Why does $\sqrt 5$ occur in manageable situations of these scenarios? [closed]

Question

1. Banach-Mazur distance between $P_5$ and $P_3$ is $d(P_5,P_3)=1+\frac{\sqrt5}2$ where $P_n$ is regular polygon in $n$ sides https://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=7968198&tag=1.

2. Shannon zero error capacity of Pentagon is $\sqrt 5$ http://web.cs.elte.hu/~lovasz/scans/theta.pdf.

3. Lovasz Theta and regular odd sided polygon agree and are algebraic for Pentagon https://en.wikipedia.org/wiki/Lov%C3%A1sz_number (similar to 2. but this resemblance is on tightness of semi-definite programming and algebraicity).

4. $5$ is minimum sum of squares of two distinct natural numbers and also appears in Hurwitz theorem https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(number_theory) and seems related to geometry https://www.jstor.org/stable/pdf/2302799.pdf.

Does the presence of $\sqrt 5$ somehow make certain things easier by inducing spectacular constraints based on symmetries and in particular interest to me why is it difficult to prove 2. and 3. for any odd number above $5$?

Are there other scenarios where $\sqrt 5$ appeared and a seemingly hard general situation becomes tame with situation at hand?

Perhaps this is coincidence however it seemed hidden reason is plausible.

The answer so far does not address the problem.

Answer 1

The geometric reason for the ubiquity of $\sqrt 5$ in problems involving a pentagon is that it is the diagonal of a $1\times 2$ rectangle (a "half-square"). This links $\sqrt 5$ to the construction of a pentagon from its side, which may be at the origin of the first three geometric observations in the OP.

The fourth observation on Hurwitz theorem is not geometric, but it does involve the golden ratio, which is the side-to-diagonal ratio in a regular pentagon and in this way brings us back to $\sqrt 5$.

Answer 2

$\def\QQ{\mathbb{Q}}$Three of these four examples involve $5$-fold symmetry. If $\zeta_5$ is a primitive $5$-th root of unity, then $\zeta_5 + \zeta_5^{-1} = \tfrac{1 \pm \sqrt{5}}{2}$. So any computations with $5$-fold symmetry are likely to include square roots of $5$.

In the same way, if $p$ is any prime which is $1 \bmod 4$, then $\sqrt{p} \in \QQ(\zeta_p)$.

To connect this to the fact that $p = a^2+b^2$, I have to work a little harder, but I can say something. Let $p \equiv 1 \bmod 4$ be prime, and $L = \QQ(i, \zeta_p)$. The Galois group of $L/\mathbb{Q}$ is $C_2 \times C_{p-1}$, so it has a quotient $C_2 \times C_4$. (I write $C_n$ for the cyclic group of order $n$.) Let $K$ be the corresponding $C_2 \times C_4$ extension of $\QQ$.

Then $K/\QQ(i)$ is a $C_4$-extension so, by Kummer's theorem, $K = \QQ(i)(\sqrt[4]{\alpha})$ for some $\alpha \in \QQ(i)$. One can show that one can take $\alpha$ of the form $(a+bi)^3 (a-bi)$ for some $a+b i \in \QQ(i)$. Then $\sqrt{\alpha} = (a+bi) \sqrt{a^2+b^2}$ is in $K$ and thus $\sqrt{a^2+b^2}$ is in $K$. If you trace through the Galois theory, the element $\sqrt{a^2+b^2}$ is in $\QQ(\zeta_p)$. So we see that, if $\QQ(\sqrt{D})$ is the unique quadratic subfield of $\QQ(\zeta_p)$, then $D$ is of the form $a^2+b^2$. Of course, this is not at all the easiest way to show that a prime which is $1 \bmod 4$ is a sum of two squares!

Answer 3

I'm not sure whether $\sqrt5$ is anywhere special, as the same relation holds for primes $p$ with form $4k+1$.

(2) (3) Let $G$ be a Paley graph with vertices in $\mathbb F_p$. $G$ is self-complementary and vertex-transitive, so $\vartheta (G)\vartheta ({\bar {G}})=p$, and it follows that $\vartheta (G)=\sqrt p$.

$G$ has Shannon capacity at least $\sqrt p$, as $\{(x,ax)|x\in\mathbb F_p\}$ is independent in $G⊠G$ if $a$ is a quadratic nonresidue.

By combining the bounds above, it follows that the Shannon capacity of $G$ is exactly $\sqrt p$.

(4) Every such $p$ is a sum of squares of two distinct natural numbers.

Answer 4

One case where the specific appearance of $5$ as a radicand involves certain inverse trigonometric functions.

Suppose we were to seek an inverse sine for numbers having the form

$$\dfrac{1\pm\sqrt{a}}{b}$$

with $a$ and $b$ posituve and rational. Letting $\theta$ be the desired inverse sine, we render

$$b^2\sin^2\theta-2b\sin\theta+(1-a)=0.$$

We multiply this by $\cos\theta$ and use the trigonometric sum-product relations to express the products in terms of functions of multiple angles. Thus \begin{gather*} \sin\theta\cos\theta=(1/2)\sin(2\theta) \\ \sin^2\theta\cos\theta=(1/2)\sin\theta\sin(2\theta) =(1/4)[\cos\theta-\cos(3\theta)] \end{gather*} and from these results

\begin{equation} -b^2\cos(3\theta)-4b\sin(2\theta)+(4(1-a)+b^2)\cos\theta=0. \tag{**}\label{starstar} \end{equation}

If we select $b=4$ to match the coefficients of the first two terms and then $\color{blue}{a=5}$ to eliminate the third term, this collapses to \begin{equation} \cos(3\theta)=-\sin(2\theta), \end{equation}

from which $5\theta=(4n-1)\pi/2$ and we extract the first quadrant root

$$\sin^{-1}\left(\dfrac{1+\sqrt5}{4}\right)=\dfrac{3\pi}{10}$$

and the fourth quadrant root

$$\sin^{-1}\left(\dfrac{1-\sqrt5}{4}\right)=-\dfrac{\pi}{10}.$$

These are the only cases where \eqref{starstar} is sufficiently simplified to render the inverse sine of a quadratic surd with nonzero rational part as a rational multiple of $\pi$.

The hidden symmetry

The origin of the number $5$ could be related to the sum $5=4+1$. It may also be related to the geometric picture below.

The blue lines intersect the circle at points corresponding to the roots of the original equation $\sin\theta=(1\pm\sqrt{a})/b$. These roots have mirror-image symmetry, but by adding a fifth root at the appropriate location along the mirror plane, corresponding to the gold line, we upgrade that to a polygonal symmetry which is fivefold — four from the original roots plus one more to complete the symmetric arrangement. The appropriate angles must then be multiples of $\pi/5$ or $\pi/10$, whose sine and cosine values must contain specifically $\sqrt5$ if they are to be quadratic surds at all.