We know that Wilson's theorem states the following :

$x$ is a prime if $(\frac {\Gamma(x)+1}{x})$ is an integer .

We can extend this to Twin primes as :

$x$ and $x+2$ is prime if $(\frac {4(\Gamma(x)+1)+x}{x(x+2)})$ is an integer.

Now my question :

what is an equivalent result to the tuples of primes ?


for given $k_0$

${\mathcal H} = (h_1, \ldots ,h_{k_0})$

and $n+ {\mathcal H } $ consists entirely of primes.

What is equivalence in terms of analog of Wilson's theorem to State the above condition ( I know it's relatively easy ) ?

(For twin primes ${\mathcal H} = (0,2)$)

Can we first prove the infinity of primes by proving the statement there are infinitely many integers of the form $(\frac {\Gamma(x)+1}{x})$ if x is an integer ; and then proceed in similar direction for tuples ? -- actually this is my main question .

( Forgive my English and type of typing I'm new here )

  • 2
    $\begingroup$ I think you have to pay attention to false positives such as $2=4×1/2$, so maybe the naive approach won't be enough $\endgroup$ Sep 3, 2019 at 18:29
  • $\begingroup$ So maybe raising each "Wilson gamma factor" to a positive power could help rule them out. $\endgroup$ Sep 3, 2019 at 18:30
  • 1
    $\begingroup$ You want to prove there's an infinity of twin primes, based on there being an infinity of integers of the form $(\Gamma(x)+1)/x$? Good luck! $\endgroup$ Sep 3, 2019 at 22:44
  • 1
    $\begingroup$ @Gerry Myerson No ,sir that's not my main concern ( that's why I added " in some sense" i.e. equivalent type) . I'm just curious is it possible or not with some additional Analysis ? $\endgroup$
    – TPC
    Sep 3, 2019 at 22:59
  • 1
    $\begingroup$ @Gerry Myerson you commented in support on that MSE post of mine . $\endgroup$
    – TPC
    Sep 3, 2019 at 23:10

1 Answer 1


I assume that all $h_i$ are even. Then $(n+h_1,n+h_2,\dots,n+h_k)$ is a prime tuple iff $$\begin{cases} h_1!m \equiv -1\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -1\pmod{n+h_k}, \end{cases} $$ where $m=(n-1)!$. The system implies $$\begin{cases} h_k!m \equiv -\frac{h_k!}{h_1!}\pmod{n+h_1},\\ \dots\\ h_k!m \equiv -\frac{h_k!}{h_k!}\pmod{n+h_k}, \end{cases} $$ which further combines into $$h_k!m \equiv - \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}\pmod{(n+h_1)\cdots (n+h_k)}.$$ That is, $(n+h_1)\cdots (n+h_k)$ divides the numerator of $$h_k!(n-1)! + \sum_{i=1}^k \frac{h_k!}{h_i!}\prod_{j=1\atop j\ne i}^k \frac{n+h_j}{h_j-h_i}.$$

Example. For twin primes $(n,n+2)$, we have $k=2$ with $h_1=0$ and $h_2=2$. Then the last expression becomes $$2(n-1)!+2\frac{n+2}2 + \frac{n}{-2} = \frac{4(n-1)!+n+4}{2},$$ and thus we want $n(n+2)\mid (4(n-1)!+n+4)$.

  • $\begingroup$ Thank you very much , sir . $\endgroup$
    – TPC
    Sep 4, 2019 at 6:14
  • $\begingroup$ What about the main question ? $\endgroup$
    – TPC
    Sep 4, 2019 at 6:16
  • 1
    $\begingroup$ Typo is corrected. I doubt this approach can help to prove infinitude of prime twins or tuples. $\endgroup$ Sep 4, 2019 at 11:13
  • $\begingroup$ @ Max Alekseyev The doubt is mutual . But I don't have very strong reason to firmly believe it . $\endgroup$
    – TPC
    Sep 4, 2019 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.