3
$\begingroup$

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 $9$.

$\endgroup$
2
  • $\begingroup$ 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). $\endgroup$ 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
    Dec 13, 2011 at 0:05

2 Answers 2

5
$\begingroup$

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 $5, 7, 17, 19, 29, 31, 43, 53, 79, 89$.

$\endgroup$
3
  • $\begingroup$ @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
    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
    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.... $\endgroup$ Dec 13, 2011 at 4:49
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.