$ c^2 = a^2 + b^2 $ factors as $ (a+ib)(a-ib) $ so let $ a+ib = (p+iq)^2 = p^2-q^2 + i(2pq) $ and $ a-ib = (p-iq)^2 = p^2-q^2 - i(2pq) $ now $ c = (p+iq)(p-iq) = p^2 + q^2 $.

In conclusion, if we pick $ p $, $ q$ then ($ p^2 - q^2 $, $ 2pq $, $ p^2 + q^2 $) forms a pythagorean triple.

Number theory notes

Expository preaching, non-rigorous proofs

## Saturday, 5 June 2010

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