1 plus Power of 2 is not Perfect Power except 9

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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:

\(\ds a\) \(=\) \(\ds 2 m 1\)
\(\ds \) \(=\) \(\ds 3\)
\(\ds 2^n\) \(=\) \(\ds 3^2 - 1\)
\(\ds \) \(=\) \(\ds 8\)
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 3\)

$\blacksquare$


Also see

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