# Triangular Fibonacci Numbers

## Theorem

The only Fibonacci numbers which are also triangular are:

$0, 1, 3, 21, 55$

## Proof

 $\displaystyle 0$ $=$ $\displaystyle \dfrac {0 \times 1} 2$ $\displaystyle 1$ $=$ $\displaystyle \dfrac {1 \times 2} 2$ $\displaystyle 3$ $=$ $\displaystyle \dfrac {2 \times 3} 2$ $\displaystyle = 1 + 2$ $\displaystyle 21$ $=$ $\displaystyle \dfrac {6 \times 7} 2$ $\displaystyle = 8 + 13$ $\displaystyle 55$ $=$ $\displaystyle \dfrac {10 \times 11} 2$ $\displaystyle = 21 + 34$

It remains to be shown that these are the only ones.

Let $F_n$ be the $n$th Fibonacci number.

From Odd Square is Eight Triangles Plus One, $F_n$ is triangular if and only if $8 F_n + 1$ is square.

It remains to be demonstrated that $8 F_n + 1$ is square if and only if:

$n \in \set{\pm 1, 0, 2, 4, 8, 10}$

So, let $8 F_n + 1$ be square.

Then:

$n \equiv \begin{cases} \pm 1 \pmod {2^5 \times 5} & : n \text { odd} \\ 0, 2, 4, 8, 10 \pmod {2^5 \times 5^2 \times 11} & : n \text { even} \end{cases}$