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Math Help - pythagorean

  1. #1
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    Post pythagorean

    The question is:- does 12 divide the product of the pythagorean triangle?
    My answer is that yes,
    since a=2mn
    12=2mn
    6=mn
    since (3,2)=1

    I don't know if this is correct, but how can I prove this.

    2)
    if (a,b)=1 and a b=c^n, show that a and b are nth powers
    Last edited by Sally_Math; April 8th 2009 at 06:08 PM.
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  2. #2
    Super Member Aryth's Avatar
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    This is for the case rs = t^2

    If r, s, and t are positive integers such that (r,s) = 1 and rs = t^2, then there are integers m and n such that r = m^2 and s = n^2.

    If r = 1 or s = 1, then this is obviously true, so we may suppose that r > 1 and s > 1. Let the prime-power factorizations of r, s, and t be

    r = p_1^{a_1}p_2^{a_2}...p_u^{a_u}

    s = p_{u+1}^{a_{u+1}}p_{u+2}^{a_{u+2}}...p_v^{a_v}

    and

    t = q_1^{b_1}q_2^{b_2}...q_k^{b_k}.

    Because (r,s) = 1, the primes occurring in the factorizations of r and s are distinct. Because rs = t^2, we have

    p_1^{a_1}p_2^{a_2}...p_u^{a_u}p_{u+1}^{a_{u+1}}p_{  u+2}^{a_{u+2}}...p_v^{a_v} = q_1^{2b_1}q_2^{2b_2}...q_k^{2b_k}

    From the fundamental theorem of arithmetic, the prime-powers occurring on the two sides of the above equation are the same. Hence, each p_i must be equal to q_j for some j with matching exponents, so that a_i = 2b_j. Consequently, every exponent a_i is even, and therefore \frac{a_i}{2} is an integer. We see that r = m^2 and s = n^2, where m and n are the integers

    m = p_1^{a_1/2}p_2^{a_2/2}...p_u^{a_u/2}

    n = p_{u+1}^{a_{u+1}/2}p_{u+2}^{a_{u+2}/2}...p_v^{a_v/2}

    Maybe this will help make a general case.
    Last edited by Aryth; April 10th 2009 at 11:57 AM.
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