# Prime numbers which divide $n^3-3n+1$

Let $$f(n)=n^3-3n+1$$. It can be proved that for any prime $$p$$ and integer $$n$$ such that $$p\mid f(n)$$ we have either $$p=3$$ or $$p\equiv\pm1\pmod 9$$ (see below).

Indeed, suppose that for prime number $$p$$ and integer $$n$$ we have $$p\mid n^3-3n+1$$. Firstly, note that if $$x=t+\frac{1}{t}$$, then $$f(x)=t^3+\frac{1}{t^3}+1=\frac{t^6+t^3+1}{t^3}.$$

Now we will consider two cases:

Case 1. In $$\mathbb{F}_p\backslash\{0\}$$ there is a $$a$$ such that $$n\equiv a+\frac{1}{a}\pmod p$$. Then, we have $$a^6+a^3+1\equiv 0\pmod p,$$ so $$a^9\equiv 1\pmod p$$. Moreover, $$x^{p-1}\equiv 1\pmod p$$. Thus, the order $$d$$ of $$a$$ in $$\mathbb{F}_p^{\times}$$ must divide $$\gcd(p-1,9)$$. If $$d\in\{1,3\}$$, then $$a^3\equiv 1\pmod p$$, so $$p=3$$. Otherwise, $$d=9$$ and $$9\mid p-1$$, as desired.

Case 2. There is no $$a\in\mathbb{F}_p$$ such that $$n\equiv a+\frac{1}{a}$$. Then, we can consider the extension $$\mathbb{F}_p(a)$$, where $$a$$ is a root of the polynomial $$x^2-nx+1=0$$ (which is irreducible in $$\mathbb{F}_p$$ due to our assumption). Note that $$|\mathbb{F}_p(a)|=p^2$$ since the degree of this extension is 2. Similarly, as in the first case we deduce that in $$\mathbb{F}_p(a)$$ $$a^6+a^3+1=0,~\text{so}~a^9=1.$$ If $$d$$ is the order of $$a$$ in $$\mathbb{F}_p(a)^{\times}$$, then $$d\mid\gcd(p^2-1,9)$$. As in the first case, if $$d\in\{1,3\}$$, then $$a^3=1$$ in $$\mathbb{F}_p(a)$$ and $$p=3$$. Otherwise, $$d=9$$, so $$9\mid p^2-1$$. Therefore, $$p\equiv\pm 1\pmod 9$$, as desired.

I am interested in elementary proof of this fact (without using field extensions, groups, etc.). Is this possible?

## 1 Answer

A "problem" with a fully elementary proof is that concepts are often pulled out of nowhere. This is already seen in your proof, since you do not explain why you consider writing $$n \equiv a + 1/a \bmod p$$. How did you decide to try to solve such a congruence? Why are you interested in a proof making no use of groups or fields?

Motivation for the congruence $$n \equiv a + 1/a \bmod p$$ comes from looking in $$\mathbf R$$. The polynomial $$x^3 - 3x + 1$$ has three real roots: $$2\cos(2\pi/9)$$, $$2\cos(4\pi/9)$$, and $$2\cos(8\pi/9)$$. In terms of the primitive 9th root of unity $$z = e^{2\pi i/9}$$, there are 6 primitive 9th roots of unity $$z$$, $$z^2$$, $$z^4$$, $$z^5 = 1/z^4$$, $$z^7 = 1/z^2$$, and $$z^8 = 1/z$$, and the three roots of $$x^3 - 3x + 1$$ are the sum of a primitive 9th root of unity and its complex conjugate: $$2\cos(2\pi/9) = z + \overline{z} = z + 1/z$$, $$2\cos(4\pi/9) = z^2 + \overline{z^2} = z^2 + 1/z^2$$, and $$2\cos(8\pi/9) = z^4 + \overline{z^4} = z^4 + 1/z^4$$. This suggests that when $$x^3 - 3x + 1$$ has a root $$n \bmod p$$, $$n \bmod p$$ should have the form $$a + 1/a$$ for a 9th root of unity $$a$$ in characteristic $$p$$. That is analogous to $$2\cos(2\pi/9) = z + 1/z$$ above. The intuition behind $$a^6 + a^3 + 1 \equiv 0 \bmod p$$ comes from $$a \bmod p$$ being a primitive 9th root of unity in characteristic $$p$$ and $$x^9 - 1$$ factoring like this: $$x^9 - 1 = (x-1)(x^2+x+1)(x^6 + x^3 + 1).$$ In this factorization, the root of $$x - 1$$ is $$1$$ and the roots of $$x^2+x+1$$ are the nontrivial cube roots of unity, so the roots of $$x^6 + x^3 + 1$$ are the primitive 9th roots of unity.

