
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.(Headbang)
2)
if $\displaystyle (a,b)=1$ and $\displaystyle a b=c^n$, show that a and b are nth powers

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 primepower factorizations of r, s, and t be
$\displaystyle r = p_1^{a_1}p_2^{a_2}...p_u^{a_u}$
$\displaystyle s = p_{u+1}^{a_{u+1}}p_{u+2}^{a_{u+2}}...p_v^{a_v}$
and
$\displaystyle t = q_1^{b_1}q_2^{b_2}...q_k^{b_k}$.
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 primepowers 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.