# 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 $b > 1$.

## 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:

 $\ds 1 + 2^n$ $=$ $\ds \paren {2 m + 1}^b$ $\ds$ $=$ $\ds \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i \paren 1^{b - i}$ Binomial Theorem $\ds$ $=$ $\ds \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i$ $\ds$ $=$ $\ds 1 + \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i$ Binomial Coefficient with Zero $\ds 2^n$ $=$ $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i$ $\ds$ $=$ $\ds 2 m \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ $m \ne 0$

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

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

But since each summand is non-negative:

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

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

We have:

 $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ $=$ $\ds \binom b 1 + \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 1}$ $\ds$ $=$ $\ds b + 2 m \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 2}$ $\ds$ $\equiv$ $\ds b$ $\ds \pmod 2$

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

In that case:

 $\ds 2^n$ $=$ $\ds \paren {2 m + 1}^2 - 1$ $\ds$ $=$ $\ds 4 m^2 + 4 m + 1 - 1$ $\ds$ $=$ $\ds 4 m \paren {m + 1}$

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

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

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

$\blacksquare$