better posed problem? (small surprise)

[Paul C. Leopardi]

Hi all,
Response below.
Best regards
On Thu, 17 Mar 2005 10:50 pm, Meeussen Wouter (bkarnd) wrote:
> it is a small surprise that,  for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) ,
> w+w^7+w^8+w^18+w^19+w^25  = 0
> no cancellation of symmetric pairs, triples or quintuples in sight,
> but w+w^7+w^19+w^25 = -w^13  (* four out of five *)
> and w^8+w^18 = -w^28 = +w^13  (* two out of three *)
>
> could be called "cancellation by missing parts" ;-))

Your identities are one way of observing that Z/30Z has subgroups isomorphic 
to Z/6Z and Z/5Z. In other words, the vertices of the regular 30-gon contains 
subsets which are the vertices of regular pentagons and regular hexagons.

The basic identity is 1+x+x^2+x^3+x^4=0, where x=w^6, which is one way of 
describing the symmetry of a regular pentagon. From this, we obtain
w+w^7+w^13+w^19+w^25=0.

A second  set of  identities is 1+y^2+y^4=0, 1+y^3=0, where y=w^5. This 
describes some of the symmetries of a regular hexagon. From this second set, 
we obtain y+y^4=0, 1+y^2-y=0, 1+w^10-w^5=0, w^8+w^18-w^13=0, and
1+w^15=0, w^13+w^28=0.
Adding identities of the first set to the second, we obtain 
w+w^7+w^8+w^18+w^19+w^25 = 0.

You will probably find similar, but possibly more complicated identities for 
any root of unity z, where z^(ab) = 1, with a, b intgers > 1. In other words,
Z/abZ has subgroups isomorphic to Z/aZ and Z/bZ, and the vertices of the 
regular ab-gon contains subsets which are the vertices of regular a-gons and 
regular b-gons.