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Working on this recent MSE question, I was led to the following conjecture :

Suppose that $n$ is an integer with at least one prime divisor $\geq 7$. Then $\Phi_n$ has at least seven non-zero coefficients.

I have checked this conjecture up to $n\leq 10^5$. It is not hard to treat the case when $n$ is of the form $p^a$ with $p\geq 7$. In the general case $n$ will be of the form $n=p^a m$ with $m$ coprime to $p$, and $\Phi_n=\Phi_{p^{a}}\Phi_m$. What is unclear to me is how non-zero coefficients are somewhat "preserved" when we multiply by $\Phi_m$ where $m$ is coprime to $p$.

UPDATE (10/18/2014) : One can assume without loss of generality that $n$ is square-free. Indeed, let $n=\prod_{k=1}^r {p_k}^{a_k}$ be the prime factorization of $n$, with prime $p_k$ and $a_k\geq 1$. Let $m=\prod_{k=1}^r p_k$ be the square-free part of $n$. If $\zeta$ is a $n$-th root of unity, then it is a primitive $n$-th root of unity iff $\zeta^\frac{n}{m}$ is a primitive $m$-th root of unity. It follows that $\Phi_{n}(X)=\Phi_{m}(X^\frac{n}{m})$, so that $\Phi_n$ and $\Phi_m$ share the same number of non-zero coefficients.

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Ewan Delanoy
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  • Are you sure this is what you want? For MSE question 962636 it seems like you'd need to classify all the ways that $w$ roots of unity can sum to zero for some small $w$ (I guess all $w \leq 6$ from your question); such a sum might come not from $\Phi_n$ itself but from some multiple. For example, $\Phi_{15}(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$ has seven nonzero coefficients, but $(x+1) \Phi_{15}(x) = x^9 - x^7 + x^6 + x^3 - x^2 + 1$ has only six. (Fortunately for you, $15$ has no prime factors $\geq 7$.) – Noam D. Elkies Oct 22 '14 at 04:28
  • @NoamD.Elkies "Are you sure this is what you want?" Yes. And all you state after your question is correct (in fact, I only need $w=6$ not $w<6$). I am aware that the conjecture in itself does not solve question 962636, but I hope that methods useful to prove this conjecture could be re-used or extended. – Ewan Delanoy Oct 22 '14 at 04:38
  • Well there are known results that describe, for small $w$, all the ways that $;w$ roots of unity can sum to zero; but your conjecture concerns the number of nonzero coefficients, not the sum of their absolute values, so it seems to call for a different approach. – Noam D. Elkies Oct 22 '14 at 04:48
  • @NoamD.Elkies What known results are you alluding to ? Forgive my ignorance. Also, I conjecture separately that no solution $n$ to question 962636 can have a divisor $\geq 7$, but I'll ask about that here only if I solve my conjecture first. – Ewan Delanoy Oct 22 '14 at 04:49
  • See the paper http://www-math.mit.edu/~poonen/papers/ngon.pdf = "The number of intersection points made by the diagonals of a regular polygon" by B.Poonen and M.Rubinstein (motivated by another geometric problem of very similar flavor). The table on page 7 (backed up by Theorem 3 on the same page) goes up to $w = 12$; the text credits to Mann 1965 [8] for $w \leq 7$, and Conway-Jones 1976 [2] for $w \leq 9$. In fact the earlier papers allowed arbitrary integer coefficients, so Mann's 1965 paper should already contain a proof of your present conjecture too. – Noam D. Elkies Oct 22 '14 at 05:45
  • @NoamD.Elkies Many thanks for all this stuff! – Ewan Delanoy Oct 22 '14 at 06:15
  • @NoamD.Elkies I don’t know if you realized it, but the result of Mann's 1965 paper (quoted in the link you gave) immediately entails a complete solution to question 962636. I'll add an answer to it when I have the time (unless someone else does before me) – Ewan Delanoy Oct 22 '14 at 06:29
  • I didn't look closely at Question 962636, but I'm not surprised that Mann's result should answer it. (He describes it in terms of yet another geometrical problem, the description of polygons with rational sides and angles.) – Noam D. Elkies Oct 22 '14 at 06:42

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Here is my proof for the conjecture. As you mentioned, we can assume that $n=pm$ so that $p$ does not divide $m$.

Suppose that the conjecture is false and $\Phi_n(x)$ has at most $p-1$ nonzero coefficients: $$ \Phi_n(x) = \sum_{\ell=1}^{p-1} a_{d_\ell} x^{d_\ell} $$ with some exponents $d_1,\ldots,d_{p-1}$ and integer coefficients $a_{d_1},\ldots,a_{d_{p-1}}$.

By the pigeonhole principle, there is some integer $u$ such that none of $d_1+u$, ..., $d_{p-1}+u$ is divisible by $p$. Take such a $u$. Let $\varepsilon=e^{2\pi i/n}$ and $\varrho=e^{2\pi i/p}=\varepsilon^m$, and consider the following expression: $$ S = \sum_{j=1}^p \varepsilon^{jum} \Phi_n(\varepsilon^{jm+1}). $$ Among the numbers $m+1,2m+1,\ldots,pm+1$ precisely one is divisible by $p$; the other ones are co-prime with $n=mp$. So, among $\varepsilon^{m+1},\ldots,\varepsilon^{pm+1}$ there are $p-1$ primitive $n$th roots of unity and one number that has lower order, namely $m$. Therefore, the numbers $\Phi_n(\varepsilon^{m+1}),\Phi_n(\varepsilon^{2m+1}),\ldots,\Phi_n(\varepsilon^{pm+1})$ are all zeros except for exactly one. Hence, $S\ne0$.

On the other hand, $$ S = \sum_{j=1}^{p} \varepsilon^{jum} \sum_{\ell} a_{d_\ell} \varepsilon^{(jm+1)d_\ell} = \sum_{\ell} a_{d_\ell} \varepsilon^{d_\ell} \sum_{j=1}^{p} \varepsilon^{jm(d_\ell+u)} = \sum_{\ell} a_{d_\ell} \varepsilon^{d_\ell} \sum_{j=1}^{p} \varrho^{j(d_\ell+u)}. $$

It is well-known that $\sum\limits_{j=1}^p \varrho^{jK}=0$ for all integers $K$ that are not divisible by $p$. Applying this to $K=d_1+u,\ldots,d_{p-1}+u$, we can see that $$ S = \sum_{\ell} a_{d_\ell} \varepsilon^{d_\ell} \sum_{j=1}^p \varrho^{j(d_\ell+u)} = \sum_{\ell} a_{d_\ell} \varepsilon^{d_\ell} \cdot 0 = 0. $$

We have proved both $S\ne0$ and $S=0$, the conjecture must be true.

In fact we proved that the exponents that occur in $\Phi_n$ form a complete residue system modulo $p$.

G.Kós
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  • This is nice! May I ask how the idea of defining this $S$ came to you ? – Ewan Delanoy Oct 24 '14 at 11:56
  • My first idea was to substitute a $p$th root of unity and then use that $1+x+\ldots+x^{p-1}$ is irreducible. If the value was 'nice' then this could prove that we have exponents in every remainder class. But expressing the value of $\Phi_n(e^{2\pi i/p})$ was too ugly. The next thing is to substitute all $p$th roots and take the sum or a linear combination in order to cancel out most of the terms except for an arithmetic progression. Next is rotating the $p$th roots by a common angle, which is substituting $\varepsilon^{jm+c}$ for every $j$. Of course there were more things that did not work... – G.Kós Oct 24 '14 at 12:41