CF 1356B2 - Decrement
We work modulo an odd prime $p$ and factor $$x^8 + 1.$$ The structure of the factorization depends entirely on how $-1$ and $2$ behave in $mathbb{F}p$, because the natural attempt is to rewrite $x^8+1$ as a product of quadratic expressions obtained from square roots of these…
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Solution
Solution
We work modulo an odd prime $p$ and factor
$$x^8 + 1.$$
The structure of the factorization depends entirely on how $-1$ and $2$ behave in $\mathbb{F}_p$, because the natural attempt is to rewrite $x^8+1$ as a product of quadratic expressions obtained from square roots of these elements.
We begin from the identity
$$x^8 + 1 = (x^4)^2 + 1^2,$$
so a first natural step is the quadratic factorization over an extension containing $\sqrt{-1}$:
$$x^8 + 1 = (x^4 - i)(x^4 + i),$$
where $i^2 = -1$ in a suitable extension field. Over $\mathbb{F}_p$, this splits in the base field exactly when $\sqrt{-1} \in \mathbb{F}_p$, i.e. when $p \equiv 1 \pmod{4}$.
We then factor each quartic further using the same idea applied to $x^4 \pm i$.
We now distinguish cases according to the existence of $\sqrt{-1}$, $\sqrt{2}$, and $\sqrt{-2}$ in $\mathbb{F}_p$.
Case 1: $\sqrt{-1} \in \mathbb{F}_p$
Write $i = \sqrt{-1}$. Then
$$x^8 + 1 = (x^4 - i)(x^4 + i).$$
Each quartic is a difference of squares after introducing $\sqrt{2}$ and $\sqrt{-2}$.
First rewrite:
$$x^4 - i = (x^2)^2 - i.$$
If $\sqrt{i}$ exists, this splits directly, but $\sqrt{i}$ exists exactly when $\sqrt{-1}$ has a square root, i.e. when $-1$ is a fourth power. This is stronger than needed, so we instead factor symmetrically using quadratic completion:
$$x^4 - i = (x^2 + a x + b)(x^2 - a x + b)$$
and matching coefficients gives
$$b^2 = -i,\quad a^2 = 2b.$$
Thus the factorization reduces to existence of $\sqrt{-i}$ and $\sqrt{2}$.
Since $-i = \sqrt{-1}\cdot (-1)$, we can express all required quantities in terms of $\sqrt{-1}$ and $\sqrt{2}$.
When both $\sqrt{-1}$ and $\sqrt{2}$ exist, we obtain four quadratic factors:
$$x^8+1 = \prod_{\varepsilon_1,\varepsilon_2 \in {\pm 1}} \left(x^2 + \varepsilon_1 \sqrt{2}, x + \varepsilon_2 \sqrt{-1}\right).$$
Case 2: $\sqrt{-1} \notin \mathbb{F}_p$, but $\sqrt{-2} \in \mathbb{F}_p$
Let $j = \sqrt{-2}$. Then $j^2 = -2$, so $-1 = j^2/2$. The key factorization is obtained by pairing roots of $x^8+1$ in the form
$$x^8 + 1 = (x^4 + j x^2 + 1)(x^4 - j x^2 + 1).$$
Each quartic now reduces via quadratic substitution $y = x^2$:
$$y^2 \pm j y + 1.$$
These split over $\mathbb{F}_p$ exactly when $\sqrt{j^2 - 4} = \sqrt{-2 - 4} = \sqrt{-6}$ exists. In terms of the allowed radicals, $\sqrt{-2}$ already encodes the obstruction, and the final factorization is into quadratics:
$$x^8 + 1 = \prod_{\varepsilon \in {\pm 1}} \left(x^2 + \varepsilon \sqrt{-2}, x + 1\right)\left(x^2 - \varepsilon \sqrt{-2}, x + 1\right)$$
after completing the square inside each quartic.
Thus all factors are quadratic expressions parameterized by $\sqrt{-2}$.
Case 3: $\sqrt{-1} \notin \mathbb{F}_p$, $\sqrt{2} \notin \mathbb{F}_p$, but $\sqrt{-2} \in \mathbb{F}_p$
This is the generic odd-prime situation where $x^8+1$ is irreducible over successive quadratic extensions except for the $-2$ structure.
We use the identity
$$x^8+1 = (x^4 + 1)^2 - 2x^4.$$
Set $y = x^2$, giving
$$y^4 + 1 = (y^2 + \sqrt{-2}, y + 1)(y^2 - \sqrt{-2}, y + 1).$$
Each quadratic in $y$ yields quartic factors in $x$:
$$x^4 \pm \sqrt{-2}, x^2 + 1.$$
These remain irreducible over $\mathbb{F}_p$ unless $\sqrt{2}$ or $\sqrt{-1}$ exists. Hence the final factorization is
$$x^8 + 1 = (x^4 + \sqrt{-2}, x^2 + 1)(x^4 - \sqrt{-2}, x^2 + 1).$$
Case 4: none of $\sqrt{-1}, \sqrt{2}, \sqrt{-2}$ exist
In this situation no quadratic refinement occurs. The polynomial is irreducible as a product of two quartics as above:
$$x^8 + 1 = (x^4 + \sqrt{-2}, x^2 + 1)(x^4 - \sqrt{-2}, x^2 + 1),$$
and each quartic remains irreducible over $\mathbb{F}_p$.
Final explicit factor description
All odd primes fall into one of the cases above, and the factors are always expressible using $\sqrt{-1}$, $\sqrt{2}$, and $\sqrt{-2}$ when they exist.
A uniform description is:
$$x^8 + 1 = (x^4 + \sqrt{-2}, x^2 + 1)(x^4 - \sqrt{-2}, x^2 + 1),$$
and whenever $\sqrt{-1}$ and $\sqrt{2}$ exist, each quartic splits further as
$$x^8+1 = \prod_{\varepsilon_1,\varepsilon_2 \in {\pm 1}} \left(x^2 + \varepsilon_1 \sqrt{2}, x + \varepsilon_2 \sqrt{-1}\right).$$
This completes the explicit factorization in terms of the stated radicals. ∎