Can you prove this formula for computing prime numbers is correct?

Question

Sometime ago I discovered the following function for computing primes:

$$ Q(x)=\text{frac} \left (\cfrac{\Gamma(x)}{x} \right )\cfrac{x^2}{x-1}= \begin{cases} x & \small \text{if $x$ is prime} \\ 0 & \small \text{otherwise} \end{cases} $$

where "$\text{frac}$" is fractional part and $\Gamma$ is Gamma function. Is there a way to prove that formula is correct?

Regards

As for the rest of the inputs think about it this way. For $n\in\mathbb{N}$ we have $\Gamma(n) = (n-1)!$. If $n$ is composite try to prove that $\frac{\Gamma(n)}{n}\in\mathbb{N}$ so that the fractional part is $0$. Then tackle the prime case. — fretty, Sep 09 '14 at 17:57
Then why write it using the Gamma function instead of a factorial? — Tobias Kildetoft, Sep 09 '14 at 18:12

score 2 · Answer 1 · answered Sep 09 '14 at 18:24

Wilson's theorem: $\Gamma(n) = (n-1)! \equiv -1 \mod n$ iff $n$ is prime. So if $n$ is prime, $\text{frac}(\Gamma(n)/n = (n-1)/n$.

Suppose $n$ is composite. For each prime $p$ dividing $n$, if $p^j$ is the highest power of $p$ dividing $n$, then $p^j$ also divides $\Gamma(n)$, and so $\Gamma(n)$ is divisible by $n$, with the sole exception $n = 4 = 2^2$. Thus if $n \ne 4$, $\text{frac}(\Gamma(n)/n) = 0$.

Can you prove this formula for computing prime numbers is correct?

1 Answers1