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Let $N$ and $n$ be positive integers with $\mathrm{GCD}(n,N)\ne1$. I want to prove the following claim:

$\Gamma\left(\frac nN\right)$, $\pi$ and the $\Gamma\left(\frac uN\right)$ ($u\in[1,N-1]$, $\mathrm{GCD}(u,N)=1$) are algebraically dependent.

Obviously; one can assume $n\in[1,N-1]$ by reflection formula for $\Gamma$. I thought to apply Gauss multiplication formula for the $\Gamma$ function, but I did not manage to prove the claim.

Thanks in advance for any answer or help.

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  • $\begingroup$ $\Gamma$ has a pole at $0$.... $\endgroup$
    – Asaf
    Commented yesterday
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    $\begingroup$ I suppose not much is known. You may check the wiki article and references therein: en.wikipedia.org/wiki/Particular_values_of_the_gamma_function $\endgroup$
    – Pietro Majer
    Commented yesterday
  • $\begingroup$ If "the" before "$\Gamma(u/n)$" means that you want to prove algebraic dependence of 3 numbers, then the classical conjecture that all relations among $\Gamma(q)$ for rational $q$ come from functional equations suggests that it might not hold even for $N=16$, $u=1$, $n=2$. If you want to write $\Gamma(n/N)$ in terms of $\pi$ and all $\Gamma(u/N)$, $(u,N)=1$, use induction on $\Omega(N/d)$ + multiplication by $p$ at $z=(n/d)/(p(N/d))$, where $d=(n,N)$ (one can avoid induction but, I guess, with more tedious calculations) $\endgroup$
    – te4
    Commented yesterday
  • $\begingroup$ I meant "all the $\Gamma(u/N)$". What is $\Omega(N/d)$? $\endgroup$
    – joaopa
    Commented yesterday
  • $\begingroup$ Thank you for the clarification. $\Omega(n)$ is the number of prime powers $p^k$, $k\ge1$, dividing $n$, i.e. $\sum_{p\mid n}\nu_p(n)$. Sorry, I found a flaw in my argument $\endgroup$
    – te4
    Commented yesterday

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