# Five Ramanujan-Nagell Numbers

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## Theorem

There exist exactly $5$ Ramanujan-Nagell numbers: positive integers of the form $2^m - 1$ which are triangular:

- $0, 1, 3, 15, 4095$

This sequence is A076046 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

## Proof

Consider the numbers of the form $2^m - 1$ which are triangular:

\(\displaystyle 2^m - 1\) | \(=\) | \(\displaystyle \frac {r \left({r + 1}\right)} 2\) | Closed Form for Triangular Numbers | ||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle 8 \left({2^m - 1}\right)\) | \(=\) | \(\displaystyle 4 r \left({r + 1}\right)\) | ||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle 2^{m + 3} - 8\) | \(=\) | \(\displaystyle 4 r^2 + 4 r\) | ||||||||||

\(\text {(1)}: \quad\) | \(\displaystyle \iff \ \ \) | \(\displaystyle 2^{m + 3} - 7\) | \(=\) | \(\displaystyle 4 r^2 + 4 r + 1\) | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({2 r + 1}\right)^2\) |

Let:

- $n = m - 3$
- $x = 2 r + 1$

and it can be seen that $(1)$ is equivalent to:

- $x^2 + 7 = 2^n$

From Solutions of Ramanujan-Nagell Equation:

- $x = 1, 3, 5, 11, 181$

Setting $r = \dfrac {x - 1} 2$ it is seen that the corresponding triangular numbers are:

- $\dfrac{\left({x - 1}\right) \left({x + 1}\right)} 8$

Thus the corresponding Ramanujan-Nagell numbers are:

- $0, 1, 3, 15, 4095$

$\blacksquare$