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If $f$ of degree $d$ is irreducible over $\mathbb{F}_p$, then $\mathbb{F}_p[x]/(f)$ is isomorphic to $\mathbb{F}_{p^d}$. Denoting $\mathbb{F}_p$ by $K$ and $\mathbb{F}_{p^d}$ by $L$, we have for $\alpha$ in $L$ the trace of $\alpha$: $Tr_{L/K}(\alpha) = \alpha + \alpha^{p} + \cdots + \alpha^{p^{d-1}}$. If $\alpha \in L$ is a root of $f$, then since $Gal(L/K)$ acts transitively on the roots of $f$, and since the Frobenius map generates the Galois group, the $\alpha^{p^j}$ for $0\leq j<d$ are distinct, and are the $d$ roots of $f$.

Suppose $f = x^3 + 2x + 1559$ is irreducible over $\mathbb{F}_p$. By Vieta's formulas, $Tr_{L/K}(\alpha)$ is the negative of the coefficient of $x^2$ in $f$, which is $0$. Therefore, in $\mathbb{F}_p$ $\alpha + \alpha^p + \alpha^{p^2} = 0$. I looked for composite $n$ that satisfy relations like: $x + x^n + x^{n^2} \equiv 0 \pmod {f,n}$ and found that for $n=26599$, we do in fact have $x + x^{26599} + x^{26599^2} \equiv 0 \pmod {f,26599}$. I'm wondering if there's an arithmetic explanation for this; I've found nothing.

Also, I searched for other semiprimes $n$ and cubics $f = x^3 + c_2 x + c_3$ with $gcd(n,c_2)=1$ and $gcd(n,c_3)=1$ and where $f$ has no roots in $\mathbb{Z}/n\mathbb{Z}$, satisfying a similar congruence. I've found none so far.

Daniel Donnelly
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David Bernier
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1 Answers1

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Nice question! This actually has a satisfying answer. Write $p = 67, q = 397$, so that $n = 26599 = pq$. We want to show that for any root $\alpha$ of $f(x) \bmod p, q$ we have $\alpha + \alpha^n + \alpha^{n^2} \equiv 0 \bmod p, q$. There are very similar coincidences at each prime that make this possible, which work out so that in each case

$$1 + \alpha^{n-1} + \alpha^{n^2-1}$$

ends up being the sum of the three cube roots of unity. I don't know how likely this is but it seems quite unlikely. The following involves a bunch of WolframAlpha computations.

Working $\bmod 397$, we have

$$f(x) \equiv (x + 67)(x + 129)(x + 201) \bmod 397$$

(the $67$ here is unrelated, as far as I can tell), so all three roots of $f(x) \bmod 397$ live in $\mathbb{F}_{397}$ (rather than in a proper extension) with order dividing $q - 1 = 396 = 2^2 \cdot 3^2 \cdot 11$. Note that $p - 1 = 66 = 2 \cdot 3 \cdot 11$ which is suggestive. We have in fact that the exact orders of the roots are

$$\text{ord}_{397}(-67) = 198, \text{ord}_{397}(-129) = 198, \text{ord}_{397}(-201) = 9$$

so they all satisfy $\alpha^{198} \equiv 1 \bmod 397$ and in all three cases $1, \alpha^{66}, \alpha^{132}$ are the three cube roots of unity $\bmod 397$.

Now, we also have $n = pq \equiv p \bmod (q - 1)$, or $n \equiv 67 \bmod 396$, hence

$$n \equiv 67 = 1 + 66 \bmod 396$$

and this is the coincidence that makes everything work, because it gives

$$n^2 \equiv (1 + 66)^2 \equiv 1 + 2 \cdot 66 = 133 \bmod 396$$

(the last term vanishes), hence

$$1 + \alpha^{n-1} + \alpha^{n^2-1} \equiv 1 + \alpha^{66} + \alpha^{132} \bmod 397$$

is the sum of the three cube roots of unity $\bmod 397$, and hence is zero!

Now we do the calculation $\bmod 67$. Here we have that

$$f(x) \equiv x^3 + 2x + 18 \bmod 67$$

is irreducible, so its roots live in $\mathbb{F}_{67^3-1}$ and so could have order as large as $67^3 - 1 = 300762 = 2 \cdot 3^2 \cdot 7^2 \cdot 11 \cdot 31$. Fortunately it doesn't turn out to be this high. We can compute with the companion matrix

$$M = \begin{bmatrix} 0 & 0 & -18 \\ 1 & 0 & -2 \\ 0 & 1 & 0 \end{bmatrix} \in GL_3(\mathbb{F}_{67})$$

to show that the characteristic polynomial of $M^{31}$ is

$$\det(t - M^{31}) \equiv t^3 - 6 \bmod 67$$

which implies that the multiplicative order of $M \bmod 67$, and hence of its eigenvalues, is

$$\text{ord}_{67}(6) \cdot 3 \cdot 31 = 3^2 \cdot 11 \cdot 31 = 3069$$

from which it follows that $1, \alpha^{1023}, \alpha^{2046}$ are the three cube roots of unity $\bmod 67$. (Strictly speaking we should also check that the order doesn't divide $3^2 \cdot 11$, which can be done with another characteristic polynomial computation.) Now our second coincidence appears: we have

$$n \equiv 2047 \equiv 1 + 2046 \bmod 3069$$

which as above gives

$$n^2 \equiv (1 + 2046)^2 \equiv 1 + 4092 \equiv 1024 \bmod 3069$$

and hence that

$$1 + \alpha^{n-1} + \alpha^{n^2-1} \equiv 1 + \alpha^{2046} + \alpha^{1023} \bmod 67$$

which as above is the sum of the three cube roots of unity $\bmod 67$ and hence is zero!

Abstracting, what we needed for this calculation to work out was that

  • $d = \text{ord}_p(\alpha)$ and $e = \text{ord}_q(\alpha)$ both needed to be divisible by $3$ so that the corresponding powers $\alpha^{\frac{d}{3}} \bmod p, \alpha^{\frac{e}{3}} \bmod q$ were nontrivial cube roots of unity, and then
  • we also needed $n \equiv 1 \pm \frac{d}{3} \bmod d$ and $n \equiv 1 \pm \frac{e}{3} \bmod e$, so that $\alpha^{n-1}$ and $\alpha^{n^2-1}$ would work out to be the two nontrivial cube roots of unity $\bmod p, q$.

The first condition is not so hard but the second condition seems like a big coincidence that needs to line up. At minimum it requires $pq-1$ to share factors with both $p-1$ and $q-1$ and hence $p-1$ and $q-1$ have to share factors with each other, which is the case here.

We can get a hint about what's going on here by looking up $n = 26599$ on the OEIS, which appears $18$ times. Several of these entries involve pseudoprime behavior; for example we learn that $26599$ is a strong pseudoprime to bases $27, 30, 43$, as well as being a Lucas pseudoprime of some sort. I don't know if any of those are directly related to what happens here but they are at least suggestive.

Qiaochu Yuan
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