Getting Pythagorean Triples with Hyperbolic Functions

Right triangle with hyperbolic function side lengths.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.


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