## 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 \(5\): \[3^2+4^2=9+16=25=5^2\]

### 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...

## 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:

**Basis:**

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!
\[\begin{align}
F_{2n+1}^2 &= (F_n^2+F_{n+1}^2)^2 \\[8pt]
&= F_n^4 + 2F_n^2F_{n+1}^2 + F_{n+1}^4 \\[8pt]
&= F_n^4 + 4F_n^2F_{n+1}^2 - 2F_n^2F_{n+1}^2 + F_{n+1}^4 \\[8pt]
&= 4F_n^2F_{n+1}^2 + (F_n^4 - 2F_n^2F_{n+1}^2 + F_{n+1}^4) \\[8pt]
&= (2F_nF_{n+1})^2 + (F_n^2 - F_{n+1}^2)^2
\end{align}\]

#### Conclusion

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 \(F_5=5\).