# 1 plus Power of 2 is not Perfect Power except 9

Jump to navigation
Jump to search

## Theorem

The only solution to:

- $1 + 2^n = a^b$

is:

- $\tuple {n, a, b} = \tuple {3, 3, 2}$

for positive integers $n, a, b$ with $b > 1$.

This is a special case of Catalan's Conjecture.

## Proof

It suffices to prove the result for prime values of $b$.

For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power.

For $n > 0$, $1 + 2^n$ is odd.

Hence for the equation to hold $a$ must be odd as well.

Writing $a = 2 m + 1$ we have:

\(\displaystyle 1 + 2^n\) | \(=\) | \(\displaystyle \paren {2 m + 1}^b\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i \paren 1^{b - i}\) | Binomial Theorem | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i\) | Binomial Coefficient with Zero | ||||||||||

\(\displaystyle 2^n\) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 m \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}\) | $m \ne 0$ |

Since all factors of $2^n$ are powers of $2$:

- $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a power of $2$.

But since each summand is non-negative:

- $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$

we must have $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is even.

We have:

\(\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}\) | \(=\) | \(\displaystyle \binom b 1 + \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b + 2 m \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 2}\) | |||||||||||

\(\displaystyle \) | \(\equiv\) | \(\displaystyle b\) | \(\displaystyle \pmod 2\) |

Therefore we must have $b = 2$, the only even prime.

In that case:

\(\displaystyle 2^n\) | \(=\) | \(\displaystyle 2 m \sum_{i \mathop = 1}^2 \binom 2 i \paren {2 m}^{i - 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 2 m \paren {2 + 2 m}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 4 m \paren {m + 1}\) |

So $m$ and $m + 1$ are powers of $2$.

The only $m$ satisfying this is $1$, giving the solutions:

- $a = 2 m + 1 = 3$
- $2^n = 3^2 - 1 = 8 \implies n = 3$

$\blacksquare$