A golden trigonometric diophantine equation

Question

After answering this question I reflected on the identity $$\cos\frac{\pi}{5}=\phi\cos\frac{\pi}{3}$$ and thought of looking for all the quadruplets of positive integers $(a,b,c,d)$ satisfying $$\cos \frac{a}{b}\pi=\phi\cos\frac{c}{d}\pi,\tag{1}$$ where both fractions are in lowest terms and smaller than $2$. Of course $a/b=c/d=1/2,3/2$, $a/b=3c/d=3/2$ and $c/d=3a/b=3/2$ work, but let us focus on non-trivial solutions, and assume none of the cosines is $0$. At first I tried using properties of $\cos$ and $\phi$, but eventually I turned to infnite products. However, it didn't work either, but I'll go through my reasoning. Hopefully someone will be able to go further, if not come up with something else; and perhaps simpler.

Using the infinite product formula for $\cos x$ we can rewrite $(1)$ as $$\require\cancel \prod_{n=1}^\infty\left(1-\frac{a^2\cancel{\pi^2}}{b^2\cancel{\pi^2}(n-1/2)^2}\right)=\phi\prod_{n=1}^\infty\left(1-\frac{c^2\cancel{\pi^2}}{d^2\cancel{\pi^2}(n-1/2)^2}\right) \\ \prod_{n=1}^\infty\frac{b^2(n-1/2)^2-a^2}{b^2(n-1/2)^2}=\phi\prod_{n=1}^\infty\frac{d^2(n-1/2)^2-c^2}{d^2(n-1/2)^2},$$ and isolating $\phi$, $$\prod_{n=1}^\infty\frac{d^2}{b^2}\frac{b^2(n-1/2)^2-a^2}{d^2(n-1/2)^2-c^2}=\phi $$ $$\prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}=\phi \\ \frac{\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)_\infty \left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)_\infty}{\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)_\infty \left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)_\infty}=\phi \tag{2} $$ $$ \lim_{n\to\infty}\frac{\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)}{\Gamma\left(n+\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(n+\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}\frac{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)}=\phi \\ \frac{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{c}{d}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{c}{d}\right)}{\Gamma\left(\displaystyle\frac{1}{2}-\displaystyle\frac{a}{b}\right)\Gamma\left(\displaystyle\frac{1}{2}+\displaystyle\frac{a}{b}\right)}=\phi \tag{3}.$$ Note that $(x)_n$ is the Pochhammer symbol. Now, unfortunately I doubt $(3)$ is more tractable than $(1)$, therefore I looked up an infinite product to substitute for $\phi$ in $(2)$, and found it here: $(2)$ is equivalent to $$\prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}=\frac{1}{2}\prod_{n=0}^\infty\frac{100n(n+1)+25}{100n(n+1)+9} \\ \prod_{n=1}^\infty\frac{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{a^2}{b^2}}{\left(n-\displaystyle\frac{1}{2}\right)^2-\displaystyle\frac{c^2}{d^2}}\frac{100n(n+1)+9}{100n(n+1)+25}= \frac{25}{18},\tag{4}$$ which at least contains no irrational numbers.

Answer 1

Suppose you put $a/b$ and $c/d$ in terms of a common denominator as $m/n$ and $k/n$, and let $\omega = \exp(\pi i/n)$, so $$\eqalign{\cos(a\pi/b) &= \cos(m \pi/n) = (\omega^m + \omega^{-m})/2\cr \cos(c\pi/d) &= \cos(k \pi/n) = (\omega^k + \omega^{-k})/2\cr}$$ The equation $\phi^2 = \phi + 1$ then becomes $$ \left(\dfrac{\omega^m + \omega^{-m}}{\omega^k + \omega^{-k}}\right)^2 = \dfrac{\omega^m + \omega^{-m}}{\omega^k + \omega^{-k}} + 1$$ or $$(\omega^{2m} + 1)^2 \omega^{2k}- (\omega^{2m} + 1)(\omega^{2k} + 1)\omega^{m+k} - (\omega^{2k} + 1)^2 \omega^{2m} = 0$$ $$ {\omega}^{2\,k+4\,m}+{\omega}^{2\,k}-{\omega}^{3\,k+3\,m}-{\omega}^{k+ 3\,m}-{\omega}^{3\,k+m}-{\omega}^{k+m}-{\omega}^{2\,m+4\,k}-{\omega}^{ 2\,m} = 0$$ Now the minimal polynomial of $\omega$ is the cyclotomic polynomial $C_n(X)$, so this will work only if $C_n(X)$ divides $$ {X}^{2\,k+4\,m}+{X}^{2\,k}-{X}^{3\,k+3\,m}-{X}^{k+3\,m}-{X}^{3\,k+m}-{ X}^{k+m}-{X}^{2\,m+4\,k}-{X}^{2\,m} $$ That's not necessarily going to mean $\cos(m\pi/n)/\cos(k\pi/n) = \phi$: the left side could actually be $\cos(jm\pi/n)/\cos(jk\pi/n)$ for some $j$ coprime to $n$, and the right side could be either $\phi$ or $-1/\phi$. But it's a necessary condition, and we can filter out the ones we want. No surprises: searching through $1 \le m,k \le 100$ I get the following results: $$ \phi = \dfrac{\cos(\pi/3)}{\cos(2\pi/5)} =\dfrac{\cos(2\pi/3)}{\cos(3\pi/5)}$$ (together with others related by the symmetries of $\cos$)