Consecutive Integers which are Powers of 2 or 3

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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:

\(\displaystyle b + 1\) \(=\) \(\displaystyle a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b + 1\) \(\equiv\) \(\displaystyle 0\) \(\displaystyle \pmod 8\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(\equiv\) \(\displaystyle 7\) \(\displaystyle \pmod 8\)

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:

\(\displaystyle a\) \(=\) \(\displaystyle 2^m\)
\(\displaystyle \) \(=\) \(\displaystyle b - 1\)
\(\displaystyle \) \(=\) \(\displaystyle 3^n - 1\)
\(\displaystyle \) \(=\) \(\displaystyle 3^{2 k} - 1\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {3^k + 1} \paren {3^k - 1}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \paren {3^k + 1} \paren {3^k - 1}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle 2^m\)
\(\displaystyle \) \(=\) \(\displaystyle 2^p \, 2^q\) 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 $1343$.


Sources