The condition $$p \equiv \pm 1 \bmod 9$$ that you are interested in is the same as $$p^2 \equiv 1 \bmod 9$$, and that very strongly suggests looking at the field $$\mathbf F_{p^2}$$ and showing its group of nonzero elements contains an element of order $$9$$ (primitive 9th root of unity in characteristic $$p$$). Then $$9 \mid (p^2-1)$$ by Lagrange's theorem. But you want to avoid group theory. I don't see how to derive $$p^2 \equiv 1 \bmod 9$$ in a nice way by working only in $$\mathbf F_p^\times$$ instead of in $$\mathbf F_{p^2}^\times$$.

If $$r$$ is one of the roots of $$x^3-3x+1$$ in $$\mathbf R$$ then the other two roots are $$r^2-2$$ and $$-r^2-r+2$$. This has an analogue in $$\mathbf F_p$$ if $$p \not= 3$$ and $$n^3 - 3n + 1 \equiv 0 \bmod p$$ for an integer $$n$$: $$x^3 - 3x + 1$$ has three different roots in $$\mathbf F_p$$, namely $$n \bmod p$$, $$n^2 - 2 \bmod p$$, and $$-n^2 - n + 2 \bmod p$$. (When $$p = 3$$ these roots are all equal.) Maybe you can get something useful by working with all three of the roots in $$\mathbf F_p$$ and not just one of the roots mod $$p$$.

[Update: since you know some field theory already, here is how I would solve this problem using field theory, rather than the method using more machinery below. The key insight, as mentioned above, is that $$x^3 - 3x + 1$$ has roots of the form $$z + 1/z$$ as $$z$$ runs over primitive 9th roots of unity. This is true in $$\mathbf C$$ and in fact it's true in every field where there are 9 ninth roots of unity: if $$z^9 = 1$$ and $$z^3 \not= 1$$ then $$z + 1/z$$ is a root of $$x^3 -3x + 1$$. For a prime $$p \not= 3$$, the polynomial $$x^9 - 1$$ has nine distinct roots in characteristic $$p$$ and only 3 of those roots are cube roots of unity, so $$x^9-1$$ has 6 roots in characteristic $$p$$ that are primitive 9th roots of unity. Pairing the 6 roots together as $$z$$ and $$1/z$$, the three sums $$z + 1/z$$ are the three roots of $$x^3 - 3x + 1$$. All 3 roots can be expressed in terms of any one root $$r$$: the roots are $$r$$, $$r^2 - 2$$, and $$-r^2 - r + 2$$, so a field containing one root of $$x^3 - 3x + 1$$ contains all three of its roots. Let $$\zeta$$ be a primitive 9th root of unity in characteristic $$p$$. It does not have to lie in $$\mathbf F_p$$, but $$\mathbf F_p(\zeta)$$ is a finite field since $$\zeta$$ is a root of $$x^6 + x^3 + 1$$. Since $$\zeta + 1/\zeta$$ is a root of $$x^3 - 3x + 1$$ and a field containing one root of this polynomial contains all three roots, your question is about showing, for $$p \not= 3$$, that $$\zeta + 1/\zeta$$ lies in $$\mathbf F_p$$ if and only if $$p \equiv \pm 1 \bmod 9$$.

("If") In a field $$K$$ of characteristic $$p$$, the elements of its subfield $$\mathbf F_p$$ are the solutions in $$K$$ to $$a^p = a$$: that polynomial equation has at most $$p$$ solutions in $$K$$ and all $$p$$ elements of $$\mathbf F_p$$ are solutions of that equation (the $$p-1$$ nonzero elements of $$\mathbf F_p$$ satisfy $$a^{p-1} = 1$$, so $$a^p = a$$, and $$a = 0$$ also fits $$a^p = a$$). If $$p \equiv \pm 1 \bmod 9$$ then $$\zeta + 1/\zeta \in \mathbf F_p$$ since $$\zeta + 1/\zeta$$ equals its own $$p$$th power: $$(\zeta + 1/\zeta)^p = \zeta^p + (1/\zeta)^p = \zeta^p + 1/\zeta^p = \zeta + 1/\zeta$$ since $$\zeta^p$$ only depends on $$p \bmod 9$$ and thus $$\zeta^p$$ is $$\zeta$$ or $$1/\zeta$$ from $$p \equiv \pm 1 \bmod 9$$.

