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).

$$\left(3 + 2\cos\left(\frac{2\pi}{2}\right)\right) = 1 = F_2^2$$

$$\left(3 + 2\cos\left(\frac{2\pi}{3}\right)\right) \left(3 + 2\cos\left(\frac{4\pi}{3}\right)\right) = 4 = F_3^2$$

$$\left(3 + 2\cos\left(\frac{2\pi}{4}\right)\right)\left(3 + 2\cos\left(\frac{4\pi}{4}\right)\right) \left(3 + 2\cos\left(\frac{6\pi}{4}\right)\right) = 9 = F_4^2$$


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?

  • 2
    $\begingroup$ Well, the square part probably comes from $k$ and $n+1-k$ giving the same value. $\endgroup$ Nov 10, 2022 at 20:51
  • 4
    $\begingroup$ This is the kind of problem that sometimes can be solved with Chebyshev polynomials. $\endgroup$ Nov 10, 2022 at 20:53

3 Answers 3


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: $$ F_n^2 = \prod_{k=1}^{n-1}\left(3 + 2\cos\left(\frac{2\pi k}{n}\right)\right). $$

  • $\begingroup$ This is a really nice solution! $\endgroup$
    – egglog
    Nov 11, 2022 at 8:52
  • 1
    $\begingroup$ Thanks! It is nice, but I didn't have a flash of inspiration. I just thought that the factor in the product looked like the squared modulus of a complex number. The obvious thing to try was $a + \omega/a,$ for real $a$ and $\omega = \exp(2\pi ik/n).$ This required $a^2 + 1/a^2 = 3.$ I nearly gave up, because I'd been expecting the golden ratio to show up in an obvious way, but the equation didn't look familiar. I doggedly solved the quadratic equation for $a^2$ anyway, and duly kicked myself when it turned out that $a = \varphi.$ $\endgroup$ Nov 11, 2022 at 16:34
  • 1
    $\begingroup$ With hindsight, knowing that the roots of the equation $x^2 + 1/x^2 = 3$ are $\{\varphi,-\varphi,1/\varphi,-1/\varphi\},$ I'm kicking myself even harder for not having solved it directly by writing: $$x^4 - 3x^2 + 1 = (x^2 - 1)^2 - x^2 = (x^2 - x - 1)(x^2 + x - 1).$$ $\endgroup$ Nov 12, 2022 at 5:42

This was a blast solving. Never in a million years would I have guessed these to be true. But the $n+1$ in the denominator really hints at what to use. I will be skipping the more well known proofs.

We start with the Chebyshev polynomials of the second kind $U_n(x)$ which satisfy the following recursion relations

$$U_0(x)=1,\ U_1(x)=2x,\ U_n(x)=2xU_{n-1}(x)-U_{n-2}(x)$$

Observe that $U_n(x)$ is a $n$-degree polynomial. Easy proof. It is true for $n=0,1$. Then use induction.

Claim 1 : $$U_n(\cos\theta)=\frac{\sin(n+1)\theta}{\sin\theta}$$

Then the roots of the polynomial $U_n(x)$ are exactly where the RHS of the above vanishes, namely $\frac{k\pi}{n+1}$, and you can produce $n$ of them.

Proposition 2 : $$U_n(x)=2^n\prod_{k=1}^n\left(x-\cos\left(\frac{k\pi}{n+1}\right)\right)$$

Proof : All that's left to prove is the fact that the leading coefficient of $U_n$ is $2^n$. It is true for $n=0,1$. By induction, $U_{n-1},\ U_{n-2}$ has leading coefficients $2^{n-1},\ 2^{n-2}$ respectively and thus $U_n$ has leading coefficient $2\cdot 2^{n-1}=2^n$. The result follows. $\ \square$

Let $F_n$ denote the $n$-th Fibonacci number, $n\ge 0$.

Theorem 3 : $$F_n=i^{-n}U_n\left(\frac{i}{2}\right)$$

Proof : Induction yet again. $$1=F_0=i^{-0}U_0\left(\frac{i}{2}\right)=1$$ $$1=F_1=i^{-1}U_1\left(\frac{i}{2}\right)=i^{-1}2\left(\frac{i}{2}\right)=1$$

Let this be true for all $0\le k\le n-1$. Then for $k=n$

\begin{align*} F_n&=F_{n-1}+F_{n-2}\\&=i^{-n+1}U_{n-1}\left(\frac{i}{2}\right)+i^{-n+2}U_{n-2}\left(\frac{i}{2}\right)\\&=i^{-n+1}2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)+i^{-n+2}U_{n-2}\left(\frac{i}{2}\right)\\&=i^{-n}\left\{2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)+i^2U_{n-2}\left(\frac{i}{2}\right)\right\}\\&=i^{-n}\left\{2\left(\frac{i}{2}\right)U_{n-1}\left(\frac{i}{2}\right)-U_{n-2}\left(\frac{i}{2}\right)\right\}\\&=i^{-n}U_n\left(\frac{i}{2}\right) \end{align*} and we are done by induction. $\ \square$

Main Result : $$F_n^2=\prod_{k=1}^n\left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$$

Proof : We first look at $U_{n}^2\left(\frac{i}{2}\right)$.

