# A new generalisation of Fermat's little theorem?

I would like to share a personal result, which I interpret as a "generalisation of Fermat's little theorem". I know that there exist already several generalisations (e.g. Euler's) but I would like to know whether the result I found is something original or not. Let me explain...

Some years ago, my researches on binary representation of numbers lead me to this sequence of naturals: 1, 2, 1, 2, 3, 6, 9, 18, 56, ... . After some researches, I found that this was already indexed as A001037 sequence in OEIS; this is known as number of "binary Lyndon words of length $n$" or "$n$-bead necklaces with beads of 2 colors". Such sequence can easily be generalised to other bases than 2. After naming $\lambda_a(n)$ the number of $n$-bead necklaces with beads of $a$ colors with $a \ge 2$, I found that $$a^n = \sum_{d|n} d \lambda_a(d)$$ For example, $$2^6 = 1\lambda_2(1) + 2\lambda_2(2) + 3\lambda_2(3) + 6\lambda_2(6) = 2 + 2 + 6 + 54 = 64$$

This is NOT an original result since you can find this relationship on the afore-mentioned OEIS page, at least.

QUESTION 1: Can someone provides references for a proof of this equation (papers, books, ...)?

Now, after defining $$\sigma_a(n) \triangleq \sum_{\substack{ d|n \\ 1<d<n}} d \lambda_a(d)$$ the previous equation can be rewritten $$a^n = a + n\lambda_a(n) + \sigma_a(n)$$ Then, using modular arithmetic, we obtain a generalisation of Fermat's little theorem: $$a^n \equiv a + \sigma_a(n) \pmod {n}$$

Note that this is indeed a generalisation since it applies on any natural $n$ and it boils down to $$a^p \equiv a \pmod {p}$$ for any prime $p$, that is the Fermat's little theorem.

This relationship looks appealing to me, at least for studying Fermat pseudoprimes: these are the composites $n$ having the "rare" property that $\sigma_a(n) \equiv 0 \pmod {n}$.

QUESTION 2: Is this generalisation of Fermat's little theorem a known result? If yes, could you please provide references?

Thanks,

PS: Please, be indulgent: this is my very first post on MO!

• Does Example 2.2(3) here suffice? – Benjamin Dickman Jun 12 '15 at 23:07
• This may be in Joe Roberts's calligraphic text on number theory. – The Masked Avenger Jun 12 '15 at 23:48
• According to Dickson's History of the Theory of Numbers, Volume 1, p. 84 (archive.org/details/historyoftheoryo01dick) this generalization of Fermat's theorem is due to Gauss, at least when $a$ is a prime. On p. 82 Dickson gives a reference for the general case to Thue in 1910, though it may be older. – Ira Gessel Jun 12 '15 at 23:54
• This is not new. Google "necklace polynomials" and consider the fact that they are integral-valued. – KConrad Jun 13 '15 at 3:50
• This result has also been attributed to Ramachandra (mathoverflow.net/questions/87048/…). – Ira Gessel Jun 13 '15 at 15:36

Here's a massive generalization, with references.

Consider a sequence $\{a_n \}_{n\ge 1}$ of integers. We'll say it satisfies the necklace congruences if $a_{n} \equiv a_{n/p} \mod n$ whenever $p \mid n$ ($p$ prime).

Here are some equivalent formulations:

• $a_{p^{k+1} m} \equiv a_{p^k m} \mod {p^{k+1}}$ for all $p,m,k$
• $\sum_{d \mid n} a_d \mu(n/d) \equiv 0 \mod n$ for all $n$
• The Mobius function can be replaced by any arithmetic function $f$ satisfying $\sum_{d \mid n} f(d) =0 \mod n, f(1) = \pm 1$.

So any such sequence gives rise to an integer sequence $\lambda_{\tilde{a}}(n) := \frac{1}{n}\sum_{d \mid n} a_d \mu(n/d)$ which satisfies $a_n = \sum_{d \mid n} d \lambda_{\tilde{a}}(n)$ by Mobius inversion. One can then define $\sigma_{\tilde{a}}(n):=\sum_{d \mid n, 1<d<n} d\lambda_{\tilde{a}}(d)$ and obtain the following generalization of Fermat's little theorem:

$a_n \equiv a_1 + \sigma_{\tilde{a}}(n) \mod n$

Now, let's go back to necklace congruences. A less obvious equivalence is the following:

Theorem: $\{a_n\}_{n \ge 1}$ satisfies the necklace congruences iff $\zeta_{a}(x):=\exp(\sum_{n\ge1} \frac{a_n}{n}x^n)$ has integer coefficients.

