In my physics research I came across a mathematical proposition (translated into the mathematical language from the physical problem) that I feel to be true, and would like to prove it:

Proposition: Consider the function $f(m)=\cos(\frac{2m+1}{N}\pi)$ where the integer $m$ satisfies $0\leq m<N$ and the integer $N>2$. The situation $f(m_1)-f(m_2)=f(m_2)-f(m_3)$ while $f(m_1)>f(m_2)>f(m_3)$ can happen if and only if $\frac{2m_2+1}{N}=\frac{1}{2}$ or $\frac{2m_2+1}{N}=\frac{3}{2}$.

The if part is very simple. However I have not been able to prove the only if part. Without loss of generality it should be sufficient to consider only the case where $\frac{2m_i+1}{N}\pi\leq \pi$ because the cosine function is symmetric w.r.t $\pi$. If I restrict $m_1-m_2=m_2-m_3$ then I can show the proposition is true only if $\frac{2m_2+1}{N}=\frac{1}{2}$. Without the restriction however I'm having no luck.

The proposition probably relies on the fact that all the $m$'s and $N$ are integers. One idea I have tried is to notice that if $\theta=\frac{2m+1}{N}$ then $\cos(N\theta)=-1$ and there is a formula to expand $\cos(N\theta)$ into a polynomial $p_N(\cos(\theta))$ of $\cos(\theta)$, hence $f(m_1)$, $f(m_2)$ and $f(m_3)$ are all roots of the equation $p_N(\cos(\theta))$+1=0. But I haven't been able to proceed from there.

Of course the proposition may not be true after all despite my intuition, but if then it'd be nice to find out for which $N$ it fails. It certainly is true for $N\leq 8$ since I have calculated all solutions. Either way the proposition would have real physical consequences.

In my physics research I came across a mathematical proposition (translated into the mathematical language from the physical problem) that I feel to be true, and would like to prove it:

Proposition: Consider the function $f(m)=\cos(\frac{2m+1}{N}\pi)$ where the integer $m$ satisfies $0\leq m<N$ and the integer $N>2$. The situation $f(m_1)-f(m_2)=f(m_2)-f(m_3)$ while $f(m_1)>f(m_2)>f(m_3)$ can happen if and only if $\frac{2m_2+1}{N}=\frac{1}{2}$ or $\frac{2m_2+1}{N}=\frac{3}{2}$.

The if part is very simple. However I have not been able to prove the only if part. Without loss of generality it should be sufficient to consider only the case where $\frac{2m_i+1}{N}\pi\leq \pi$ because the cosine function is symmetric w.r.t $\pi$. If I restrict $m_1-m_2=m_2-m_3$ then I can show the proposition is true only if $\frac{2m_2+1}{N}=\frac{1}{2}$. Without the restriction however I'm having no luck.

The proposition probably relies on the fact that all the $m$'s and $N$ are integers. One idea I have tried is to notice that if $\theta=\frac{2m+1}{N}$ then $\cos(N\theta)=-1$ and there is a formula to expand $\cos(N\theta)$ into a polynomial $p_N(\cos(\theta))$ of $\cos(\theta)$, hence $f(m_1)$, $f(m_2)$ and $f(m_3)$ are all roots of the equation $p_N(\cos(\theta))$+1=0. But I haven't been able to proceed from there.

Of course the proposition may not be true after all despite my intuition, but if then it'd be nice to find out for which $N$ it fails. It certainly is true for $N\leq 8$ since I have calculated all solutions. Either way the proposition would have real physical consequences.