Pythagorean Triples in the Fibonacci Sequence

Some Definitions

Pythagorean Triple

You might have heard of Pythagoras' theorem; \[a^2+b^2 = c^2\] It states that for a right angled triangle with hypotenuse length \(c\), the sum of the squares of the two shorter sides \(a\) and \(b\) equals the square of the hypotenuse. A Pythagorean triple is simply three integers which satisfy this equation, such as \(3,4\) and \(5\): \[3^2+4^2=9+16=25=5^2\]

3-4-5 Triangle

Fibonacci Sequence

The Fibonacci sequence is well known for lots of reasons. It is easy to generate, starting with \(0,1\), simply add the previous two terms to get the next: \[0,1,1,2,3,5,8,13,21,34,55,89,...\] The \(n^{th}\) Fibonacci number has the symbol \(F_n\), for example \(F_0 = 0, F_7 = 13\) and \[F_{n+1}=F_n+F_{n-1}\]

The Hypothesis

Starting with \(F_5\), every second Fibonacci number is the length of the hypotenuse of a right angled triangle, or in other words the longest length (\(c\) above) in a Pythagorean Triple.

It's worth noting that it's not super easy to find Pythagorean triples off the top of your head, so this is quite an impressive statement! Let's prove it's true...

Fibonacci 7 Triangle
Fibonacci 9 Triangle

The Proof

Firstly, we need some tools

To prove the hypothesis, we need two other mathematical statements to be true: \[ F_{2n} = F_n(F_{n+1} + F_{n-1}) \\[8pt] F_{2n+1} = F_{n+1}^2+F_n^2 \] Well we really only need the second, but they rely on each other to be proven by induction like this:
For \(n=1\), \[ F_{2n} = F_2 = 1 = 1(1+0) = F_1(F_2+F_0) \\[8pt] F_{2n+1} = F_3 = 2 = 1^2 + 1^2 = F_2^2+F_1^2 \] Induction Step
Suppose \(F_{2k} = F_k(F_{k+1} + F_{k-1})\) and \(F_{2k+1} = F_{k+1}^2+F_k^2\) are true for some \(k \geq 1\). Using this assumption for \(n=k\), and rules about how terms in the Fibonacci are related, at \(n=k+1\) we have \[\begin{align} F_{2(k+1)} &= F_{2k+2} = F_{2k+1}+F_{2k} \\[8pt] &= F_{k+1}^2+F_k^2 + F_k(F_{k+1} + F_{k-1}) \\[8pt] &= F_{k+1}(F_k+F_{k+1}) + F_k(F_k+F_{k-1}) \\[8pt] &= F_{k+1}F_{k+2} + F_kF_{k+1} \\[8pt] &= F_{k+1}(F_{k+2} + F{k}) \end{align}\] and also, \[\begin{align} F_{2(k+1)+1}&= F_{2k+3} = F_{2k+2}+F_{2k+1} \\[8pt] &= F_{k+1}(F_{k+2} + F{k}) + F_{k+1}^2+F_k^2 \\[8pt] &= F_{k+1}(F_{k+1}+2F_k) + F_{k+1}^2+F_k^2 \\[8pt] &= (F_{k+1}^2+2F_kF_{k+1} + F_k^2) + F_{k+1}^2 \\[8pt] &= (F_{k+1}+F_k)^2+F_{k+1}^2 \\[8pt] &= F_{k+2}^2+F_{k+1}^2 \end{align}\] Proven
So the statements hold for \(n=1\), and assuming they hold for \(n=k\), they hold for \(n=k+1\) and we have proven they are true by induction! Phew! Armed with these statements, we can now quite easily show that for all odd \(n\), the original hypothesis is true.

Proving the Hypothesis

If \(n\) is an integer, then \(2n+1\) is an odd integer. As we're interested in odd terms of the Fibonacci sequence greater than \(F_5\), we can say \(n\geq 2\) and look at \(F_{2n+1}\).
We know, using our new tools, that \[ F_{2n+1}^2 = (F_n^2+F_{n+1}^2)^2 \] We're aiming at something which looks like \[ F_{2n+1}^2 = a^2 + b^2 \] It turns out, we can make this work! \[\begin{align} F_{2n+1}^2 &= (F_n^2+F_{n+1}^2)^2 \\[8pt] &= F_n^4 + 2F_n^2F_{n+1}^2 + F_{n+1}^4 \\[8pt] &= F_n^4 + 4F_n^2F_{n+1}^2 - 2F_n^2F_{n+1}^2 + F_{n+1}^4 \\[8pt] &= 4F_n^2F_{n+1}^2 + (F_n^4 - 2F_n^2F_{n+1}^2 + F_{n+1}^4) \\[8pt] &= (2F_nF_{n+1})^2 + (F_n^2 - F_{n+1}^2)^2 \end{align}\]


So, for any \(n \geq 2\), if we define \[\begin{align} a &= 2F_nF_{n+1} \\[8pt] b &= F_n^2 - F_{n+1}^2 \\[8pt] c &= F_{2n+1} \end{align}\] Then \(a^2+b^2=c^2\) and we have arrived at the desired result, true for every other Fibonacci number from \(F_5=5\).