulo $p$, then the second is a difference of two squares.

If neither $2$ nor $3$ is a quadratic residue modulo $p$, then their product $6$ is a quadratic residue modulo $p$, and so the third is a difference of two squares. Hence in all three cases, the polynomial is reducible.

answered Mar 29 '16 at 17:40

Dietrich Burde

130,978

Very nice ${}{}{}{}{}$ – Asinomás Mar 29 '16 at 17:43
For a more general result see here. – Dietrich Burde Mar 29 '16 at 17:44
So then it is always the product of two polynomials of degree $2$? – Asinomás Mar 29 '16 at 17:49
One question:::what was your aim while doing the factorisation? Did it came up just like that by brute-force method or you had some specific goal while doing it – Learnmore Mar 30 '16 at 02:22
The aim was to cover all primes (this method is often applied in number theory). – Dietrich Burde Mar 30 '16 at 13:42

How to show the following polynomial is reducible

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