# 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:

\(\ds a^n\) | \(=\) | \(\ds 2^m - 1\) | ||||||||||||

\(\ds \) | \(\equiv\) | \(\ds -1\) | \(\ds \pmod 4\) | as $m > 1$ |

$a$ is immediately seen to be odd.

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

Now:

\(\ds 2^m\) | \(=\) | \(\ds a^n + 1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {a + 1} \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a^{n - k - 1}\) | 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.

$\blacksquare$