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:
| \(\ds b + 1\) | \(=\) | \(\ds a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b + 1\) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod 8\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds b\) | \(\equiv\) | \(\ds 7\) | \(\ds \pmod 8\) |
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:
| \(\ds a\) | \(=\) | \(\ds 2^m\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds b - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3^n - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3^{2 k} - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {3^k + 1} \paren {3^k - 1}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {3^k + 1} \paren {3^k - 1}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2^m\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2^p \, 2^q\) | for some $p, q \in \Z_{\ge 0}$ such that $2^p = 2^q + 2$ |
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.
$\blacksquare$
Historical Note
The result Consecutive Integers which are Powers of 2 or 3 was demonstrated rigorously by Levi ben Gershon, under the name Leo Hebraeus, in his De numeris harmonicis of $1343$.