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Math Help - Conjecture to establish if a number is not prime

  1. #1
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    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.
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  2. #2
    Super Member PaulRS's Avatar
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    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
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