A novel approach for primality test

Asked Dec 31 '24 at 14:45

Active Dec 31 '24 at 14:53

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$p$ is a prime number, we have the following sum $S(p)$:

$$S(p) = \sum_{k=1}^{p-1} \left( k^{p-1} \mod p \right) \equiv p - 1 \pmod{p}$$

Definitions

$k^{p-1} \mod p $ represents the remainder when $k^{p-1}$ is divided by $p$

$\sum_{k=1}^{p-1} \left( k^{p-1} \mod p \right)$ is the sum of these remainders for $k$ ranging from 1 to $p-1$

The conjecture states that this sum is congruent to $p - 1$ modulo $p$ for all prime numbers $p$.

This conjecture suggests that for prime numbers, the sum of these modular powers results in a value that is congruent to $p - 1$

My question is: is it this formula already known?

edited Dec 31 '24 at 14:53

Anne Bauval

asked Dec 31 '24 at 14:45

Craw Craw

1

It is an obvious corollary of the well-known https://en.wikipedia.org/wiki/Fermat%27s_little_theorem – Anne Bauval Dec 31 '24 at 14:53
Weird to include the binary mod operator when you are just writing a congruence. You can simplify this to be either: $$\sum_{k=1}^{p-1} k^{p-1}\equiv p-1\pmod p$$ or $$\sum_{k=1}^{p-1} (k^{p-1}\bmod p)=p-1$$ – Thomas Andrews Dec 31 '24 at 15:03
See also Giuga's Conjecture. – Gerry Myerson Dec 31 '24 at 16:58
As mentioned here (and many other places), this is an open problem known as Giuga;s conjecture, Please delete this duplicate question. – Bill Dubuque Dec 31 '24 at 18:10

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