We have

\begin{align*} U_n^2\left(\frac{i}{2}\right)&=2^{2n}\left[\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]^2\\&=\left[2^n\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\left[2^n\prod_{k=1}^n\left(\left(\frac{i}{2}\right)-\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\\&=\left[\prod_{k=1}^n\left(i-2\cos\left(\frac{k\pi}{n+1}\right)\right)\right]\left[\prod_{k=1}^n\left(i-2\cos\left(\frac{k\pi}{n+1}\right)\right)\right] \end{align*}

Now to calculate this, pair up the $j$-th term of the first product with the $n+1-j$-th term of the second product. This produces terms like $$\left(i-2\cos\left(\frac{j\pi}{n+1}\right)\right)\left(i-2\cos\left(\frac{(n+1-j)\pi}{n+1}\right)\right)=\left(i-2\cos\left(\frac{j\pi}{n+1}\right)\right)\left(i+2\cos\left(\frac{j\pi}{n+1}\right)\right)=-\left(1+4\cos^2\left(\frac{j\pi}{n+1}\right)\right)$$

Plugging this back, we have


Using Theorem 3, we get


Now using the standard cosine double angle identity, we get $$F_n^2=\prod_{k=1}^n\left(3+2\cos\left(\frac{2k\pi}{n+1}\right)\right)$$ and we are done. $\ \square$

Some random musings:

This seems generalizable in quite a few directions. What if we use $\sin$ instead of $\cos$? Do we still get something that has some recurrence? What if we used Chebyshev polynomials of the first kind? What kind of identities will that lead to?

More algebraically, does this suggest some deeper connection between cyclotomic extensions and quadratic extensions (Since Fibonacci numbers are related to $\mathbb Z[\sqrt 5]$ and we have a bunch of products of cosines)? [Kronecker-Weber is a step in this direction, but I am not sure how it related honestly]. Or more intriguing, is there a relation between cyclotomic extensions and Chebyshev polynomials?

I am not learned enough to even try and make a guess. Maybe people in the comment can guide me?

I just now saw the comment by Thomas Andrews mentioning Chebyshev polynomials. Interesting how we came up with the same idea. The $n+1$ denominator suggested this approach. I would be interested in learning how they came up with it though.

  • $\begingroup$ $ \prod \limits_{k=1}^{n} \left( 3 + 2{\color{red}{\sin}}\left(\frac{2\pi}{n+1}k\right) \right) $ produces this OEIS. Interesting. $\endgroup$
    – GohP.iHan
    Nov 11, 2022 at 5:19

I'll provide a general way to compute $$ \prod_{k=1}^{n} \left(a + b \cos\left(\frac{2 k\pi}{n}\right)\right) $$ that gives your identity with slight modification (notice that I changed $n+1$ to $n$), and for $a \geq b$ (in the end, I'll remove this assumption).

First, we will interpret the terms in the product as lengths from uniformly distributed points on a certain circle to the origin. Let's assume that we've found $r$ such that $a/b = (r^2 + 1) / 2r$ (this is where we need $a \geq b$. Then from the identity $(1 + r\cos\theta)^2 + (r\sin\theta)^2 = (r^2 + 1) + 2r \cos \theta$, $a + b\cos(2k\pi / n)$ is a (constant multiple of) square of the distance from the origin to $(1 + r\cos(2k\pi / n), r\sin(2k\pi /n ))$. Hence the product becomes (a constant multiple of) multiplication of the square of the distances from the origin to these points. In view of complex numbers, these points can be regarded as a zeros of the equation $(z-1)^n = r^n$, and the product of the zeros become $1 + (-r)^n$ and the norm of the square of it is $(1 + (-r)^n)^2$, which is the same as the product of square of the distances.

In case of your original problem, I think $r = (3 - \sqrt{5})/2$ works and this will give a proof of your identity, combined with the formula for $F_n$ (in terms of golden ratio). Also, this approach will leads to a formula for the above product, and the formula also should hold for $a < b$ case since it is a continuous function in $a$ and $b$.


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