<s>[INST] Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$. Reduction into linear factors  $\mathbb{Z}_{17}[x]$:
This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so
$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv x^4+1$ mod 17.
Reduction over  $\mathbb{Z}_{11}[x]$:
This one doesn't have such an easy solution, as neither of $y^2 \equiv -1$ mod 11 or $x^4 \equiv -1$ mod 11 have solutions.
I've tried  $x^4+1 = (x^2- \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ but $x^2 \equiv 2$ mod 11 also has no solutions so there's no easy substitution here.
I think that this approach is not going to work here, so I need something new. Any suggestions?
 [/INST] An abstract algebra argument, without much computation.
Claim: The polynomial $x^4+1$ splits over $\mathbb F_{p^2}$ for any odd (*) prime $p$.
We can see this because  $F_{p^2}^\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.
This means that $x^4+1$ can't be irreducible in $\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.
(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.
Specific solutions:
We can make this explicit by looking at the "standard" complex $4$th roots of $-1$:
$$\pm\frac{\sqrt 2}2 \pm\frac{\sqrt{2}}{2}i=\frac{1}{2}(\pm \sqrt{2}\pm\sqrt{-2})$$
If $p\equiv 1\pmod 8$ then $2$ and $-2$ have square roots in $\mathbb Z_p$, so $x^4+1$ has four roots.
If $p\equiv -1\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\equiv 2\pmod p$. Then write:
$$x^4+1 =  \left(x^2 -ax + 1\right)\left(x^2+ax+1\right)$$
If $p\equiv 3\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\equiv -2\pmod p$, we get:
$$x^4+1 = \left(x^2-bx-1\right)\left(x^2+bx-1\right)$$
Finally, if $p\equiv 5\pmod 8$ then neither of $\pm 2$ is a square, but $-1$ is a square. Letting $c^2\equiv -1\pmod p$, we see that the roots are $(\pm 1 \pm c)\sqrt{2}/2$ and the result is:
$$x^4+1 = (x^2-c)(x^2+c)$$
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