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How do I prove that $(2mn, m^2 - n^2, m^2 + n^2)$ is true for $m>n>0$?

Since $m^2 + n^2$ is the hypotenuse, I applied the Pythagoren theorem: $(2mn)^2 + (m^2 - n^2)^2 = (m^2 +n^2)^2$ and simplified it so that I would get $(m^2 + n^2)^2 = (m^2 + n^2)^2$ but I wasn't able to prove anything. How do I continue from here?

Thanks!

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    For the record: I very much prefer that people I managed to prod into coming up with the answer(s) on their own write up what they did as an answer. :)2011-10-08

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Realizing that I made a silly mistake, I reevaluated the equation and:

Given the Pythagorean Triple $(2mn, m^2-n^2,m^2+n^2)$ and $m>n>0$, we can apply the Pythagorean theorem to see if they are equal:

$(2mn)^2+(m^2-n^2)^2 = (m^2+n^2)^2$

$4m^2n^2 + (m^4-2m^2n^2+n^4) = (m^2+n^2)^2$

$m^4+2m^2n^2+n^4=(m^2+n^2)^2$

$(m^2+n^2)^2 = (m^2 + n^2)^2$

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HINT $\ $ By difference of squares $\rm \ (m^2+n^2)^2 - (m^2-n^2)^2 =\ (2\:m^2)\ (2\:n^2)\: =\: (2\:m\:n)^2 $

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    That was cool. Thanks!2011-10-09