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The Wendt binomial circulant determinant $W_n$ can be defined quite simply as a resultant: $$ W_n = \operatorname{res}(x^n-1, (x+1)^n-1). $$ Truer to its name, one may also define it as the determinant $\det(A)$ of the circulant matrix with entries $A_{i,j} = \binom{n}{\lvert i-j\rvert}$.

The Wendt determinant was of interest historically to number theory because of its connection to Fermat's last theorem. The sequence is available on the OEIS as A048954, beginning as follows: $$1, -3, 28, -375, 3751, 0, 6835648, -1343091375, \dotsc$$

I have recently become interested in some of the prime factors of the Wendt determinant, a list of which is available online. Specifically, I am wondering: for which $n$, relatively prime to 6*, is $W_n$ divisible by $3$? I am interested in any result that gives a sufficient condition for $W_n$ to not be divisible by 3.

The small $n$, relatively prime to $6$, for which $W_n$ is divisible by $3$ are multiples of 13, 121, 671, and 757 (note that $W_m$ divides $W_n$ if $m$ divides $n$). I was not successful in finding this sequence or any other related sequence in the OEIS.

* I ask for relatively prime to $6$ for some technical reasons. Every sixth entry is zero, and also every even entry is known to be divisible by three. I am also interested in which of the even entries is twice divisible by $3$, ie divisible by $9$.

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  • $\begingroup$ It might be worth having a look at Greg Fee and Andrew Granville, The prime factors of Wendt's binomial circulant determinant, Math. Comp. 57 (1991), no. 196, 839–848, MR1094948 (92f:11183). $\endgroup$
    – Gerry Myerson
    Commented Dec 12, 2011 at 21:50
  • $\begingroup$ @Gerry Myerson: Thank you for your comment. I have found that paper as well as a few others (Ford & Jha, Helou, Simalarides). Although the titles are promising, they appear to focus on large prime factors. In any case, I could not resolve my question, nor manage to find any applicable results. $\endgroup$
    – aorq
    Commented Dec 13, 2011 at 0:05

2 Answers 2

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The resultant $W_n$ is a multiple of $3$ if and only if the two polynomials $x^n-1$ and $(x+1)^n-1$ share a common irreducible factor when considered as polynomials in $({\mathbb Z}/3{\mathbb Z})[x]$.

Suppose now that $n>3$ is an odd prime. Factoring out the obviously unique factors $x-1$ and $x$, respectively, we see that $3\mid W_n$ if and only if $(x^n-1)/(x-1) = \Phi_n(x)$ and $((x+1)^n-1)/x = \Phi_n(x+1)$ share a common irreducible factor.

Suppose further that $n$ is an odd prime for which $3$ happens to be a primitive root (mod $n$). Then $\Phi_n(x)$ is irreducible in $({\mathbb Z}/3{\mathbb Z})[x]$ (see for example the Corollary on page 2); in particular, $\Phi_n(x)$ shares no common irreducible factor with $\Phi_n(x+1)$.

This gives a sufficient condition for $W_n$ not to be a multiple of $3$: if $n$ is a prime for which $3$ is a primitive root. There should be infinitely many such $n$, but unfortunately we can only prove this under the assumption of a generalized Riemann hypothesis. The first few such $n$ are $5, 7, 17, 19, 29, 31, 43, 53, 79, 89$.

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  • $\begingroup$ @Greg Martin: Thanks for your analysis! It is, of course, a correct sufficient condition. Unfortunately, I seem to have encountered a common problem when posting a question on MO: how much of the specific issue to mention. It so happens that I'm looking at $W_n$ for divisors $n$ of $3^d-1$, which is to say, numbers $n$ such that $3$ is not a primitive root modulo $n$. I suppose in more detail what I've been trying to understand is why for some numbers, like $20$, $W_n$ is not divisible by 3, while for others like $121$, it is, despite the common fact that $3$ generates neither $\mathbb Z/n$. $\endgroup$
    – aorq
    Commented Dec 13, 2011 at 2:22
  • $\begingroup$ (As indicated in the footnote in the question, "$W_{20}$ divisible by $3$" is to be interpreted as "$W_{20}/3$ divisible by $3$ because $20$ is even.) If you can think of any other sufficient conditions, particularly those appropriate precisely when $3$ is not a primitive root, let me know. Thanks! $\endgroup$
    – aorq
    Commented Dec 13, 2011 at 2:48
  • $\begingroup$ In general, if $k$ is the multiplicative order of $3$ modulo $n$, then the $n$th cyclotomic polynomial factors (mod $3$) into a product of irreducibles all of degree $k$. The smaller $k$ is, the more likely there is of the coincidence that one of those factors $f_i(x)$ is equal to one of the others $f_j(x+1)$ shifted; that will cause $W_n$ to be a multiple of $3$. Unfortunately you're looking exactly at the situation where the order is forced to be small (a divisor of $d$)! - making these coincidences more likely. But I have no idea how to detect the coincidences other than trial and error.... $\endgroup$
    – Greg Martin
    Commented Dec 13, 2011 at 4:49
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An extensive study of Wendt determinants is done in "Elimination : Résultants et Sous-résultants, le cas d'une variable", by François Apéry and Jean-Pierre Jouanolou (Hermann edit., 2006). See Exercises 8.9.5. pp.229-221 and their solutions pp.411-415. Exercise 59 (p. 221) is about a conjecture by Helou and Terjanian.

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