<s>[INST] Irreducible polynomial which is reducible modulo every prime 
How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?

For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$  gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$?
 [/INST] If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+1=X^4-a^2=(X^2+a)(X^2-a).$$
If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we have
$$X^4+1=(X^2+1)^2-(bX)^2=(X^2+bX+1)(X^2-bX+1) $$
If $p$ is odd and neither $-1$ nor $2$ is a square, then their product $-2$ is a square, say $-2=c^2$. (Without using anything even remotely as deep as quadratic reciprocity, this follows immediately from the fact that $\Bbb F_p^\times$ is a cyclic group of even order). Then we have
$$ X^4+1=(X^2-1)^2-(cX)^2=(X^2-cX-1)(X^2+cX-1)$$
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