SOLUTION: show that (m^2-n^2, 2mn, m^2+n^2) is a phythagoream triple
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Question 619109: show that (m^2-n^2, 2mn, m^2+n^2) is a phythagoream triple
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Those three expressions form a Pythagorean triple if and only if  + (2mn)^2 = (m^2 + n^2)^2)
(we can show that m^2 + n^2 > 2mn by the AM-GM inequality, m^2 + n^2 > m^2 - n^2 is obvious, given that m,n > 0).
Here, just show that LHS equal the RHS.
 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4)
This is true, so the three expressions form a Pythagorean triple.
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