# Natural Deductions

Expository preaching, non-rigorous proofs

## Saturday, 5 June 2010

### Pythagorean Triples

$c^2 = a^2 + b^2$ 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