Thread: Conjecture to establish if a number is not prime

1. Conjecture to establish if a number is not prime

If the quotient of a/b has a period of "c" figures and (b-1)/c results to be a non-integer number, then "b" is NOT a prime number. "a" and "b" are two integers, primes between them.

For instance, from 2/57 results the quotient 0,035087719298245614 which has a period of 18 figures and because the result of (57-1)/18 is 3,1 which is a non-integer, then 57 is not prime.

I have to mention that this conjecture doesn't determine if “b” is prime but if “b” is not prime, since it does not succeed in finding all numbers that are not prime.
The non-prime numbers that the conjecture does not find on an interval between 3 and 5000 (from which all multiples of 2 and 5 have been excluded) are: 9 - 33 - 91 - 99 - 259 - 451 - 481 - 561 - 657 - 703 - 909 - 1233 - 1729 - 2409 - 2821 - 2981 - 3333 - 3367 - 4141 - 4187 - 4521 - …

The Carmichael numbers are a subset of the whole non-prime numbers which the conjecture can not identify.
I would appreciate any suggestions or demonstrations regarding this conjecture. Thank you all in advance.

2. Let us show that if p is prime with (p;10)=1 then $\frac{1}{p}$ has a period $\delta(p)$ such that $\delta(p)|(p-1)$ (1)

Indeed by Fermat's Little Theorem $10^{p-1}-1\equiv{0}(\bmod.p)$ thus there is a positive integer $s$ such that $s\cdot{p}=10^{p-1}-1$

Then $\frac{1}{p}=\frac{s}{10^{p-1}-1}$

Let $\epsilon(n)$ be the number of digits of n.

Then if $\epsilon(s)-(p-1)=k$ we have that $\frac{10^{k}}{p}=\frac{q}{10^{p-1}-1}$ where $\epsilon(q)=p-1$ (2)

Now remember that if we have: $\frac{a}{99...9 }$ where in the denominator we have t nines and $\epsilon(a)=t$ it can be expressed with a decimal expansion that has a period length of t digits, but this doesn't mean there are no digits repeated, it only means that the minimun period divides t

Thus back in (2) we have that $\frac{10^{k}}{p}=\frac{q}{10^{p-1}-1}$ may be expressed with a period of p-1 digits thus the period of $\frac{1}{p}$ divides $p-1$

This means that (1) is a necessary condition.

The proof for the case $\frac{a}{p}$ where p is prime (p;10)=1, (p;a)=1 has little difference