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?


  • $\begingroup$ Are you asking if the product can equal one, or stating that? $\endgroup$ Jun 25, 2010 at 18:42
  • $\begingroup$ +1 for the weird product. It's quite unlikely to have both $a$ and $b$ different and rational, at least for reasonably large $n$. $\endgroup$ Jun 26, 2010 at 3:49

4 Answers 4


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.

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    $\begingroup$ Nice one David. Conjecture: the OP knew the question was equivalent to FLT! $\endgroup$ Jun 26, 2010 at 16:10
  • $\begingroup$ Thanks for the typo fix. Yeah, I can't imagine coming up with this by accident. $\endgroup$ Jun 26, 2010 at 16:20
  • $\begingroup$ I was duped! +1 all around. $\endgroup$ Jun 26, 2010 at 16:22
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    $\begingroup$ And they said FLT was a useless result ;-) $\endgroup$ Jun 26, 2010 at 16:51
  • $\begingroup$ You all figured it out. Congratulations! I was actually hoping someone would give a different answer than quoting Wiles and Taylor's theorem. Notice that the points on the ellipse that intersect the rays generate a subfield of the complex numbers if their product equals one. And if a and b are rational, only two if n is odd (or four if n is even) of the points on the ellipse have rational coordinates. Is it possible to use this fact to generate a contradiction and prove FLT? $\endgroup$ Jun 27, 2010 at 21:07

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


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 ;-)


(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 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.

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    $\begingroup$ I didn't check Will's claim that the $n=6$ problem is equivalent to the cubic he gives above, but I can solve $4a^3b^2=3a^2+b^2$. Clumping the b^2 terms together one gets $(4a^3-1)b^2=3a^2$ so one is seeking rationals $a$ with $3a^2(4a^3-1)$ a square. Now $a^2$ is a square so we just want $3(4a^3-1)$ a square; this is an elliptic curve, and multiplying up by 144 and setting $x=12a$ we get $y^2=x^3-432$, which is $X_0(27)$ and IIRC birational to the Fermat cubic, which has rank 0 and torsion leading to $a=1$. In particular the question for $n=6$ is basically equivalent to FLT for $n=3$. $\endgroup$ Jun 26, 2010 at 9:15
  • $\begingroup$ Because of the coincidence above, there's a non-zero chance that the question for general even $n$ is equivalent to FLT for $n/2$. This is just a stab in the dark though. $\endgroup$ Jun 26, 2010 at 9:17
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    $\begingroup$ $2a^3b^3=a^2+b^2$ does not have nontrivial solutions either. Indeed, the equation implies that $(a^2+b^2)/2ab$ is a square of a rational, hence $(x^2+1)/2x=y^2$ for $x=a/b$ and $y$ rational. The discriminant of this as of a quadratic equation in $x$ is $4(y^4-1)$. It is never a square of a rational unless equals zero (this quickly reduces to the equation $x^4-y^4=z^2$ over integers). $\endgroup$ Jun 26, 2010 at 11:56
  • $\begingroup$ So you're doing the $n=8$ case by writing the relevant curve as a degree two cover of the curve Fermat used to prove no non-triv solutions to $x^4+y^4=z^4$. So there is a chance that the problem for even $n$ is equivalent to FLT after some cunning re-arrangement. $\endgroup$ Jun 26, 2010 at 13:27

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