Phew!

So why is this useful? Well, going back to my first claim, for any two positive integers, m and n, m^2 + n^2 is the hypotenuse of a pythagorean triple.

Wait a second. F(n) and F(n+1) are both integers. So, by my lemma, F(n)^2 + F(n+1)^2 is the hypotenuse of a pythagorean triple!

I’ve just proved that F(n)^2 + F(n+1)^2 = F(2n+1), so we can plug anything in for n and get F(2n+1) as this value. 2n+1 represents all odd numbers, so it would seem that this holds for all odd numbers greater than 0.

There is one small wrinkle: In our lemma, we stipulated3 that m ≠ n. This means that we can’t plug in 1 for n, because F(1) = F(2).

Plugging in n = 2 has no such problems. F(2) = 1 and F(3) = 2, and indeed 1^2 + 2^2 = 5 = F(5).

Because the Fibonacci sequence is strictly increasing after F(2), no successive terms will be equal, so our fact holds: for all odd n, n ≥ 5, F(n) is the hypotenuse of a pythagorean triple:4