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Let $X_n^c=\{\cos\left((4k-c)\frac{\pi}{2n}+\frac{\pi}{4}\right): k=0, 1, \dots, n-1\}$ where $c\in\{0, 1, \ldots, \lfloor\frac{n}{2}\rfloor\}$ and $n$ is any positive integer greater than 3. I want to know, for a given $n$ under what condition(s) there exists (at least) two different $c$ and $c'$ such that we have $X_n^c=X_n^{c'} $.

I have asked this question in stackexchange on 12 August, but didn't get the result yet.

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    $\begingroup$ Please provide a link to your corresponding math.SE question. -- This helps to avoid duplication of answers. $\endgroup$
    – Stefan Kohl
    Aug 25, 2015 at 15:28

1 Answer 1

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It's easy to see that $X_n^c = X_n^{c+4}$ for any $n,c$. This implies that the set of suitable $n$ contains all integers $\geq 8$, for which we can take $c=0$ and $c'=4$.

Smaller $n$ can be tested manually:

$n=4:$ none

$n=5,7:\quad c=0, c'=1$

$n=6:\quad c=0, c'=2$

So, among $n>3$ only $n=4$ does not have distinct $c,c'\in\{0,1,\dots,\lfloor n/2\rfloor\}$ with $X_n^c = X_n^{c'}$.

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