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Math Help - Primes and Factorial problem.

  1. #1
    Junior Member
    Joined
    Sep 2010
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    Primes and Factorial problem.

    Let k be a positive integer and let m = 4k + 3.
    a Show that the even numbers 2k +2, 2k +4, . . . , 4k, 4k +2 are congruent
    modulo
    m to the negatives of the odd numbers 2k + 1, 2k 1, . . . , 3, 1,
    and deduce that
    2^
    k+1(k+1)(k+2) (2k+1) = (congruent)

    (
    1)^k+1(2k+1)(2k1) 31 (mod m).

    [Done]


    (
    ii) Use Part (i) and the fact that 2^k(k!) is the product of the even numbers
    from 2 to 2
    k to deduce that 2^2k+1(2k +1)!=(1)^k+1(2k+1)!(mod m)


    [ Done ]

    Assume now that
    m is prime. Use Part (ii) to show that if k is odd then
    2^0.5(m-1)=2^2k+1=1 mod m

    I have no idea...

    Thanks

    Do i use my original m=4k-3 and make m-1 = 4k +2 so 0.5(m-1) = 2k+1

    now I from part ii) that 2^2k+1 = 2^0.5(m-1) = (-1)^k+1

    k = 1,3,5 ...

    hence we get (-1)^k+1 = 1 at all times... btw if your wondering what happend to (2k+1)! i just divided it by both side.. because I can.

    Thanks

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  2. #2
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