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Math Help - equation in Z/kZ (k prime)

  1. #1
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    equation in Z/kZ (k prime)

    hi everybody
    i'm interested in solving the folowing 'equation' whitout spending hours to do so!
    I know some of you have percular skill and some may be lucky as well :I feel i could spend hours into this job whithout getting through just digging the wrong place (i remember i did it (or thought i did it once) in a few minutes but i can't put the hand on it )
    so after this trying to be short and polite introduction this is the stuff
    (i must had i have found the relation i want to proove by trying a lots of rather small exemple whith 'excel')

    so k is a prime number
    p is a odd number (it could be prime) and we want to proove in Z/kz:

    (1+c^p=(c+1)^p and not c=0 and not c=-1)=> c^(p-1)=(c+1)^(p-1) =1
    or c^(p+1)=(c+1)^(p+1)=1

    thanks for your help
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  2. #2
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    Quote Originally Posted by SkyWatcher View Post

    (1+c^p=(c+1)^p and not c=0 and not c=-1)=> c^(p-1)=(c+1)^(p-1) =1
    or c^(p+1)=(c+1)^(p+1)=1

    thanks for your help
    I am not sure what you are asking. It is messy.

    Maybe the following theorem will help you.

    Theorem: If F is a field of prime charachteristic p then:
    (\alpha + \beta)^{p^n} = \alpha^{p^n} + \beta^{p^n} for all \alpha, \beta \in F.
    Last edited by ThePerfectHacker; June 7th 2007 at 09:16 AM.
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  3. #3
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    i was interested in solving the equation (c+1)^p-(c^p)-1=0 modulo k (k prime)

    so as i have an excel sheet that display me for the p(ème) column the residu modulo k of x^p i was able to see that for p odd and for k not to big (because my excel is not infinite!!) always one of the two relation mentioned above was true and i could find no exeption (except for the trivial case c=0 or c=-1)
    i tried a lot of k (not all)
    i tried to make the demonstration and i eventualy find one but unfortunatly i loosed the paper on wich it was writen (and so i can not make sure it was not just waste!) i feel i could make a lot of try before finding it
    so i'm hoping the problem is less messy (my english is lamentable) and that some of you are more aware and methodic than me

    another point is i cannot edit my post so next time i will reflect more before writing (i'm speeking about another thread)
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  4. #4
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    Quote Originally Posted by SkyWatcher View Post
    i was interested in solving the equation (c+1)^p-(c^p)-1=0 modulo k (k prime)
    (c+1)^p - c^p - 1 = pc+p(p-1)/2 \cdot c^2 + .... + pc^{p-1}

    Factor,
    pc( 1 + (p-1)/2\cdot c + ... + c^{p-1} ) \equiv 0 (\bmod k)
    So p = k is one solution.
    c\equiv 0 (\bmod k) is another.

    But I do not see how to solve:
    1 + (p-1)/2\cdot c + ... + c^{p-1} \equiv 0 (\bmod k)
    I do not even think you can find a nice formula that will solve this.
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  5. #5
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    up

    Quote Originally Posted by SkyWatcher View Post
    i was interested in prooving ((c+1)^p-(c^p)-1=0 mod k))=>((c+1)^(p+1) = c^(p+1) = 1 (mod k))or ((c+1)^(p-1) = c^(p-1) = 1 (mod k))
    quite difficult to rewrote it all almost correctly! ( i should have add : k and p even , c and c+1 prime with k , the equation is writen on Z/kZ)!

    any one have seen that or would had a suggestion ? !!!

    (by the way thanks to ThePerfectHacker for trying to anwser the latest version of this question wich was i must admit very confused!)


    Last edited by SkyWatcher; November 16th 2007 at 11:26 PM. Reason: mistake
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