Prove the following: $$\large\sum_{n=1}^{231}\sec^{4}\left(\frac{2\pi n}{231}\right)=949167681$$
Trying to add the sum to itself doesn't do much and I cannot telescope it easily, and I got lost using trigonometric identities but the $\frac{2\pi}{N}$ component reminds me of the roots of unity, so what do I do?
First observe that the roots of $z^n - 1 = 0$ are $\cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right)$ for $k = 1, \ldots, n$.
Writing $z = x + iy$ and using the binomial theorem, we have $$\Re\ (x + iy)^n = \Re\ \sum\limits_{l=0}^n i^l{n \choose l}x^{n-l}y^l = \sum_{l\text{ even}}^n (-1)^{l/2}{n \choose l}x^{n-l}y^l$$
Therefore, substituting $x = \cos\left(\frac{2k\pi}{n}\right)$ and $y = \sin\left(\frac{2k\pi}{n}\right)$, we have $$\sum_{l\text{ even}}^n (-1)^{l/2}{n \choose l}\cos^{n-l}\left(\frac{2k\pi}{n}\right)\sin^l\left(\frac{2k\pi}{n}\right) = \Re\ \left[\cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right)\right]^n = 1$$
Using the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ and substituting $x_k = \cos\left(\frac{2k\pi}{n}\right)$ we obtain $$\sum_{l\text{ even}}^n (-1)^{l/2}{n \choose l}x_k^{n-l}\left(1 - x_k^2\right)^{l/2} - 1 = 0$$
So we see that $\cos\left(\frac{2k\pi}{n}\right)$ for $k = 1, \ldots, n$ are all the roots of the polynomial $$P(x) = \sum_{l\text{ even}}^n (-1)^{l/2}{n \choose l}x^{n-l}\left(1 - x^2\right)^{l/2} - 1$$
Now if $P(x) = a_nx^n + \ldots + a_0$ has roots $x_1, \ldots, x_n$ (possibly with multiplicity) then Newton's identities and Vieta's formulas allow us to write $$x_1^4 + \ldots + x_n^4$$ $$= e_1^4 - 4e_1^2e_2 + 4e_1e_3 + 2e_2^2 - 4e_4$$ $$= \left(-\frac{a_{n-1}}{a_n}\right)^4 - 4\left(-\frac{a_{n-1}}{a_n}\right)^2\frac{a_{n-2}}{a_n} + 4\left(-\frac{a_{n-1}}{a_n}\right)\left(-\frac{a_{n-3}}{a_n}\right) + 2\left(\frac{a_{n-2}}{a_n}\right)^2 - 4\frac{a_{n-4}}{a_n}$$ $$ = \frac{a_{n-1}^4}{a_n^4} - \frac{4a_{n-1}^2a_{n-2}}{a_n^3} + \frac{4a_{n-1}a_{n-3} + 2a_{n-2}^2}{a_n^2} - \frac{4a_{n-4}}{a_n}$$ $$ = \frac{a_{n-1}^4 - 4a_na_{n-1}^2a_{n-2} + 4a_n^2a_{n-1}a_{n-3} + 2a_n^2a_{n-2}^2 - 4a_n^3a_{n-4}}{a_n^4}$$
Then if none of the roots are zero, $x^nP(1/x) = a_0x^n + \ldots + a_n$ has roots $1/x_1, \ldots, 1/x_n$ and so we have $$\frac{1}{x_1^4} + \ldots + \frac{1}{x_n^4} = \frac{a_1^4 - 4a_0a_1^2a_2 + 4a_0^2a_1a_3 + 2a_0^2a_2^2 - 4a_0^3a_4}{a_0^4}$$
Note that when $n$ is odd, $\cos\left(\frac{2k\pi}{n}\right) \neq 0$ for any $k$, so none of the roots of $P(x)$ are zero.
