Let $N$ be an odd integer $\geq 3$.
I am interested in finding a lower bound for $$\min \vert{\cos(\frac{2\pi k}{N})}+\cos(\frac{2\pi l}{N})-\cos(\frac{2\pi m}{N})-\cos(\frac{2\pi n}{N})\vert,$$ where $k,l,m,n \in \{0,\dots, N-1\}$ and $\{k,l\}\not\subset \{m,n,N-m, N-n\}$ (without a comparable assumption the above quantity is trivially zero).
Numerically, one sees (at least for reasonably large $N\sim 100$) that this quantity is strictly greater than zero, and that it decreases with increasing $N$ (although not in a monotone fashion). Does anybody have an idea of how one could go about lower bounding this quantity? I am also interested in finding an argument for why it is strictly greater than zero.
Thanks!
You could think about it geometrically: $$\cos\big(\tfrac{2\pi k}{N}\big)+\cos\big(\tfrac{2\pi l}{N}\big)-\cos\big(\tfrac{2\pi m}{N}\big)-\cos\big(\tfrac{2\pi n}{N}\big)=\Re\big(e^{\frac{2\pi i k}{N}}+e^{\frac{2\pi i l}{N}}-e^{\frac{2\pi i m}{N}}-e^{\frac{2\pi i n}{N}}\big)$$ so you are trying to choose four vertices of a regular $N$-gon inscribed in the unit circle and minimise the sum of two $x$-coordinates minus the sum of two others (the absolute value is irrelevant as you can always swap your choices of $\{k, l\}$ and $\{m, n\}$). When $N$ is odd, as you require, say $N=2r+1$, you can choose, for example $k=0$, $l=r-1$, $m=2$, $n=r-2$. Then a lower bound is $$1+\cos\big(\tfrac{2(r-1)\pi}{2r+1}\big)-\cos\big(\tfrac{4\pi}{2r+1}\big)-\cos\big(\tfrac{2(r-2)\pi}{2r+1}\big).$$ Using trig identities, this comes to $$2\sin^2\big(\tfrac{2\pi}{2r+1}\big)-2\sin\big(\tfrac{\pi}{2r+1}\big)\sin\big(\tfrac{4\pi}{2r+1}\big).$$ As $N \to \infty$, using the first two terms of the Maclaurin series for $\sin$, this is approximately $\frac{12\pi^4}{N^4}$.
I think this is the lower bound, but I don't (yet) have a proof of this, though this does seem to be the combination of $\{k, l, m, n\}$ where the quadratic Maclaurin terms vanish and the quartic ones have minimum coefficient, so gives the minimum asymptotically at least.