We know cyclotomic polynomials $\Phi_{2^kp^rq^m}(x)$ have coefficients in $\{0,\pm1\}$.

  1. What is the largest degree $f_{d,n}(x)=\frac{x^{2^{k}p^{r}q^{m}}-1}{\Phi_d(x)}$ with $\{0,\pm1\}$ coefficients where $1<d<2^{k}p^{r}q^{m}$ with $d|2^{k}p^{r}q^{m}$ holds? (bound by $a$ for given integer $a\in\Bbb Z$ is not needed).

  2. What properties of $d$ give only $\{0,\pm1\}$ coefficients to $f_{d,n}(x)$ at given $2^{k}p^{r}q^{m}$?

  • 1
    $\begingroup$ So, you've changed it from a product of two general cyclotomic polynomials to a quotient of two very special cyclotomic polynomials, right? Maybe you should change the title, to reflect the change in the body. Also, I assume you want $d$ dividing $2^kp^rq^m$. $\endgroup$ – Gerry Myerson Jan 15 '18 at 18:13
  • $\begingroup$ yes $d>1$ and $d|2^kp^rq^m$ that makes most sense to me. $\endgroup$ – T.... Jan 15 '18 at 18:23
  • $\begingroup$ @GerryMyerson it was slightly easier to specify as quotient rather than product. $\endgroup$ – T.... Jan 15 '18 at 18:46

A trivial answer (to 2) is just that if $f(x),g(x) \in \mathbb{Z}[x]$ have degrees $m,n$ and coefficients in $\{-1,0,1\}$ then $f(x)g(x) = \sum_{k \leq m+n} x^k \sum_{i+j=k;i \leq m;j \leq n} f_i g_j$ and so the $x^k$ coefficient is bounded by the number of terms in the inner sum, which is $1+\min(k,m)-\max(0,k-n)$, from which you can bound the whole polynomial.

  • $\begingroup$ but cancellations can happen $\endgroup$ – T.... Jan 15 '18 at 12:01
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    $\begingroup$ but it is still a bound, like you asked $\endgroup$ – Doris Jan 15 '18 at 12:09
  • $\begingroup$ No I actually did not intend that. $\endgroup$ – T.... Jan 15 '18 at 16:59

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