I was studying gcd then I encountered this sum $(1).$

A conjecture:

If $(1)=1$ for any values of $N\ge3$, then N is a prime number.


$$f(N)=\frac{1}{N^{1-s}(N-1)}\sum_{j=1}^{N}(-1)^jj^s\frac{{N \choose j}}{gcd(j,N)}=1\tag1$$

where gcd is the greatest common divisor and $s\ge1$

If this sum is correct, can we uses it to test for prime numbers or look for a largest prime number?

  • 3
    $\begingroup$ I don't think that a summation involving $N$ values for a large value of $N$ is an efficient way to test for prime numbers. $\endgroup$ – Peter Foreman Mar 27 at 9:36

This sum is lacking efficient usability. There is no largest prime, only largest known. So, no it can't be used for values that don't exist. The sum can be made to look as follows:$$\begin{eqnarray}\frac{N^{s-1}\sum_{j=1}^{N}(-1)^jj^s\frac{{N \choose j}}{gcd(j,N)}}{(N-1)}=1\tag2\\{N^{s-1}\sum_{j=1}^{N}(-1)^jj^s\frac{\frac{N!}{j!(N-j)!}}{(N-1)gcd(j,N)}}=1\tag3\\{N^{s}\sum_{j=1}^{N}(-1)^jj^s\frac{\frac{(N-1)!}{j!(N-j)!}}{(N-1)gcd(j,N)}}=1\tag4\\{N^{s}\sum_{j=1}^{N}(-1)^jj^s\frac{\frac{(N-2)!}{j!(N-j)!}}{gcd(j,N)}}=1\tag5\\{N^{s}\sum_{j=1}^{N}(-1)^jj^{s-1}\frac{\frac{(N-2)!}{(j-1)!(N-j)!}}{gcd(j,N)}}=1\tag6\\\sum_{j=1}^{N}(-1)^jj^{s-1}\frac{(N-2)!}{gcd(j,N)(j-1)!(N-j)!}=\frac{1}{N^s}\tag7\\\end{eqnarray}$$

It still isn't useful as written though.


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