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September 2

cyclotomic polynomials

For an arbitrary integer m, is there a cyclotomic polynomial with m as a coefficient of some power of x?(I don't trust ai's answer). 2) For an arbitrary integer m, are there infinitely many such cyclotomic polynomials with m as a coefficient? 3) Given an arbitrary finite set of integers, is there a cyclotomic polynomial that has all integers in that set as coefficients? Rich (talk) 04:50, 2 September 2026 (UTC)

2. Not an answer to your question per se, but note that if $p$ is a prime and $r$ is a positive integer coprime to $p$ (equivalently, a positive integer not divisible by $p$), then $\Phi _{p^{a}r}(x)=\Phi _{pr}(x^{p^{a-1}})$, which means that if $\Phi _{q}(x)$ has coefficient $m$, then any $q'$ with the same radical as $q$ will have a corresponding cyclotomic polynomial that also $m$ as a coefficient. Since there are infinitely many such $q'$, if $m$ is a coefficient in one cyclotomic polynomial, then it is a coefficient in infinitely many cyclotomic polynomials. GalacticShoe (talk) 14:25, 2 September 2026 (UTC)

1. While I haven't been able to find a copy online to view, apparently Jiro Suzuki's On coefficients of cyclotomic polynomials (1987), published in the Proceedings of the Japan Academy, Series A, Mathematical Sciences 63, proves that every single integer appears as a coefficient in some cyclotomic polynomial. By the point earlier, this implies that every single integer appears as a coefficient in infinitely many cyclotomic polynomials. GalacticShoe (talk) 14:41, 2 September 2026 (UTC)

OEIS sequence A013594 may be relevant here. --RDBury (talk) 23:27, 2 September 2026 (UTC)

thank you both. For my 3rd question, instead of finite set I should have said finite list or finite multiset, to include the possibility of an arbitrary integer appearing several times as a coefficient in the same polynomial.Rich (talk) 00:47, 3 September 2026 (UTC)

For enormously large nonprime degrees with enormously many divisors, do these polynomials start getting 1s and -1s as coefficients rarely? Rich (talk) 00:52, 3 September 2026 (UTC)

Suzuki's construction essentially works for (3), actually infinitely many such cases. Choose t bigger than any coefficient in your target set, and choose an increasing sey of odd primes p1,...,pt so that p1+p2 > pt, and let n be the product of the pi. Then modulo x^(pt+1), $\phi _{n}(x)\equiv (1+x+\cdots +x^{p_{t}})(1-\sum _{i=1}^{t}x^{p_{i}})$. So the k-th term is 1-(number of primes in the list less than k). Thus the coefficient is 1-r for each k between p_r and p_{r+1}. Using $\phi _{2n}(x)=\phi _{n}(-x)$ is then sufficient to get the r-1 as well. This argument works for prescribed multisets of minimum multiplicities too. All forms rely on the PNT in short intervals. Sławomir Biały (talk) 08:06, 3 September 2026 (UTC)

September 3

How to build an elliptic curve having a multiple of a given semiprime order without factorizing it?

everything is in the title, though I need having the embedding degree between 3 and 10 million. ~2026-47724-31 (talk) 11:15, 3 September 2026 (UTC)

This can be done by constructing a supersingular curve over F_p with p+1 elements, where p is a prime that is -1 mod N. Not sure about the embedding degree. Sławomir Biały (talk) 12:47, 3 September 2026 (UTC)

Sounds like solving a root modulo N without its factorisation? ~2026-47781-23 (talk) 17:58, 3 September 2026 (UTC)

Finding a prime p that is congruent to -1 modulo N does not require factoring N. Sławomir Biały (talk) 05:01, 4 September 2026 (UTC)

but then the embedding degree is smaller than 3? ~2026-47781-23 (talk) 14:16, 5 September 2026 (UTC)

How to generate ordinary elliptic curve pairings defined over finite rings of unknown order?

I was noticing something: unlike on semiprimes, it's quite easy to generate a point of known small order on a curve defined over a semiprime without knowing its factorization using global torsion. The idea is to transfer this small order to the underlying semiprime ring using a pairing method derived from the Weil pairing. Knowing whether a given specific small order exists can be derived using congruences modulo N.

As you know, the problem is then to avoid a degenerate output from Miller's algorithm. The problem is that the only method I know to avoid this without using the factorization is to use a distortion map that works only with supersingular elliptic curves. But using a supersingular curve caps the order to gcd(p+1,q+1) which is unknown most of the time or just equal to 2 Taking order 2 results in generating outputs equal to 1 or −1. Useless for GCD factoring.

So, any idea of a pairing mechanism that would work over ordinary elliptic curves using embedding degree 1? Possibly using a distortion map that works over ordinary elliptic curves in such a case? Or a pairing that works with simple final exponentiation, such as cubing or squaring? ~2026-47781-23 (talk) 18:52, 3 September 2026 (UTC)

For embedding degree 1, there are distortion maps: . For the remaining questions, have you looked at this, seems relevant: . Sławomir Biały (talk) 14:15, 4 September 2026 (UTC)

The problem is getting them working when the factorization is unknown. ~2026-47781-23 (talk) 14:14, 5 September 2026 (UTC)