Possible Pythagorean relation with Golden Ratio $ \phi^2+e^2 \approx \pi^2$

Question

While study Numerics and playing with famous constants ($e$, $\pi$, Golden ratio) I came across the following relation

$$ \color{blue}{1.6^2+2.7^2 = 9.85\approx 3.14^2}$$

This is nothing special but patently, by first oder approximation one glimpses that,

$$ e \approx 2.7,~~~~~~\phi\approx1.6~~~~~~\text{and}~~~~~\pi\approx3.14$$ where $\phi =\frac{1+\sqrt5}{2}$ is the well known Golden ratio. Putting this in the previous relation we get

$$ \color{blue}{\phi^2+e^2 \approx \pi^2}.$$

Although it is hopeless to get the perfect relation $ \color{blue}{\phi^2+e^2 =\pi^2},$ which could be an amazing Pythagorean relation between famous constants $\color{red}{e,\phi} $ and $\color{red}{\pi}$. I believe there is a chance that there exists a well know constant $\color{red}{\delta}$ (of this kind) such that we have the following approximation.

$$ \color{blue}{\phi^2+e^2 +\delta^2\approx \pi^2}.\tag{I}\label{Eq}$$ Which is also a Pythagorean relation in three dimension.

But since $ \color{blue}{\phi^2+e^2 >\pi^2}$, the best idea to find such constant is obviously to consider $\delta $ satisfying the relation.

$$\color{red}{\pm i\delta= \sqrt{\phi^2+e^2-\pi^2}\approx 0.37079062365 }$$

My Question: Which Well known constant could be suitable in other to improve this Pythagorean relation in $\eqref{Eq}$?

I am thankful to all your propositions. Please I would like if it is possible some very close approximation like here, Proving that: $9.9998\lt \frac{\pi^9}{e^8}\lt 10$? or here How to prove that: $19.999<e^\pi-\pi<20$?

Answer 1

With a similar approach to the other question,

1. $\phi^2=\phi+1 = [2;1,1,1,\ldots]$ has simple convergents;
2. $e^2=[7; 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1,\ldots]$ has simple convergents too;
3. $\pi^2$ is related to the solution of the Basel problem through creative telescoping: $$\pi^2 = \sum_{n\geq 1}\frac{18}{n^2 \binom{2n}{n}} $$ and the last series is pretty fast-convergent.

4. It follows that $$ \phi^2=\frac{1597}{610}\pm 2\cdot 10^{-6} $$ $$ e^2 = \frac{12288}{1663}\pm 2\cdot 10^{-7}$$ $$ \pi^2 = \frac{16929464521}{1715313600}\pm 2\cdot 10^{-6}$$ hence $\phi^2+e^2$ is very close to $10$, which on its turn is not very far from $\pi^2$ ($\pi\approx\sqrt{10}$ was already known to Babylonians). According to the inverse symbolic calculator we have

   $$ \phi^2+ e^2 \approx \pi^2+\delta^2 $$ with $$ \delta = G\left[\zeta\left(\frac{1}{2}\right)+\Gamma\left(\frac{11}{12}\right)\right] $$ and $G$ being Catalan's constant.

Answer 2

$\delta = \sqrt{e^2 + \phi^2 - \pi^2}$ is presumably a transcendental number (though AFAIK there's no proof that it isn't rational). A good rational approximation of it is $33/89$. I don't know if you'd call that a "well known constant".