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