When is the Wendt binomial circulant determinant divisible by 3?

2011-12-12T20:40:40Z

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 $9$.

Answer by Greg Martin for When is the Wendt binomial circulant determinant divisible by 3?

2011-12-13T01:38:39Z

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

When is the Wendt binomial circulant determinant divisible by 3? - MathOverflow

When is the Wendt binomial circulant determinant divisible by 3?

Answer by Greg Martin for When is the Wendt binomial circulant determinant divisible by 3?