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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?


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$(m^2+n^2)^2=(2mn)^2+(m^2-n^2)^2$ –  pedja Oct 8 '11 at 7:13
In short: you've neglected $m^2-n^2$ in squaring. –  J. M. Oct 8 '11 at 7:26
@Juan, I've edited your answer at two places, so that you can see where the exponent 2 was missing. If you have a look now, you'll see that you in fact already proved it. –  Martin Sleziak Oct 8 '11 at 7:37
@JuandelaJohn You could consider writing your calculation as an answer and accepting that. If the other users who helped you with their hints don't want to expand the hints to a full answer, it is very much ok for you to do so yourself. That way the community can check out and give you feedback on the presence of any lingering unclear steps (if any) in your argument. You also get piece of mind (and some rep). Also the automated system is relieved of the duty to periodically 'bump' your question to the front page as 'unanswered'. That kind of bumping reduces our enjoyment of the site :-) –  Jyrki Lahtonen Oct 8 '11 at 9:44
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. :) –  J. M. Oct 8 '11 at 15:29

2 Answers 2

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:

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

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


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

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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! –  Juan de la John Oct 9 '11 at 2:34

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