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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...).

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closed as not a real question by Denis Serre, Franz Lemmermeyer, Chandan Singh Dalawat, Andrés E. Caicedo, Andreas Blass Nov 10 '12 at 14:05

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I think it is pretty unclear what the real question is. Anyway, $Z^4 - Z^3 + Z^2 - Z + 1$ is the tenth cyclotomic polynomial and does not have the shape as $P$, neither does $Z^A(Z-1)(Z^4 - Z^3 + Z^2 - Z + 1)=Z^A(Z^5 - 2Z^4 + 2Z^3 - 2Z^2 + 2Z - 1)$. – Peter Mueller Nov 9 '12 at 12:53
Cyclotomic polynomials have abelian Galois groups and roots on the unit circle. – Franz Lemmermeyer Nov 9 '12 at 14:46
up vote 2 down vote accepted

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 $q,r,s=2,5,8$

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

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Thanks. One counterexample suffices :-) (Although I would have preferred if there were none...) – Hauke Reddmann Nov 10 '12 at 12:03
That is all the same example. In general $(Z^k-1)(Z^{k-1}+1)(Z+1).$ – Aaron Meyerowitz Nov 10 '12 at 12:25

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