# Fibonacci Pythagorean Triples

**OMGOMGOMGOMGOMGOMGOM...it’s him......Matt Parker..........sjd;lfkjasdlfjsfas ^{1}**

**breathe.....in....out....in.....out**

**/fangirl**

**I haven't heard from him since, so I'll just assume that he has accepted my clearly superior argument, and moved on with life. **

**Instead of gloating over my victory, I'm going to share an interesting fact about the Fibonacci sequence that I think we can all appreciate. The fact that this property doesn’t apply to the Lucas numbers is entirely coincidental. **

**Here we are:**

**For all odd integers n, n ≥ 5, F(n) is the hypotenuse of a pythagorean triple.**

**Think about that for a second. Every other number in a sequence generated by breeding rabbits results in the hypotenuse of an integer-sided right triangle.
Now, you can't just take my word for it, this is a math blog after all. We'll need a proof:**

**To start off, I'm going to make an intermediate claim (or lemma): for any two positive integers, m and n, (m>n), the triple (m^2 - n^2, 2mn, m^2 + n^2) is a pythagorean triple.**

**This is easy enough to verify. Just plug it into the pythagorean theorem (below). If you're interested in how this result comes about, watch 3blue1brown's video on the topic.**

**Depending on your disposition towards symbol shuffling, the next part requires either a pleasing or an arduous amount of working out.**

**Keeping in mind the fibonacci identity ^{2} F(2n) = F(n+1)F(n) + F(n)F(n-1), we’ll prove that F(n)^2 + F(n+1)^2 = F(2n + 1) using strong induction.**

**Basis step: n = 1**

**Now assume that this holds for n = 1, 2, 3,…., k, and prove it holds for k + 1. To save space, note that by expanding F(n), then squaring it, you get F(n)^2 = F(n-2)^2 + 2F(n-1)F(n-2) + F(n-1)^2. **

**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 stipulated ^{3} 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}**

**And just like that, Pythagoras appears out of nowhere. **

^{1}if you zoom in, you can see he actually is scribbling about my stuff

^{2}For the thoroughly pedantic and the pedantically thorough, here is a link to a proof of this identity.

^{3}This should make sense, as the resulting "right" triangle would have one side with length 0

^{4}There’s actually another pattern here that I’ll leave you to untangle