pythagorean

**Sally_Math** · April 8th 2009, 05:38 PM

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

**Aryth** · April 9th 2009, 08:01 PM

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.

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