Triangular Fermat Number

Theorem
The only one Fermat number which is triangular is $3$.

Proof
Let $F_n$ be a Fermat number which is triangular.

We use the weaker form of Divisor of Fermat Number:
 * Every divisor of $F_n$ is of the form $k \times 2^{n + 1} + 1$

as this formulation is valid for all $n \in \N$.

From Closed Form for Triangular Numbers:
 * $F_n = \dfrac {m \paren {m + 1} } 2$

Either $m$ or $m + 1$ is even.

Suppose $m$ is even.

Then:

As the is always odd, we must have $2^n = 1$.

This forces $n = 0$, with $F_n = 3$.

We check that $3$ is indeed a triangular number.

Suppose $m + 1$ is even.

Then:

As $k \times 2^{n + 2} + 3 > 1$ and is odd, $k \times 2^{n + 2} + 3$ cannot be a power of $2$.

So $m + 1$ cannot be even, and we have exhausted all possibilities.