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Consider the sequence

$$ a(n) = \prod_{u^n=1,u \neq 1}( (1+u)^n+1) $$

Some terms are: $$ 1,1,0,9,121,2704,118336, 4092529,0,97734390625, \ldots $$

Alonso del Arte asks:

Question: What are the multiples of $3$ such that

$$ a(3k) =0 $$

I tried some factorization of cyclotomic polynomials without success. May be true for all odd $k$ ???

EDIT: Another simple property of the sequence is

(hope this may please the negative voter (???))

$$ a(p) \equiv 1 \pmod{p} $$

for any prime $p>3$

since

$$ a(n) (2^n+1) $$ is the determinant of a circulant matrix with first line $$ 3,\binom{n}{1}, \ldots,\binom{n-1}{n} $$

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Why does this question deserve negative feedback? –  Daniel Parry May 14 '11 at 22:38
    
@Daniel: faq, first line, first paragraph –  Franz Lemmermeyer May 15 '11 at 18:37
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2 Answers

up vote 9 down vote accepted

Suppose that $n=3m$, where $m$ is odd, and $u=e^{2\pi i/3}$. Then $$ (1+u)^n+1 = ((1+u)^3)^m+1 = (-1)^m+1=0, $$ so $a(n)=0$.

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nice ! I do not see that... –  Luis H Gallardo May 13 '11 at 23:03
4  
Reminds me of the old joke about how to simplify the expression $(x-a) (x-b) \ldots (x-z)$. –  Terry Tao May 15 '11 at 18:37
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The complex number $a(n)$ is the resultant of the polynomials $P=(X^n-1)/(X-1)$ and $Q=(X+1)^n+1$; similarly, $(2^n+1)a(n)$ is the resultant of $X^n-1$ and $(X+1)^n+1$. Since these polynomials have integer coefficients, their resultant is a rational integer.

The resultant of two polynomials vanishes whenever they have a common root. So $a(n)=0$ if and only if there exists a $n$th root of unity $u$ such that $(u+1)^n+1=0$. This implies that $u$ and $u+1$ are both roots of unity, in particular they belong to the unit circle, so that necessarily $u=e^{2\pi i/3}$ or $u=e^{-2\pi i/3}$, and $u+1=e^{\pm i\pi/3}$. If $u$ is a $n$th root of unity, one gets $3|n$; if $(u+1)^n=-1$, one obtains that $n/3$ is odd. Conversely, if $n=3m$ with $m$ odd, $u=e^{2\pi i/3}$ satisfies $u^n=1$, $u\neq 1$, and $(u+1)^n=-1$, hence $a(n)=0$.

Since the two polynomials $P$ and $Q$ above are monic, their resultant vanishes mod $p$ if and only if they have a common root when considered as polynomials modulo $p$. If $n=p$ is prime, then $X^n-1=(X-1)^p$, so $1$ is the only root of $P$, with multiplicity $p-1$; it follows that $$ a(n)\equiv ((1+1)^p+1)^{p-1}\equiv (2^p+1)^{p-1}\equiv 3^{p-1} \pmod p.$$ If, moreover, $p\neq 3$, then $a(n)\equiv 1\pmod p$.

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