We want to compute the first $5$ terms of $P(x)$.
$$P(x) = \sum_{l\text{ even}}^n (-1)^{l/2}{n \choose l}x^{n-l}\left(1 - x^2\right)^{l/2} - 1 $$ $$= \sum_{l\text{ odd}}^n (-1)^{(n-l)/2}{n \choose l}x^l\left(1 - x^2\right)^{(n-l)/2} - 1$$ $$ = (-1)^{(n-1)/2}nx\left(1 - x^2\right)^{(n-1)/2} + (-1)^{(n-3)/2}{n \choose 3}x^3\left(1 - x^2\right)^{(n-3)/2} - 1 + O(x^5)$$ $$ = (-1)^{(n-1)/2}nx\left[1 - \frac{n-1}{2}x^2 + O(x^4)\right] + (-1)^{(n-3)/2}{n \choose 3}x^3\left[1 + O(x^2)\right] - 1 + O(x^5)$$ $$ = -1 + (-1)^{(n-1)/2}nx - (-1)^{(n-1)/2}\frac{n(n-1)}{2}x^3 + (-1)^{(n-3)/2}{n \choose 3}x^3 + O(x^5)$$ $$ = -1 + (-1)^{(n-1)/2}nx + (-1)^{(n-3)/2}\left[{n \choose 3} + {n\choose 2}\right]x^3 + O(x^5)$$
This shows that $a_0 = -1$, $a_1 = (-1)^{(n-1)/2}n$, $a_2 = 0$, $a_3 = (-1)^{(n-3)/2}\left[{n \choose 3} + {n\choose 2}\right]$ and $a_4 = 0$.
Plugging in the formula above, we conclude that $$\sum_{k=1}^n\sec^4\left(\frac{2k\pi}{n}\right) = \sum_{k=1}^n\frac{1}{\cos^4\left(\frac{2k\pi}{n}\right)} = a_1^4 + 4a_1a_3 = n^4 - 4n\left[{n\choose 3} + {n\choose 2}\right]$$ which is valid when $n$ is odd.
Using Pascal's triangle identity, we can further simplify ${n\choose 3} + {n\choose 2} = {n+1\choose 3} = \frac{n(n^2-1)}{6}$, so we get $$\sum_{k=1}^n\sec^4\left(\frac{2k\pi}{n}\right) = \frac{n^2(n^2 + 2)}{3}\text{ when }n\text{ is odd.}$$
This formula gives $949167681$ for $n = 231$.
This answer will show that the summation in the question is an integer. But we consider a little more general case: replacing $231$ by any fixed odd $p\ge3$. Although this answer does not solve your concerned equality, it uses the roots of unity at the end. The main idea comes from the Post.
First, since $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$, \begin{align*} \sec\left(\frac{2n\pi}{p}\right) =\frac{2}{e^{i\frac{2n\pi}{p}}+e^{-i\frac{2n\pi}{p}}} =\frac{2e^{-i\frac{2n\pi}{p}}}{1+e^{-i\frac{4n\pi}{p}}} =\frac{2x_n}{1+x_n^2}, \end{align*} where we denote $x_n:=e^{-i\frac{2n\pi}{p}}$. It follows that \begin{align*} \sum_{n=1}^{p-1}\sec^4\left(\frac{2n\pi}{p}\right) &=16\sum_{n=1}^{p-1}\frac{x_n^4}{(1+x_n^2)^4}\\ &=16\sum_{n=1}^{p-1}\frac{x_n^4}{(1+x_n^2)^4}\cdot\frac{\left(x_n^{2(p-1)}-x_n^{2(p-2)}+\cdots-x_n^2+1\right)^4}{\left(x_n^{2(p-1)}-x_n^{2(p-2)}+\cdots-x_n^2+1\right)^4}\\ &=16\sum_{n=1}^{p-1}\frac{\left(x_n^{2p-1}-x_n^{2p-3}+\cdots-x_n^3+x_n\right)^4}{\underbrace{\left(x_n^{2p}+1\right)^4}_{=16}}\\ &=\sum_{n=1}^{p-1}\left(\sum_{j=1}^p(-1)^{j-1}x_n^{2j-1}\right)^4\\ &=\sum_{n=1}^{p-1}\sum_{j,k,l,m=1}^p(-1)^{j+k+l+m-4}x_n^{2(j+k+l+m)-4}\\ &=\sum_{j,k,l,m=1}^p(-1)^{j+k+l+m-4}\sum_{n=1}^{p-1}x_n^{2(j+k+l+m)-4}, \end{align*} where the inner sum in the last equality is \begin{align*} \sum_{n=1}^{p-1}x_n^{2(j+k+l+m)-4} =\sum_{n=1}^{p-1}e^{-i2n\pi\frac{2(j+k+l+m)-4}{p}} =\left\{ \begin{array}{ll} p-1 & p\,|\,2(j+k+l+m)-4, \\ -1 & \mbox{otherwise.} \end{array} \right. \end{align*}
