Is there any ellipse with nonzero rational minor and major axis lengths $a$ and $b$ such that the circumference of the ellipse is rational too? (or the weaker variant: ... such that the circumference is algebraic?)
Note that the surface area $S=\pi ab$ is of course always transcendental in such a situation.
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3$\begingroup$ This is probably related to the transcendence of periods of elliptic curves defined over the rationals, which you can find, e.g., in Baker's book Transcendental number theory. I am not sure it is exactly the same thing though, you'd have to look at some old book on elliptic integrals. $\endgroup$– Felipe VolochCommented Jun 12, 2013 at 17:05
1 Answer
No, there is no such ellipse. This is exactly theorem $6.5$ of Alan Baker's book TRANSCENDENTAL NUMBER THEORY, as pointed out by Felipe Voloch.
Let $\omega$ be a primitive period of a $\wp$-function with algebraic invariants $g_2, g_3$ and let $\eta=2\zeta(\omega /2)$ be the associated quasi-period of the Weierstrass $\zeta$-function satisfying $\zeta'(z) = -\wp (z)$. Then we have
Theorem $6.5$. Any linear combination of $\omega$ and $\eta$ with algebraic coefficients, not both zero, is transcendental.
Because $\omega$ and $\eta$ can be represented as elliptic integrals of the first and second kinds respectively, Theorem 6.5 implies that the circumference of any ellipse with algebraic axes-lengths is transcendental.