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This is a continuation of the question

Factorization of polynomial with prime coefficients from earlier today.

As in the linked question we are interested in the possibility of a polynomial of the form $$ f(x)=p_0+p_1x+\cdots+p_{n-1}x^{n-1} $$ having a common factor with $x^n-1$ in the ring $\Bbb{Z}[x]$. Here the coefficients $p_i$ are constrained to be distinct primes.

Gerry Myerson and I provided examples of such polynomials $f(x)$ as answers to the linked question. All those examples took advantage of low degree factors of $x^n-1$, and were based on $f$ having a root of unity of order $d, d\mid n, d<n$, as a zero.

This leaves open the more difficult possibility of whether such a polynomial $f(x)$ can have as a zero a primitive $n$th root of unity. In other words,

is it possible to find a polynomial $f(x)$ of the prescribed form such that $\gcd(f(x),\Phi_n(x))\neq1$?

Preliminary thoughts:

  • I'm not sure this is possible. A proof of impossibility of this is obviously a great answer, but may be asking a lot.
  • As $\Phi_n(x)$ is irreducible, we need $\Phi_n(x)$ to be a factor of $f(x)$.
  • To have enough elbow room to search for $f(x)=g(x)\Phi_n(x)$ we probably want $\Phi_n(x)$ to have a degree a lot less than $n$.
  • However, selecting $n$ to be a prime power is a non-starter. If $n=p^m$, then we must have $\deg g(x)<p^{m-1}$, and the coefficients of $g(x)$ will just repeat in the product $g(x)\Phi_n(x)$.
  • In case it makes a difference, I would relax the constraints and allow a negative prime as a coefficient of $f$ as long as $p_i\neq\pm p_j$ whenever $i\neq j$.
Jyrki Lahtonen
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  • wikipedia has "The cyclotomic polynomial $\Phi _{n}(x)$ may be computed by (exactly) dividing $x^{n}-1$ by the cyclotomic polynomials of the proper divisors of n previously computed recursively by the same method" –  Jul 19 '17 at 11:29

1 Answers1

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If you allow negative primes, you have examples like $$3-11x+29x^2-97x^3+89x^4-71x^5$$ for $n=6$.

ADDED IN EDIT

How about $$5+13x+17x^2+7x^3+11x^4+19x^5$$ again with $n=6$?

Angina Seng
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    Lovely! Using a way to write $100$ as a sum of two primes in three different ways, and adjusting the signs. – Jyrki Lahtonen Jul 19 '17 at 12:24
  • One may be able to add a multiple of $1+x+\cdots +x^5$ to this to get six positive prime coefficients, but I lack the time and inclination to check this right now. – Angina Seng Jul 19 '17 at 12:35
  • That general approach certainly sounds promising (because we should also have quite a bit of flexibility in selecting a similar multiple of $x^2-x+1$). I will leave the question open for a couple of days in case different ideas come up. – Jyrki Lahtonen Jul 19 '17 at 12:41
  • Mind you, the method of adding a multiple of $1+x+\cdots+x^5$ to your solution won't produce only positive primes as coefficients, because, when taking the signs into account, the coefficients fall into three distinct residue classes modulo six. You probably need a solution without $\pm3$, but I won't have the time to look for one either. – Jyrki Lahtonen Jul 19 '17 at 13:12
  • The second suggestion is, of course, perfect. – Jyrki Lahtonen Jul 19 '17 at 18:36
  • @JyrkiLahtonen I think a real hard problem would be for prime $n$, for example $n=7$. Even cases allow, as we see, always some kind of symmetry. – Widawensen Jul 20 '17 at 11:39
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    @Widawensen There are no solutions for prime $n$. The thing to look at is other odd $n$. – Angina Seng Jul 20 '17 at 11:42
  • @LordSharktheUnknown There are no solutions ? What is the explanation ? – Widawensen Jul 20 '17 at 11:44
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    @Widawensen The polynomial must be a polynomial multiple of the cyclotomic polynomial. – Angina Seng Jul 20 '17 at 11:45
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    The "add a multiple of $1+x+\cdots+x^5$" idea works if you start by using $5+19=7+17=11+13$ to form $5-7x+11x^2-19x^3+17x^4-13x^5$ and then add 26 throughout to get $31+19x+37x^2+7x^3+43x^4+13x^5$. – Gerry Myerson Jul 20 '17 at 23:45
  • @Widawensen As I said, a prime power (including primes) will never work. I spent half an hour with Mathematica looking for an example with $n=15$. I learned a few things not to do, but not enough to find a candidate polynomial where I could start applying Lord Shark's trick of adding a multiple of $1+x+\cdots+x^{14}$ to it. For there to be any hope you need to find a multiple of $\Phi_{15}(x)$ such that its coefficients don't cover too many residue classes modulo a small prime. If I find the time I may continue that search. Accepting this for now. – Jyrki Lahtonen Jul 22 '17 at 06:42