Triangular Fibonacci Numbers

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Theorem

The only Fibonacci numbers which are also triangular are:

$0, 1, 3, 21, 55$

This sequence is A039595 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


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:

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



Sources