On duplicate representations as $2^x + 3^y$ for nonnegative integers $x$ and $y$

TL;DR

This paper proves that only five positive integers can be expressed as the sum of a nonnegative power of 2 and a nonnegative power of 3 in more than one way, confirming a conjecture from the OEIS.

Contribution

The paper establishes the exact count of integers with multiple representations as sums of powers of 2 and 3, including the case with zero exponents, using elementary methods.

Findings

Exactly five positive integers have multiple representations.

The case with positive exponents is known from Bennett's theorem.

Elementary methods suffice to prove the case with zero exponents.

Abstract

We prove a conjecture posted in the Online Encyclopedia of Integer Sequences, namely that there are exactly five positive integers that can be written in more than one way as the sum of a nonnegative power of 2 and a nonnegative power of 3. The case for both powers being positive follows from a theorem of Bennett. We use elementary methods to prove the case where zero exponents are allowed.

Equations

$$5$$

$$2^x+3^y=2^a+3^b,$$

$$2^x-3^b=2^a-3^y.$$

$$-1=2^3-3^2=2-3,5=2^5-3^3=2^3-3,13=2^8-3^5=2^4-3.$$

$$11=2^3+3=2+3^2,35=2^5+3=2^3+3^3,259=2^8+3=2^4+3^5.$$

$$-1=2^3-3^2=2-3.$$

$$2^3+3=2+3^2=11.$$

$$5=2^5-3^3=2^3-3,$$

$$2^5+3=2^3+3^3=35.$$

$$13=2^8-3^5=2^4-3$$

$$2^8+3=2^4+3^5=259.$$

$$v_2(3^n-1)=\{\begin{array}{ll}1,& \text{if 2∤n;}\\ 2+v_2(n),& \text{if 2∣n.}\end{array}$$

$$v_3(2^n-1)=\{\begin{array}{ll}0,& \text{if 2∤n;}\\ 1+v_3(n),& \text{if 2∣n.}\end{array}$$

$$5=2^2+1=2+3,17=2^4+1=2^3+3^2.$$

$$2^x+1=2^a+3^b,$$

$$2^a(2^s-1)=3^b-1.$$

$$2^x-3^b=1.$$

$$2^2+1=2+3=5.$$

$$2^s-1=\frac{3^c-1}{2}\cdot \frac{3^c+1}{4}.$$

$$z=\frac{3^c+1}{4},$$

$$2^s-1=2z^2-z.$$

$$(4z-1{)}^2=2^{s+3}-7.$$

$$(2^{t+2}-4z+1)(2^{t+2}+4z-1)=7.$$

$$2^{t+2}-4z+1=1,2^{t+2}+4z-1=7;$$

$$2^{t+2}=4z=4.$$

$$2^4+1=2^3+3^2=17.$$

Full text

1 Introduction

In the Online Encyclopedia of Integer Sequences (OEIS, Sloane [3]), sequence A004050 comprises the integers of the form $2^{x}+3^{y}$ for nonegative integers $x$ and $y$. On this sequence’s entry in the OEIS, it was remarked as a conjecture in September 2012 that only five of these integers can be so expressed in two different ways.

In fact, sequence A085634 lists those very integers representable both as $2^{x}+3^{y}$ and $2^{a}+3^{b}$, with $x$, $y$, $a$, and $b$ nonnegative integers and $x>a$. The five elements listed are

[TABLE]

On the entry in the OEIS for sequence A085634, it was remarked in February 2005 that if $n$ is in the sequence and $n>259$, then $n>10^{4000}$. In this note, we render this lower bound vacuously true by proving the conjecture: indeed, the five numbers listed above are the only elements of A085634.

We thus assume that

[TABLE]

where $x$, $y$, $a$, and $b$ are nonnegative integers, such that (without loss of generality) $x>a$ (whence $y<b$).

