Cyclotomic polynomial question

2012-11-09T12:14:58Z

To avoid case distinction overload, I also call (say) $Z^3-Z^4$ cyclotomic. Just divide out the $Z=0$ solutions in the following if they offend.
In the following, all exponents are assumed to be positive integers.
Assume that $P=Z^a+Z^b+Z^c-Z^d-Z^e-Z^f$ is cyclotomic. The "standard" form (since $Z-1$ factors out immediately) would be something like $P'=Z^A(Z-1)(1\pm{Z}^B+Z^{2B})$. But note that, say, $P''=(Z-1)(1+Z^3+Z^4+Z^7+Z^8+Z^9+Z^{10})$ telescopes to the same form as $P$. Surely, $P''$ is not cyclotomic, but it was just a random example anyway :-) So: Is there a cyclotomic $P$ than can NOT be written in the form $P'$? OR even some $1\pm{Z}+Z^n, n>2$ that is cyclotomic (although to this my instinct says, no way - the proof is probably an one-liner in complex analysis...).

Answer by Aaron Meyerowitz for Cyclotomic polynomial question

2012-11-10T05:54:25Z

To start with your second question, cyclotomic polynomials have a central symmetry or antisymmetry, If $f(x)$ is a cyclotomic polynomial of degree $N$ then $f(Z)=\pm Z^nf(\frac1Z).$ So $Z^n\pm Z+1.$ would only be cyclotomic if $n=2$

In your construction $P'$ you could replace $Z-1$ by $Z^k-1$. Looking at $(Z-1)\Phi_q\Phi_r\Phi_s$ which avoid being of this form one finds numerous cases. One is $q,r,s=2,5,8$

$$Z^{10}+Z^9+Z^6-Z^4-Z-1=$$ $$(Z-1)(Z+1)(Z^4+Z^3+Z^2+Z+1)(Z^4+1)=$$ $$(Z-1)(Z^9+2Z^8+2Z^7+2Z^6+3Z^5+3Z^4+2Z^3+2Z^2+2Z+1)=$$ $$(Z^2-1)(Z^8+Z^7+Z^6+Z^5+2Z^4+Z^3+Z^2+Z+1)=$$ $$(Z^5-1)(Z^5+Z^4+Z+1)$$

Cyclotomic polynomial question - MathOverflow [closed]

Cyclotomic polynomial question

Answer by Aaron Meyerowitz for Cyclotomic polynomial question