Problem related to the construction of a heptadecagon.

Question

If we have a list of the $17$ roots of unity of a 17 sided polygon such as here:

$A=\zeta +\zeta^{9} +\zeta^{13} + \zeta^{15} + \zeta^{16} + \zeta^{8} + \zeta^{4} + \zeta^{2} $

$B=\zeta^{3} +\zeta^{10} +\zeta^{5} + \zeta^{11} + \zeta^{14} + \zeta^{7} + \zeta^{12} + \zeta^{6} $

$C=\zeta +\zeta^{13} +\zeta^{16} + \zeta^{4} $

$D=\zeta^{9} +\zeta^{15} +\zeta^{8} + \zeta^{2} $

$E= \zeta^{3} + \zeta^{5} + \zeta^{14} + \zeta^{12}$

$F= \zeta^{10} + \zeta^{11} + \zeta^{7} + \zeta^{6}$

$G= \zeta + \zeta^{16} = \zeta + \zeta^{-1}$

$H= \zeta^{13} + \zeta^{4}$

The question that I am stuck on how:

How would one show that $CD=EF=-1$ and that $AB=-4$ so that $A$ and $B$ are roots are roots of $x^{2}+Ax-4=0$?

I am lost at where to even start. Thanks.

Answer 1

To render $CD=-1$, simply multiply term by term. Putting in $(\zeta^m)(\zeta^n)=\zeta^{m+n}$ and $\zeta^{17}=1$ we have:

$CD=\zeta^1\zeta^9+\zeta^1\zeta^{15}+\zeta^1\zeta^8+...+\zeta^4\zeta^2$

$=\zeta^{10}+\zeta^{16}+\zeta^{9}+\zeta^{3}+\zeta^{22}+...+\zeta^{6}$

$=\zeta^{10}+\zeta^{16}+\zeta^{9}+\zeta^{3}+\zeta^{5}+...+\zeta^{6}$

where, in the last line, each exponent from $1$ to $16$ inclusive appears exactly once. Thereby the product is $\zeta+\zeta^2+...+\zeta^{16}=-1$.

The product $EF=-1$ is proven the same way. For $AB$, the term-by-term expansion is more tedious,involving $64$ terms, but each exponent from $1$ to $16$ inclusive again appears the same number of times, in this case four. Thus $(4)(-1)=-4$.

Answer 2

I think I have solved my own question. At least partially.

I noticed that if you sum $A$ and $B$, the result is $\left(\frac{-1+\sqrt{17}}{2}\right)$ and $\left(\frac{-1-\sqrt{17}}{2}\right)$resepctively.

Then, taking the product of $AB$, we can easily see that $AB=-4$.

Could anyone verify if this is correct?

Answer 3

Oh Milton P, where do we start ? ($F=?$, the obvious $4$ terms, $A,B$ don't satisfy that ?)

The first crucial observation is \begin{eqnarray*} 1+ \underbrace{ \zeta+ \zeta^2+ \cdots + \zeta^{16}}_{ \text{ $16$ terms}} =0. \end{eqnarray*} So $A+B=-1$, for $CD$ you need to calculate the $16$ terms and verify they give the $-1$ above.

Calculating $AB$ will give $64$ terms & I really don't fancy writing them down ... but they will finally give something "nice" ($-4$ probably).

In the end you will have a hierarchy of $3$ quadratics that will give $\zeta=?$. You need to write these into your question.

At the moment, your question is missing a lot of required details & it is probably best to delete it & repost with afore mentioned details.

Good luck $\ddot \smile$