Equivalently,

[TABLE]

This brings us to sequence A207079 in the OEIS, which is described in its entry as “the only nonunique differences between powers of 3 and 2.” It is given as a finite sequence of five elements, namely 1, 5, 7, 13, and 23. It is commented that the finiteness of this sequence is due to Bennett [1], who, in fact, proved a more general result, from which the finiteness of A207079 follows directly. In his article, he gives a clear, precise history of the general problem of determining the number of solutions to the exponential Diophantine equation $|a^{x}-b^{y}|=c$, and we learn that the finiteness of the specific sequence A207079 was first proved in 1982. We state here, as a lemma, the special case of Bennett’s result that applies most directly to (2).

Lemma 1**.**

(Bennett)* There are precisely three integers of the form $2^{x}-3^{b}$, with $x$ and $b$ natural numbers, that are also expressible as $2^{a}-3^{y}$, with $a$ and $y$ natural numbers such that $x>a$. They are*

[TABLE]

These are, respectively, the only two such representations for these three integers. All other integers have either a unique such representation, or none at all.

We apply Bennett’s result to the cases of (1) and (2) where $x$, $y$, $a$, and $b$ are all positive integers. This leaves us with the special case when $y=0$; clearly (1) and (2) are impossible if $a=0$. We prove the special case $y=0$ by elementary methods, except for the one instance where we apply Lemma 1 to deduce that 1 has only the single representation $1=2^{2}-3$ (although it is not difficult to prove this fact independently).

2 The case when $\boldsymbol{y>0}$

Theorem 2**.**

There are precisely three solutions to (1) when $y>0$. They are

[TABLE]

Proof.

Let $c=2^{a}-3^{y}$ in (2). By Lemma 1, if $c\notin\{-1,5,13\}$, then $x=a$, which contradicts the hypothesis $x>a$. Otherwise, $c\in\{-1,5,13\}$.

Suppose $c=-1$. By Lemma 1, we have the two representations, as in (2),

[TABLE]

Thus, $x=3$, $b=2$, $a=1$, and $y=1$. This produces

[TABLE]

Suppose $c=5$. Similarly,

[TABLE]

thus producing

[TABLE]

Suppose $c=13$. Similarly,

[TABLE]

produces

[TABLE]

∎

3 The case when $\boldsymbol{y=0}$

For a prime $p$ and a natural number $n$, we write $p\|n$ if $p\mid n$ but $p^{2}\nmid n$. We denote the $p$-valuation of $n$ by $v_{p}(n)$: i.e., $v_{p}(n)=k$ if $p^{k}\|n$.

Lemma 3**.**

If $n$ is a natural number then

[TABLE]

Lemma 4**.**

If $n$ is a natural number then

[TABLE]

Lemmata 3 and 4 follow easily from Theorems 94 and 95, Nagell [2].

Theorem 5**.**

There are precisely two solutions to (1) when $y=0$. They are

[TABLE]

Proof.

We are given

[TABLE]

where $x$, $a$, and $b$ are natural numbers, where $x>a$. Let $s=x-a$. Thus,

[TABLE]

It is necessary by Lemma 4 that $s$ is odd, as, by (4), $3\nmid 2^{s}-1$.

First, suppose $b$ is odd. Then Lemma 3 implies $2\|3^{b}-1$, hence, by (4), $a=1$. Thus, by (3),

[TABLE]

Thus, by Lemma 1, $x=2$ and $b=1$. This produces the equation

[TABLE]

It remains to let $b$ be even. Then $a=2+v_{2}(b)$ by Lemma 3. Suppose $2^{2}\mid b$. Then $3^{4}-1\mid 3^{b}-1$, hence $5\mid 3^{b}-1$. Then (4) implies $5\mid 2^{s}-1$, hence $4\mid s$, a contradiction as $s$ is odd. Therefore $2\|b$ and $a=3$. Writing $b=2c$ for an odd natural number $c$, we have by (4)

[TABLE]

Letting

[TABLE]

then $z$ is a natural number by Lemma 3, and we obtain the quadratic in $z$,

[TABLE]

Completing the square yields

[TABLE]

Writing $s=2t+1$ yields the difference of squares factorization

[TABLE]

Therefore

[TABLE]

thus,

[TABLE]

Therefore $t=0$, $z=1$; thus, $c=1$. Hence $s=1$ and $b=2$. Recalling $a=3$, we have $x=4$. This produces the equation

[TABLE]

∎