("Only if") The field extension $$\mathbf F_p(\zeta)/\mathbf F_p(\zeta + 1/\zeta)$$ has degree at most $$2$$ since $$\zeta$$ is a root of $$x^2 - (\zeta + 1/\zeta)x + 1$$. If $$\zeta + 1/\zeta$$ lies in $$\mathbf F_p$$, then $$\mathbf F_p(\zeta)/\mathbf F_p$$ has degree at most 2, so $$\zeta \in \mathbf F_{p^2}$$ (there is exactly one quadratic extension field of $$\mathbf F_p$$ and it has size $$p^2$$). That implies $$\zeta^{p^2-1} = 1$$, since $$\mathbf F_{p^2}^\times$$ is a multiplicative group of order $$p^2-1$$. Since $$\zeta$$ has multiplicative order 9 (it is a primitive 9th root of unity!), from $$\zeta^{p^2-1} = 1$$ we must have $$9 \mid (p^2-1)$$, so $$p^2 \equiv 1 \bmod 9$$ and thus $$p \equiv \pm 1 \bmod 9$$.

This concludes my update to this answer.]

I do not see a fully elementary proof for what you seek, but I do see a natural proof if the reader knows algebraic number theory. First of all, the motivation for knowing that all the roots of $$x^3-3x+1$$ are expressible in terms of one of them comes from Galois theory since $$x^3 - 3x+1$$ is irreducible over $$\mathbf Q$$ with a square discriminant $$81$$. That the discriminant is $$81$$ tells us by algebraic number theory that the only ramifying prime in $$\mathbf Q(r)$$ is 3. The Kronecker-Weber theorem tells us the abelian Galois extension $$\mathbf Q(r)/\mathbf Q$$ has to lie in some cyclotomic field, and we are led to check how $$\mathbf Q(r)$$ is related to the 9th cyclotomic field because the only primes at which the $$m$$th cyclotomic field ramifies are factors of $$m$$. So $$\mathbf Q(r)$$ should lie in a 3-power cyclotomic field. In fact, $$\mathbf Q(r)$$ is the real subfield of $$\mathbf Q(\zeta_9)$$, where $$\zeta_9 = e^{2\pi i/9}$$. The connection between prime splitting in Galois extensions of $$\mathbf Q$$ and Frobenius elements in Galois groups over $$\mathbf Q$$ tells us $$p$$ splits in $$\mathbf Q(r)$$ if and only if $$p \equiv \pm 1 \bmod 9$$ because under the natural isomorphism of $${\rm Gal}(\mathbf Q(\zeta_9)/\mathbf Q)$$ with $$(\mathbf Z/9\mathbf Z)^\times$$, the real subfield $$\mathbf Q(r) = \mathbf Q(\zeta_9 + 1/\zeta_9)$$ corresponds to the subgroup $$\pm 1 \bmod 9$$. This, to me, explains why $$x^3 - 3x + 1$$ has a root mod $$p$$ if and only if $$p=3$$ or $$p \equiv \pm 1 \bmod 9$$ in a very conceptual way. Of course it does not at all qualify as an "elementary proof" making no use of abstract algebra. Sorry.

• Actually, I am familiar only with basic notions of the field theory (like finite extensions, but not Galois theory). I came up with this proof because I recalled that polynomial $t^3-3t$ arises when one needs to compute $a^n+a^{-n}$ using $b=a+a^{-1}$, so for me this was like a miracle. (But connection becomes more clear if we note that for $a=e^{i\alpha}$ we have $a^n+a^{-n}=2\cos n\alpha$.) Of course, it's useful to understand the "nature" of this, so thank you for that. – richrow Aug 23 '20 at 19:04
• The polynomials $f_n(t)$ for which $x^n + x^{-n} = f_n(x + 1/x)$ are the Chebyshev polynomials (up to a scaling renormalization by a power of $2$), so $f_2(t) = t^2 - 2$ and $f_3(t) = t^3 - 3t$. That $f_3(t)$ has terms similar to the nonconstant terms in $x^3 - 3x + 1$ is, I think, an instance of the "strong law of small numbers" (i.e., a coincidence). You are not going to find $t^3 - 3t - 3$ has properties like that of $t^3 - 3t + 1$, for example. – KCd Aug 24 '20 at 18:21
• Yes, it was a coincidence. Considering $f(x+x^{-1})$ works because we obtain $\frac{x^6+x^3+1}{x^3}$ or $\Phi_9(x)/x^3$ (and it is known that for prime $p\mid\Phi_n(a)$ we have $p\mid n$ or $p\equiv 1\pmod n$). As far as I understand we obtain cyclotomic polynomial because roots of $f$ are $2\cos 2\pi k/9$, where $\gcd(k,9)=1$. By the way, I think that we can construct a polynomial $g_n(x)$ with similar properties, i. e. if $p\mid g_n(a)$, then $p\mid n$ or $p^2\equiv 1\pmod n$. – richrow Aug 25 '20 at 7:35
• Actually, yes, we just to need to choose $g_n$ such that $g_n(x+x{-1})=\Phi_n(x)/x^{\varphi(n)/2}$ or simply $g_n(x)=\prod\limits_{(k,n)=1, 0<k<n/2}(x-2\cos 2\pi k/n)$. – richrow Aug 25 '20 at 7:36