Suppose that $p$ is prime and $q$ is an even number divides $p-1$, such that $q<\frac{p-1}{q}$ and $u$ has order $q$ modulo $p$. Let $S$ be the subgroup of $Z^*_p$ consisting of the powers of $u$. Thus $|S|=q$ and suppose $S=\{s_i|1\leq i \leq q \}$. Write $\epsilon=e^{\frac{2\pi i}{p}}$. Let $j$ be fixed number. Can we prove that there is just one $t$, $1\leq t \leq q$, such that $\epsilon^{s_t}$ is equal to $\epsilon^{2s_i+s_j}$ for $1\leq i \leq q$?
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1$\begingroup$ Counterexample: $p=239$, $q=14$, $s_j=1$. Then $(s_i,s_t)$ can be $(-1,-1)$ or $(100,201)$ or $(141,44)$. I note that the problem can be reformulated as asking whether $\text{gcd}(x^q-1,(2x+1)^q-1)=x+1$ in $\mathbf{F}_p[x]$. $\endgroup$– Michael ZieveCommented Jun 24, 2016 at 13:32
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$\begingroup$ Can we certain $i,t$, such that $\epsilon^{s_t}=\epsilon^{2s_i+s_j}$? $\endgroup$– shamym shamimCommented Jun 27, 2016 at 11:40
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$\begingroup$ @MichaelZieve Not sure if it's relevant, but apropos $\gcd\bigl(f(x)^n-1,g(x)^n-1\bigr)$ in $\mathbb F_p[x]$, there's my article Common divisors of $a^n-1$ and $b^n-1$ over function fields, New York Journal of Math. 10 (2004), 37-43. $\endgroup$– Joe SilvermanCommented Jun 27, 2016 at 12:41
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$\begingroup$ @Joe: I think your article isn't relevant, due to the condition $q<(p-1)/q$ in the current question. $\endgroup$– Michael ZieveCommented Jun 28, 2016 at 0:12
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$\begingroup$ I cant understand what is relation between $\epsilon^{s_t}=\epsilon^{2s_i+s_j}$ and gcd$(x^q-1,(2x+1)^q-1)=x+1$ ? $\endgroup$– shamym shamimCommented Jul 7, 2016 at 14:23
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