1 plus Perfect Power is not Power of 2

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Theorem

The equation:

$1 + a^n = 2^m$

has no solutions in the integers for $n, m > 1$.

This is an elementary special case of Catalan's Conjecture.


Proof

Aiming for a contradiction, suppose there is a solution.

Then:

\(\displaystyle a^n\) \(=\) \(\displaystyle 2^m - 1\)
\(\displaystyle \) \(\equiv\) \(\displaystyle -1\) \(\displaystyle \pmod 4\) as $m > 1$

$a$ is immediately seen to be odd.

By Square Modulo 4, $n$ must also be odd.


Now:

\(\displaystyle 2^m\) \(=\) \(\displaystyle a^n + 1\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + 1} \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a^{n - k - 1}\) Sum of Two Odd Powers

The latter sum has $n$ powers of $a$, which sums to an odd number.

The only odd divisor of $2^m$ is $1$.

However, if the sum is $1$, we have:

$a^n + 1 = a + 1$

giving $n = 1$, contradicting our constraint $n > 1$.

Hence the result by Proof by Contradiction.

$\blacksquare$