I sent the following question to another forum more than a week ago but haven't got any responses. The moderator of that forum suggested that I pose the following question here:


Suppose we have an ellipse $x^2/a^2 + y^2/b^2 = 1$ (centered at the 
origin). Let $n>4$. There are $n$ rays going out of the origin, at angles 
$0, 2\pi/n, 4\pi/n, 6\pi/n,...,2\pi(n-1)/n$. Let $(x\_1,y\_1),...,(x\_n,y\_n)$ be 
the points of intersection of the rays and the ellipse. The product 
from $k=0$ to $n-1$ of $(x\_k)^2 + (y\_k)^2$ is equal to one. Can $a$ and $b$ be 
rational? Note that this is obviously possible if $a=b=1$, since then the 
ellipse becomes a circle of radius 1. But what about if $a$ is not equal 
to $b$? Can $a$ and $b$ still be rational? 


Craig 



The product of n radii in an ellipse
So, let's finish this. Starting from Qiaochu's formula,


$$\prod\_{k=0}^{n-1} (r^2 \cos^2 (2 \pi k/n) + s^2 \sin^2 (2 \pi k)/n))=1$$.


Each factor is


$$\left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) + i s \sin(2 \pi k/n) \right) \left(\vphantom{\frac{r}{2}} r \cos (2 \pi k/n) - i s \sin(2 \pi k/n) \right) =$$ 
$$\left( \frac{r+s}{2} e^{2 \pi i k/n} + \frac{r-s}{2} e^{-2 \pi i k/n} \right) \left( \frac{r+s}{2} e^{2 \pi i k/n} - \frac{r-s}{2} e^{-2 \pi i k/n} \right).$$ 


Putting $u=(r+s)/2$, $v=(r-s)/2$ and $\zeta=e^{2 \pi k/n}$, we have 
$$\prod\_{k=0}^{n=1} (u + v \zeta^{2k}) (u - v \zeta^{2k})$$
If $n$ is odd, this is $(u^n+v^n)(u^n - v^n)= (u^{2n} - v^{2n})$. If $n$ is even, I get $(u^{n/2} - v^{n/2})^2 (u^{n/2} + v^{n/2})^2 = (u^n-v^n)^2$. So either $u^{2n} - v^{2n}=1$ or $u^n-v^n=1$ or $u^n-v^n=-1$. These are all forms of FLT.


