## Getting Pythagorean Triples with Hyperbolic Functions In addition to parametrizing rational points on the unit circle, the matrix method, and one-fold origami, here’s a way to get the Pythagorean triples using the hyperbolic sine, cosine, and tangent.

It begins with the fundamental identity $\cosh^2 z - \sinh^2 z = 1$,

which looks like the Pythagorean identity $a^2 + b^2 = c^2$ if you divide by $a^2$ and rearrange terms: $(c/a)^2 - (b/a)^2 = 1$.

That is, given a Pythagorean triple $(a, b, c)$ (where $a^2 + b^2 = c^2$), if we define $z > 0$ by the condition $c/a = \cosh z$, then we will also necessarily have $\sinh z = b/a$. Conversely, if $\cosh z$ and $\sinh z$ are both rational for $z > 0$, then we can get a Pythagorean triple by scaling $(1, \sinh z, \cosh z)$.

But, how do we find such $z$ where both $\cosh z$ and $\sinh z$ are rational numbers?

We exploit the fact that $\tanh z$ is a rational number if and only if $\cosh(2z)$ and $\sinh(2z)$ are rational numbers too. This fact follows from a few hyperbolic function identities:

• $\tanh z = \frac{\cosh(2z) - 1}{\sinh(2z)}$ (aka “half-angle formula”)
• $\cosh(2z) = \cosh^2 z + \sinh^2 z$
• $\sinh(2z) = 2\sinh z \cosh z = 2 \cosh^2 z \tanh z$

The first identity shows that $\tanh z$ is a rational number if $\cosh(2z)$ and $\sinh(2z)$ are, and the other identities, combined with the fundamental identity $\cosh^2 z - \sinh^2 z = 1$ (disguised in the forms $1 - \tanh^2 z = 1/\cosh^2 z$ and $1/\tanh^2 z - 1 = 1/\sinh^2 z$) show the converse.

This means that every Pythagorean triple arises by scaling $(1, \sinh(2z), \cosh(2z))$, where $\tanh z$ is a positive rational number. Suppose $\tanh z = a/b$, where a and b are positive integers. (Note that ab, because $-1 < \tanh z < 1$.) Using the identities above, we find that $\sinh(2z) = \frac{2ab}{b^2-a^2}$ $\cosh(2z) = \frac{b^2 + a^2}{b^2 - a^2}$

And this quickly leads us to rediscovering the Pythagorean triple parametrization $(b^2 - a^2, 2ab, b^2 + a^2)$, 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. 