The product of n radii in an ellipse 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 
 A: (Not an answer.)
Let $r = \frac{1}{a}, s = \frac{1}{b}$.  Reindex your points to $(x_0, y_0), ... (x_{n-1}, y_{n-1})$ and let $z_k^2 = x_k^2 + y_k^2$.  Then $x_k = z_k \cos \frac{2\pi k}{n}, y_k = z_k \sin \frac{2\pi k}{n}$, and the intersection condition becomes
$$z_k^2 \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$
Together with the condition that $\prod_{k=0}^{n-1} z_k = 1$, it follows that the desired conditions can be stated as
$$\prod_{k=0}^{n-1} \left( r^2 \cos^2 \frac{2\pi k}{n} + s^2 \sin^2 \frac{2\pi k}{n} \right) = 1.$$
This is likely to be a hard Diophantine equation to solve in general.  For $n = 3$, for example, the equation is
$$r^2 (r^2 + 3s^2)^2 = 16.$$
The curve $r(r^2 + 3s^2) = 4$ is an elliptic curve, and in general one must use computer algebra to rule out the existence of rational points on such curves.  In this particular case we might be able to get away with some argument using unique factorization in $\mathbb{Z}[\omega]$, but this strategy will fail in general for the same reason it fails in Fermat's Last Theorem.
A: A bit long for a comment, just checking the formulation of the problem. For $n=6$ I get $$ 4 a^3 b^2 = 3 a^2 + b^2. $$ Then for $n=8$ I get $$ 2 a^3 b^3 = a^2 + b^2 .$$ These are likely to be the two easiest. Anyway, there are sometimes methods for ruling out the existence of rational points on these curves other than the automatic $(a=1, \; b=1)$ and the rather artificial $(a=0, \; b=0).$  But this really does appear to be a different problem for each $n,$ meaning your question is a large project. 
A: 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.
A: OK so I bet my bottom dollar that this question is Fermat's Last Theorem in disguise. Very clever by the OP! Take for example $n=10$. Qiaochu's formula above, when expanded out and simplified, becomes
$$r^{20} + 40s^2r^{18} + 620s^4r^{16} + 4600s^6r^{14} + 16150s^8r^{12} + 23000s^{10}r^{10} + 15500s^{12}r^{8} + 5000s^{14}r^6 + 625s^{16}r^4 - 65536=0.$$
We seek solutions in positive rationals. This degree 20 polynomial on the LHS factors into three irreducibles; two of them are visibly never zero for positivity reasons, and the third is
$$(r+s)^5+(r-s)^5-32.$$
Finding a zero of this polynomial in positive rationals other than $r=s=1$ is equivalent to finding a counterexample to FLT for $n=5$. Something like this will surely work in general but I'm not going to do the algebra; the hint is that $x=(r+s)/2$ and $y=(r-s)/2$ and then if $x^m+y^m=1$ the ellipse will work out with $n=2m$. I am not saying I've proved this but I wouldn't be surprised if it were easy; I'll pass the baton and go back to being a Dad ;-)
