Is it possible to find an ellipse whose perimeter and area are integer values? I guess that such ellipse doesn't exist, but I haven't fully proved that.
Obviously if both the semiaxes are algebraic, the area is ruled out, but the same is for the perimeter, as stated here.
The ellipse's perimeter is $ 4aE(e) $, where $a$ is the major semiaxis, the function $ E(x) $ is the complete elliptic integral of second kind, and $ 0 < e < 1 $ is the eccentricity defined as $ e = \sqrt{1- \frac {b^2}{a^2}} $ with $ b $ as the minor semiaxis
So to find an integer value for the perimeter, I need the inverse function of $ E(x) $, but contrary to the complete elliptic integral of the first kind, there is no known special function for that, but just series (here and here some references), so, because I wanted a closed form, ising the results of this theorem due to Selberg and Chowla, if k is a singular modulus then $ E(k) $ can be evaluated in closed form in terms of $ π $ and values of Gamma function at rational points (here some examples). So, if the perimeter is a positive integer $ n $ then $4aE(e) = n \Rightarrow a = \frac {n}{4E(e)} $
$ b = \sqrt{a^2(1-e^2)} = \sqrt{\frac{(1-e^2)n^2}{(4E(e))^2}} $
and the area is $ A = πab = \frac{π\sqrt{(1-e^2)}n^2}{(4E(e))^2} $
I tried with some values and I see that there is no perfect cancellation of the powers of $ π $ and the algebraic factors, so the Area can't be an integer, but I haven't a real demonstration of that.
Edit: these ellipses do exist and are infinitely many! Thanks to @heropup for showing that with a numerical example. I was too focused on finding one with parameters expressible with close form, so if f we restrict on those kind of parameters (here for some examples), is it possible to find one?
For $a = (\pi b)^{-1}$ and $a > b > 0$, then $0 < b < \pi^{-1/2}$. The area of such an ellipse is unity, and the perimeter is given by $$P(b) = \frac{4}{\pi b} \int_{\theta=0}^{\pi/2} (1 - (1-\pi^2 b^4) \sin^2 \theta)^{1/2} \, d\theta,$$ which proportional to the complete elliptic integral of the second kind. Then a plot of $P(b)$ on this interval shows a monotonically decreasing function, with $\lim_{b \to 0^+} P(b) = \infty$ and $P(\pi^{-1/2}) = 2 \sqrt{\pi} \approx 3.54491$. So there exists $b$ such that $P(b) = 4$. Newton's method or some other numerical approach (e.g., recursive bisection) yields the value $$b \approx 0.37334332668583542872022803191598896262977170372452 \ldots$$ for which the perimeter is $4$. Obviously, a closed form solution is not tractable. The corresponding value of the semimajor axis is $$a \approx 0.85259294443381110278147340714140156428522698936557 \ldots.$$
A computer can easily compile a table of such values.