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Thread: pythagorean

  1. #1
    Junior Member
    May 2008

    Post pythagorean

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

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

    if $\displaystyle (a,b)=1$ and $\displaystyle a b=c^n$, show that a and b are nth powers
    Last edited by Sally_Math; Apr 8th 2009 at 06:08 PM.
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  2. #2
    Super Member Aryth's Avatar
    Feb 2007
    This is for the case $\displaystyle rs = t^2$

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

    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

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

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


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

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

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

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

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

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

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