Reducibility of polynomials modulo p

Question

How do we show that $X^4-10X^2+1$ is reducible modulo every prime $p$?

I've managed to show it for all primes less than 10, for primes greater than 10 we have $X^4+(p-10)X^2+1$. Where do I go from here?

Answer 1

Write

$$\begin{align} x^4-10x^2+1&=(x^2-1)^2-2(2x)^2\\ &=(x^2+1)^2-3(2x)^2\\ &=(x^2-5)^2-6\cdot 2^2 \end{align}$$

Now if $2$ is a quadratic residue $\pmod{p}$ then the first identity makes our polynomial the difference of two squares.

If $3$ is a quadratic residue $\pmod{p}$ then the second identity makes our polynomial the difference of two squares.

Remember that $$\left(\frac2p\right)\left(\frac3p\right)=\left(\frac6p\right).$$

So if $2$ and $3$ are not quadratic residues $\pmod{p}$ then $6$ is and the third identity makes our polynomial the difference of two squares.

Answer 2

Convince yourself that a factorization must be into quadratics so consider such a set up with undetermined coefficients, which you will compute. Now, if $2$ or $3$ are quadratic residues modulo $p$, then you should be able to easily compute what such a factorization must be. A fact from elementary number theory now says that if neither $2$ nor $3$ were quadratic residues, then $6$ must be. Can you compute the factorization in this case?