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Wilson's theorem states $n \in \mathbb N$ is prime iff $(n-1)! \equiv -1\pmod n$. The $\Gamma$-function extends the usual factorial to complex numbers.

What are the complex numbers such that $\Gamma(z)+1 = nz$ , $n \in \mathbb Z$?

Eisenstein or Gaussian primes don't necessarily satisfy the requirement, take for example $2+\omega$ and $5+12i$ respectively.

What I've tried:

Let $z=a+ib$. From the definition of the $\Gamma$-function, we have

$\Gamma(z)=(z-1)\Gamma(z-1)$.

$\Gamma(a+ib)=(a-1+ib)\Gamma(a-1+ib)$

$=(a-1+ib)(a-2+ib)\Gamma(a-2+ib)$

$=(a-1+ib)(a-2+ib)\cdots(a-k+ib)\Gamma(a-k+ib)$

$=\Gamma(ib)\prod_{k=0}^{a-1}{(k+ib)}$

Now, turning to the imaginary-$\Gamma$, a brick wall I ran into...

$$\Gamma(ib)= \int_0^{\infty}\frac{t^{-1+ib}}{e^t}\mathrm{d}t$$

... and cannot evaluate.

Questions

  1. How do we evaluate $\Gamma(a+ib)$?
  2. How should we go about solving for $z$ once 1. is done?

Computational 'evidence'

Wolfram|Alpha thinks these $z$ exist, infact they seem plentiful. I'm not sure if approximation is muddling the results, but I doubt it.

I'm using solve Gamma(a+ib) + 1= n(a+ib) and plugging in values of $a,b,n$.

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    I assume you mean complex numbers which are not natural numbers?2012-08-26
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    If you want to generalize Wilson's theorem to rings of integers in number fields this is not the way to go about doing it. For a Gaussian integer $a + bi$, for example, it is much more natural to look at the product of the nonzero elements in $\mathbb{Z}[i]/(a + bi)$.2012-08-26
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    @Qiaochu Yuan My motive was not Wilson related so much as exploring the Gamma function, being new to it and all. Perhaps 'primality' is an unsuitable term? To Alex, I'll try to be careful =P $a+bi$ where $b$ is non-zero.2012-08-26
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    Your $z$ almost certainly are plentiful - but they're (likely to be) all transcendental. I don't even believe an explicit formula for $\Gamma(i)$ is known (Alpha has it as $-0.1549498\ldots+0.4980\ldots i$), much less $\Gamma(a+bi)$ for even integral $a,b$.2015-12-15
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    (Once you use the gamma function and complex arguments, alpha has no constraints about the integrality of your parameters, so it treats your $a$ and $b$ as arbitrary real numbers rather than integers. Under those circumstances it's no surprise that there are numerous solutions, just as there's surely a real solution (but not an integral one) to the equation $\Gamma(x)+1=4x$.2015-12-15

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