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Let $Q$ be an ellipse with integer-length axes $a$ and $b$: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \;.$$ The perimeter of $Q$ is given by the complete elliptic integral of the 2nd kind, $E(\;)$: $4 a E(\epsilon)$ where the eccentricity $\epsilon = \sqrt{1-(b/a)^2}$. Is the following known?

Q. Is the perimeter of an ellipse with integer-length axes $a$ and $b$ always irrational? If not, for which $a,b$ is it rational?


           Ellipse21
           $a,b=2,1$. Perimeter is approximately $9.051752335744220113037500073330267105760842885887$.

Answered by paul Monsky in the comments: The perimeter is transcendental! Established by Theodor Schneider in 1937.

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    $\begingroup$ If you collapse an ellipse down to a line, then you get a rational perimeter immediately :-) $\endgroup$
    – Suvrit
    Commented Nov 23, 2013 at 3:24
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    $\begingroup$ You should check out Schneider's article, "Arithmetische Untersuchungen elliptischer Integrale" , Math. Ann. 113 (1937),1-13. It may have what you're looking for, and in fact show that the perimeter is transcendental. $\endgroup$
    – paul Monsky
    Commented Nov 23, 2013 at 4:02
  • $\begingroup$ Springer link to Schneider paper: here. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 23, 2013 at 12:09
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    $\begingroup$ Joseph: As an explicit English language reference to the ellipse problem, Waldschmidt gives Schneider's book:"Transcendental numbers" Annals of Mathematical Studies #16, Princeton University Press. $\endgroup$
    – paul Monsky
    Commented Nov 23, 2013 at 14:47
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    $\begingroup$ From the Schneider paper: "Der Wert eines elliptischen Integrals erster oder zweiter Gattung mit algebraischen Koeffizienten und zwischen algebraischen Grenzen ist transzendent." $\endgroup$
    – Joseph O'Rourke
    Commented Nov 23, 2013 at 14:57

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Answered by paul Monsky in the comments: The perimeter is transcendental! Established by Theodor Schneider in 1937:


 Math.Ann.

 Satz I
Roughly (Those with better knowledge of Deutsch, please correct!):

The value of an elliptic integral of the first or second kind with algebraic coefficients and between algebraic limits is transcendental.

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    $\begingroup$ I understand that Wustholz has the following result. Any period of a meromorphic differential on a Riemann surface, when both are defined over a number field, is 0 or transcendental. (I don't know if he handled incomplete integrals between algebraic limits as well). $\endgroup$
    – paul Monsky
    Commented Nov 25, 2013 at 4:11

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