It begins with the fundamental identity
which looks like the Pythagorean identity if you divide by and rearrange terms: .
That is, given a Pythagorean triple (where ), if we define by the condition , then we will also necessarily have . Conversely, if and are both rational for , then we can get a Pythagorean triple by scaling .
But, how do we find such where both and are rational numbers?
We exploit the fact that is a rational number if and only if and are rational numbers too. This fact follows from a few hyperbolic function identities:
- (aka “half-angle formula”)
The first identity shows that is a rational number if and are, and the other identities, combined with the fundamental identity (disguised in the forms and ) show the converse.
This means that every Pythagorean triple arises by scaling , where is a positive rational number. Suppose , where a and b are positive integers. (Note that a < b, because .) Using the identities above, we find that
And this quickly leads us to rediscovering the Pythagorean triple parametrization , where, to be primitive, we require that a and b be relatively prime integers not both odd.
If you’re wanting to brush up on your trig identities, find the analogous derivation that uses sine, cosine, and tangent.