Consecutive Integers which are Powers of 2 or 3

Theorem
The only pairs of consecutive positive integers which are powers of $2$ or $3$ are:
 * $\tuple {1, 2}$, $\tuple {2, 3}$, $\tuple {3, 4}$, $\tuple {8, 9}$

Proof
Let $a$ and $b$ be two arbitrary consecutive positive integers.

Both Integers Powers of $2$
The only powers of $2$ that differ by $1$ are $2^0$ and $2^1$, which gives us the pair $\tuple {1, 2}$.

Both Integers Powers of $3$
The powers of $3$ are:
 * $1, 3, 9, \ldots$

and so trivially there are no two consecutive positive integers which are powers of $3$.

The Remaining Case
There is no need to consider the case where $a$ or $b$ is $1$, as that has already been investigated.

So, let $a = 2^m$ and $b = 3^n$ where both $m$ and $n$ are greater than $0$.

From Powers of 3 Modulo 8, $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$.

For $m \ge 3$ we have that:
 * $2^m \equiv 0 \pmod 8$

while:
 * $2^1 \equiv 2 \pmod 8$
 * $2^2 \equiv 4 \pmod 8$

Let $a = b + 1$.

Then we have the cases:
 * $b = 1, a = 2$

which has already been covered, and:
 * $b = 3, a = 4$

which gives us the pairs $\tuple {3, 4}$.

Otherwise, for $a \ge 8$, we have:

and we have demonstrated that there are no such $b$, as $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$

Let $b = a + 1$.

Let $b = 3^n$ where $n$ is odd.

Then $b \equiv 3 \pmod 8$

which means:
 * $b \equiv 2 \pmod 8$

leading to the pair $\tuple {2, 3}$.

No further pairs are possible for $b = 3^n$ with $n$ odd.

Let $b = 3^n$ where $n$ is even.

Thus:

The only such $p$ and $q$ are $2^p = 2$ and $2^q = 4$

This means that the only possible pair fulfilling this condition is $\tuple {8, 9}$.

Thus all such possible pairs have been found.