When is the Wendt binomial circulant determinant divisible by 3? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:36:46Z http://mathoverflow.net/feeds/question/83279 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83279/when-is-the-wendt-binomial-circulant-determinant-divisible-by-3 When is the Wendt binomial circulant determinant divisible by 3? aorq 2011-12-12T20:40:40Z 2011-12-13T02:48:54Z <p>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}\$.</p> <p>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 <a href="http://oeis.org/A048954" rel="nofollow">A048954</a>, beginning as follows: \$\$1, -3, 28, -375, 3751, 0, 6835648, -1343091375, \dotsc\$\$</p> <p>I have recently become interested in some of the prime factors of the Wendt determinant, <a href="http://www.numericana.com/data/wendt.htm" rel="nofollow">a list of which</a> is available online. Specifically, I am wondering: <strong>for which \$n\$, relatively prime to 6*, is \$W_n\$ divisible by \$3\$?</strong> I am interested in any result that gives a sufficient condition for \$W_n\$ to <em>not</em> be divisible by 3.</p> <p>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.</p> <p>* 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\$.</p> http://mathoverflow.net/questions/83279/when-is-the-wendt-binomial-circulant-determinant-divisible-by-3/83306#83306 Answer by Greg Martin for When is the Wendt binomial circulant determinant divisible by 3? Greg Martin 2011-12-13T01:38:39Z 2011-12-13T01:38:39Z <p>The resultant \$W_n\$ is a multiple of \$3\$ <a href="http://en.wikipedia.org/wiki/Resultant#Properties" rel="nofollow">if and only if</a> 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]\$. </p> <p>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.</p> <p>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 <a href="http://math.columbia.edu/~pugin/Teaching/USemBlog_files/CycloRed.pdf" rel="nofollow">the Corollary on page 2</a>); in particular, \$\Phi_n(x)\$ shares no common irreducible factor with \$\Phi_n(x+1)\$.</p> <p>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\$.</p>