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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, Andres Caicedo, Andreas Blass Nov 10 '12 at 14:05

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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
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Cyclotomic polynomials have abelian Galois groups and roots on the unit circle. –  Franz Lemmermeyer Nov 9 '12 at 14:46
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1 Answer 1

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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