Proof: Write, $\zeta_a$ formally as $\prod_{n\ge 1} (1-x^n)^{-\frac{b_n}{n}}$. The $b_n$'s are uniquely determined, and in fact (by taking logarithmic derivative) it can be seen that they are $b_n = \sum_{d \mid n} a_d \mu(n/d)$. Now one direction is immediate and the other requires an inductive argument. $\blacksquare$

(Reference: Exercise 5.2 (and its solution) in Richard Stanley's book "Enumerative Combinatorics, vol. 2")

As Sergei remarked, for any integer square matrix $A$, the sequence $a_n := \text{Tr}(A^n)$ satisfies the necklace congruences. This is immediate from the last theorem, as the corresponding "zeta" function $\zeta_a$ is just:

$\exp( \sum_{n \ge 1} \frac{\text{Tr}((Ax)^n)}{n}) = \exp (\text{Tr} (- \ln (I -Ax)))=\det(I-Ax)^{-1} \in \mathbb{Z}[x]$

In particular, by taking a 1 on 1 matrix, we recover your observation.

Another generalization is $a_n = [x^n]f^n(x)$, where $f \in \mathbb{Z}[[x]]$. By taking the linear polynomial $f(x)=1+ax$ we recover your example, by taking $f(x)=(1+x)^m$ we get $a_n = \binom{nm}{n}$.

The really interesting feature is that this notion has a local version. Specifically, theorems of Dieudonne-Dwork and Hazewinkel give the following result:

Theorem: Given $\{ a_n \}_{n\ge 1} \subseteq \mathbb{Q}_{p}$ ($p$-adic numbers), the following are equivalent:

• $\zeta_{a}(x) := \exp(\sum_{n\ge 1} \frac{a_n }{n}x^n)\in\mathbb{Z}_{p}[x]$
• $\exp( \sum_{n \ge 1} p \frac{a_n - a_{n/p}}{n} x^n) \in 1+px\mathbb{Z}_{p}[x]$ (I defnite $a_{n/p}=0$ if $p\nmid n$)
• $\sum_{n \ge 1} \frac{a_n - a_{n/p}}{n} x^n) \in \mathbb{Z}_{p}[x]$
• $p \mid n \implies a_n \equiv a_{n/p} \mod {n\mathbb{Z}_{p}}$

Applying this simultaneously for all $p$, we recover the previous theorem. A reference for this theorem is "A Course in p-adic Analysis" by Alain M. Robert (the theorems are stated there in a much more general context).

• Impressive, Ofir! You show that the Fermat's little theorem can be generalised even further than for $a^n$, which is a special case of a sequence satisfying the "necklace congruence", as you defined. You provide other general instances of such sequences, which indeed can be instantiated as $a^n$. However, I am still wondering whether you answered my questions or not... I must admit that some parts remain obscure for me, due to my lack of background. – Pierre Denis Jun 14 '15 at 22:06
• Small typo: You wrote $\sum_{d|b}$ after the bullet points. Should be $\sum_{d|n}$ ! – Pierre Denis Jun 14 '15 at 22:11
• I am trying to understand the very last paragraph. I cannot even understand the syntax after "in the congruences...". Can you please check and/or rephrase? Thanks. – Pierre Denis Jun 14 '15 at 22:19
• @PierreDenis First of all, thanks for pointing out the typo (corrected now). My last paragraph was indeed not very clear, and I deleted it. What I do have to say is that (as far as I can tell), calculating the $\sigma_{a}(n)$ requires knowing the factorization of $n$, unless $n=1$ or $n=p$ (which are the non-interesting cases), and it is probably easier to study the equation $a^n \equiv a \mod n$ directly. But I'm not an expert in this field at all. – Ofir Gorodetsky Jun 15 '15 at 14:52
• Indeed, $\sigma_a(n) \bmod n$ doesn't bring any new information compared to $(a^n-a) \bmod n$. However, my idea was to look at residues term by term, i.e. the $(d \lambda_a(d)) \bmod n$. I have at least one result so far for Fermat pseudoprimes with just two prime factors (aka semiprimes): I think that I can prove that for any distinct primes $p,q$, $a^{pq} \equiv a \pmod{pq} \iff a^p \equiv a \pmod{q} \wedge a^q \equiv a \pmod{p}$. OK, probably another known result...! – Pierre Denis Jun 15 '15 at 22:44

In this paper (in Russian) : http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mp&paperid=238&option_lang=eng there are discussion of this result, including theorem of Gauss, generalization to matrices by V.I.Arnold and so on.

• Thanks Sergei but I don't understand Russian, unfortunately. I've tried to guess by looking at the formulas but I don't see something similar to the congruence I gave above. Could you please give me the page number / formula number in the paper so I can use a translator on the right part? – Pierre Denis Jun 13 '15 at 18:09
• @PierreDenis, equation (12) in section 3 is the main point for you. It says $\sum_{d|m} \mu(m/d)a^d \equiv 0 \bmod m$ for any $a \in \mathbf Z$ (of course it suffices to check only for $1 \leq a \leq m-1$). He notes Gauss proved it for prime $a$ and other mathematicians proved it in general by several mathematicians in the 1880s. – KConrad Jun 13 '15 at 20:40
• Thanks for spotting. This is indeed a generalisation of Fermat's little theorem. However, as I wrote to Ira Gessel (see comments above), I am looking for references that do not use Möbius function or the like, i.e. without any negative terms, as shown in my post. – Pierre Denis Jun 14 '15 at 9:42