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Thread: Number Theory problem

  1. #1
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    Number Theory problem

    hello number theorists ,

    I'm beginner and I have troubles with this lemma which i belive it is true ,

    let $\displaystyle p$ a prime number greater than $\displaystyle $3$$, and let $\displaystyle P$ be the set of primes less than $\displaystyle p$, show that there exist $\displaystyle p_1 , p_2 ... ,p_n$ from $\displaystyle P$, such that $\displaystyle p$ divides $\displaystyle \prod_{i=1}^{i=n} p_i -1$

    Thanks in ADvance
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by shostakovich View Post
    hello number theorists ,

    I'm beginner and I have troubles with this lemma which i belive it is true ,

    let $\displaystyle p$ a prime number greater than $\displaystyle 3$, and let $\displaystyle P$ be the set of primes less than $\displaystyle p$, show that there exist $\displaystyle p_1 , p_2 ... ,p_n$ from $\displaystyle P$, such that $\displaystyle p$ divides $\displaystyle \prod_{i=1}^{i=n} p_i -1$

    Thanks in Advance
    What is this problem for? The context of the problem may make it easier to solve.
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  3. #3
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    well , the original problem says : let $\displaystyle A$ a set of prime numbers , such that for all finite subset $\displaystyle S$ of $\displaystyle A$ , and $\displaystyle S<> A$, we have: all prime divisors of : $\displaystyle (\prod_ {p\in S}p )-1$ are also in $\displaystyle A$, Show that $\displaystyle A$ is the set of prime numbers
    I had the idea to use induction and the lemma above
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe this will help you!
    I find similarity between the problems... Euclid's Proof of the Infinitude of Primes (c. 300 BC)
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by shostakovich View Post
    well , the original problem says : let $\displaystyle A$ a set of prime numbers , such that for all finite subset $\displaystyle S$ of $\displaystyle A$ , and $\displaystyle S<> A$, we have: all prime divisors of : $\displaystyle (\prod_ {p\in S}p )-1$ are also in $\displaystyle A$, Show that $\displaystyle A$ is the set of prime numbers
    I had the idea to use induction and the lemma above
    What does $\displaystyle S<>A $ mean?
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  6. #6
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    well it means that S is different of A, sorry I'm not good at latex

    Actually I found the solution in the internet , it is very elegant , I will share it here as soon as I can
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by shostakovich View Post
    well it means that S is different of A, sorry I'm not good at latex

    Actually I found the solution in the internet , it is very elegant , I will share it here as soon as I can
    Can't wait! This problem is stumping me for some reason.
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  8. #8
    Super Member Bacterius's Avatar
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    Scratch that I didn't see your next post. Hang on ...
    Yeah, I like this problem. I'll be thinking about it, feel free to post the solution you found
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  9. #9
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    The original conjecture sounds wrong. For instance, if you take p = 5, Then P = {3}, which implies 5 | 3 - 1.
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  10. #10
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    TO Jamix , what about $\displaystyle 2*3-1$? , anyway here is the solution

    ( I apologize , I forgot to mention that A contains at least 3 elements )

    First of all we shall prove that A is infinite , indeed suppose it is finite for sake of contradiction, consider $\displaystyle S= \prod_{p\in A}p$, let $\displaystyle a\in A $, by conditions $\displaystyle \frac{S}{a} -1$ has all prime divisors in A, but $\displaystyle \frac{S}{a}-1$ is prime with each element of $\displaystyle A$ Except $\displaystyle a$, we conclude that $\displaystyle \frac{S}{a}-1= a^k$ for some positive integer $\displaystyle k$ , that is $\displaystyle S= a^{k+1}+a$,

    now it is easy to prove that $\displaystyle 2,3,5$ are in A , in particuler $\displaystyle S= 2^{a}+2$ and $\displaystyle S= 3^{b}+3$ , for some positive integers $\displaystyle a,b$ since $\displaystyle S \geq 2*3*5= 30,$ we conclude that $\displaystyle a \geq 4$, looking $\displaystyle mod 16$ we get $\displaystyle S= 2 [16]$ that is $\displaystyle 3^b=-1 [16] $ which is is a contradiction ,

    let q a prime number , since A is infinite , there exist at least q-1 prime number which have the same residue in $\displaystyle F_q$, we can suppose that this residue isn't 0 , let $\displaystyle c_1,c_2,.......c_{q-1}$ those primes , hence by Fermat little theorem , we have $\displaystyle c_1c_2....c_{q-1}=1[q]$ and we are done
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  11. #11
    MHF Contributor chiph588@'s Avatar
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    I wonder if the original lemma you posted is false or not.
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  12. #12
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    Quote Originally Posted by shostakovich View Post
    TO Jamix , what about $\displaystyle 2*3-1$? , anyway here is the solution

    ( I apologize , I forgot to mention that A contains at least 3 elements )

    First of all we shall prove that A is infinite , indeed suppose it is finite for sake of contradiction, consider $\displaystyle S= \prod_{p\in A}p$, let $\displaystyle a\in A $,
    I thought $\displaystyle S$ was suppose to be a finite subset of $\displaystyle A$ which contains only primes?

    In any case, if you define $\displaystyle S$ as you did above, couldn't you just conclude that $\displaystyle A$ must have infinitely many primes based on the fact that $\displaystyle S - 1$ doesn't share any primes in common with $\displaystyle A$?
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  13. #13
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    Quote Originally Posted by jamix View Post
    I thought $\displaystyle S$ was suppose to be a finite subset of $\displaystyle A$ which contains only primes?
    2 is a prime indeed.
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  14. #14
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    I made a mistake, but that doesn't have anything to do with my last post.
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  15. #15
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    Quote Originally Posted by jamix View Post
    I thought $\displaystyle S$ was suppose to be a finite subset of $\displaystyle A$ which contains only primes?

    In any case, if you define $\displaystyle S$ as you did above, couldn't you just conclude that $\displaystyle A$ must have infinitely many primes based on the fact that $\displaystyle S - 1$ doesn't share any primes in common with $\displaystyle A$?
    WEll read carefully the conditions , S is different of A

    to chiph588@ : I'm not sure , may be some one good with programming can verify big value with computer !
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