Date: Tuesday, December 17, 1996 7:39 PM

From: Alex Bogomolny

You can check that for any m > n the following is true:

(m^{2}-n^{2})^{2} + (2mn)^{2} = (m^{2}+n^{2})^{2}

so that the numbers (m^{2}-n^{2}), 2mn, and (m^{2}+n^{2}) always form a Pythagorean
triple.

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