Triangular Fermat Number

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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:

\(\ds \exists k \in \N: \, \) \(\ds \frac m 2\) \(=\) \(\ds k \times 2^{n + 1} + 1\) Divisor of Fermat Number
\(\ds \leadsto \ \ \) \(\ds m + 1\) \(=\) \(\ds 2 \paren {k \times 2^{n + 1} + 1} + 1\)
\(\ds \) \(=\) \(\ds k \times 2^{n + 2} + 3\)
\(\ds \exists j \in \N: \, \) \(\ds m + 1\) \(=\) \(\ds j \times 2^{n + 1} + 1\) Divisor of Fermat Number
\(\ds \leadsto \ \ \) \(\ds k \times 2^{n + 2} + 2\) \(=\) \(\ds j \times 2^{n + 1}\)
\(\ds \leadsto \ \ \) \(\ds k \times 2^{n + 1} + 1\) \(=\) \(\ds j \times 2^n\)

As the left hand side 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:

\(\ds \exists k \in \N: \, \) \(\ds \frac {m + 1} 2\) \(=\) \(\ds k \times 2^{n + 1} + 1\) Divisor of Fermat Number
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds 2 \paren {k \times 2^{n + 1} + 1} - 1\)
\(\ds \) \(=\) \(\ds k \times 2^{n + 2} + 1\)
\(\ds F_n\) \(=\) \(\ds \frac {m \paren {m + 1} } 2\)
\(\ds \leadsto \ \ \) \(\ds 2^{2^n} + 1\) \(=\) \(\ds \paren {k \times 2^{n + 2} + 1} \paren {k \times 2^{n + 1} + 1}\)
\(\ds \) \(=\) \(\ds k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} } + 1\)
\(\ds \leadsto \ \ \) \(\ds 2^{2^n}\) \(=\) \(\ds k^2 \times 2^{2 n + 3} + k \paren {2^{n + 2} + 2^{n + 1} }\)
\(\ds \) \(=\) \(\ds k \times 2^{n + 1} \times \paren {k \times 2^{n + 2} + 3 }\)

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.

$\blacksquare$


Sources

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