Reducibility of polynomials modulo p 
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?
 A: 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.
A: 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? 
A: A different route. The polynomial $p(x)=x^4-10x^2+1$ is in many an exercise constructed as the minimal polynomial over $\Bbb{Q}$ of the number $\sqrt2+\sqrt3$. Consequently its zeros (in $\Bbb{R}$) are $\pm\sqrt2\pm\sqrt3$.
Therefore the splitting field of $p(x)$ over $\Bbb{F}_p$ is also gotten as
$\Bbb{F}_p[\sqrt2,\sqrt3]$. Because up to isomorphism there is only a single quadratic extension of $\Bbb{F}_p$, namely $\Bbb{F}_{p^2}$, we see that irrespective of choice of $p$ we have $\sqrt2,\sqrt3\in\Bbb{F}_{p^2}$. Therefore $\pm\sqrt2\pm\sqrt3\in \Bbb{F}_{p^2}$. Thus their minimal polynomials over $\Bbb{F}_p$ are at most quadratic. Those minimal polynomials are necessarily factors of $p(x)$ modulo $p$.
