I was experimenting with products of the form
$$\prod_{k=1}^{n} \left( a + b\cos(ck) \right)$$
when I found that the expression
$$\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$$
seems to always return a perfect square. (verified numerically).
$$\left(3 + 2\cos\left(\frac{2\pi}{2}\right)\right) = 1 = F_2^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{3}\right)\right) \left(3 + 2\cos\left(\frac{4\pi}{3}\right)\right) = 4 = F_3^2$$
$$\left(3 + 2\cos\left(\frac{2\pi}{4}\right)\right)\left(3 + 2\cos\left(\frac{4\pi}{4}\right)\right) \left(3 + 2\cos\left(\frac{6\pi}{4}\right)\right) = 9 = F_4^2$$
$$\dots$$
Furthermore, it appears these squares are the squares of Fibonacci numbers. ($F_1 = F_2 = 1$, $F_n = F_{n-1} + F_{n-2})$
I've tried to prove that this expression always gives a perfect square by considering the identity (Cassini's identity)
$$F_{n}^2 = F_{n+1}F_{n-1} + (-1)^{n-1}$$
and using induction to show that the above product satisfies this relation, and since the base cases ($F_1, F_2…$) are that of the Fibonacci sequence then this would prove that the expression is the square of Fibonacci numbers. However, this approach led nowhere.
Is there a clever way to show that the above expression is related to Fibonacci numbers?
Best Answer
Changing notation slightly (my $n$ is your $n + 1,$ and my $F_n$ is your $F_{n + 1},$ if that's not too confusing a way to put it!), let $\omega$ be a primitive $n$th root of unity. Then $$ \sqrt{5}F_n = \varphi^n - \frac{(-1)^n}{\varphi^n} = \frac{(\varphi^2)^n - (-1)^n}{\varphi^n} = \prod_{k=0}^{n-1}\frac{\varphi^2 + \omega^k}{\varphi} = \prod_{k=0}^{n-1}\left(\varphi + \frac{\omega^k}{\varphi}\right), $$ therefore $$ 5F_n^2 = \prod_{k=0}^{n-1}\left(\varphi + \frac{\omega^k}{\varphi}\right)\left(\varphi + \frac{\omega^{-k}}{\varphi}\right) = \prod_{k=0}^{n-1}\left(\varphi^2 + \frac1{\varphi^2} + 2\cos\left(\frac{2\pi k}{n}\right)\right). $$ But $$ \varphi^2 = \left(\frac{\sqrt5 + 1}{2}\right)^2 = \frac{3 + \sqrt5}{2}, \quad \therefore \ \frac1{\varphi^2} = \frac{3 - \sqrt5}{2}, \quad \therefore \ \varphi^2 + \frac1{\varphi^2} = 3. $$ Cancelling out the factor for $k = 0,$ which is equal to $5,$ we get: $$ F_n^2 = \prod_{k=1}^{n-1}\left(3 + 2\cos\left(\frac{2\pi k}{n}\right)\right). $$