# If $a_1,b_1,a_2,b_2\in\mathbb{N};\ (a_1,b_1)\neq (a_2,b_2),$ show that $\frac{1}{2^{a_1}}+\frac{1}{3^{b_1}}\neq\frac{1}{2^{a_2}}+\frac{1}{3^{b_2}}.$

I came up with this question whilst thinking about how to answer this question.

If $$(a_1,b_1) \neq (a_2,b_2)\$$ where $$\ a_1, b_1, a_2, b_2 \in\mathbb{N},\$$ show that $$\frac{1}{2^{a_1}} + \frac{1}{3^{b_1}} \neq \frac{1}{2^{a_2}} + \frac{1}{3^{b_2}} .$$

Attempt: Suppose $$a_i, b_i\in\mathbb{N}$$ and

$$\frac{1}{2^{a_1}} + \frac{1}{3^{b_1}} = \frac{1}{2^{a_2}} + \frac{1}{3^{b_2}}.$$

Suppose further that $$a_2 > a_1.$$ Then, $$\frac{1}{2^{a_2}} < \frac{1}{2^{a_1}},$$ so $$\frac{1}{3^{b_2}} - \frac{1}{3^{b_1}} > 0 \implies b_1 > b_2.\$$

Next, we have:

$$\frac{1}{2^{a_1}} - \frac{1}{2^{a_2}} = \frac{1}{3^{b_2}} - \frac{1}{3^{b_1}},\ \implies \frac{2^{a_2-a_1} - 1}{2^{a_2}} = \frac{3^{b_1-b_2} - 1}{3^{b_1}}$$

$$\implies 3^{b_1}\left( 2^{a_2-a_1} - 1 \right) = 2^{a_2}\left( 3^{b_1-b_2} - 1\right) \quad (*)$$

Since $$a_2 > a_1$$ and $$b_1 > b_2,$$ every term in $$(*)$$ is a positive integer. In particular, the LHS of $$(*)$$ is odd and the RHS of $$(*)$$ is even, a contradiction. The shows that $$a_2 \not> a_1.$$

If we instead suppose $$a_1>a_2$$ then we get a similar contradiction (the same, but with $$a_1 \leftarrow\rightarrow a_2$$ and $$b_1\leftarrow\rightarrow b_2.)$$ Therefore, $$a_1 = a_2,\ \implies b_1=b_2,\$$ as desired; that is, we proved the contrapositive of the proposition.

1. Is this correct?
2. Is this the standard/easiest approach to answer this question?

I also thought of starting with, "Suppose $$(a_1,b_1) \neq (a_2,b_2).$$ Then, $$2^{a_1}3^{b_1} \neq 2^{a_2}3^{b_2},\$$ by the fundamental theorem of algebra." This feels relevant, but I'm not sure where to take this.

• I did exactly the same as you after looking only at your title. Except that I wrote "wlog, $a_2>a_1$" instead of "Suppose further that". Commented Jun 14, 2023 at 12:39
• @AnneBauval thanks for your comment. Yes I did that at first too. But I wanted to make the proof more explicit so it can be more easily understood by everyone (esp. people who don't understand about WLOG). Commented Jun 14, 2023 at 12:49
• This is admittedly overkill, but are you familiar with $p$-adic valuations? The claim follows immediately from the non-archimedean triangle inequality (applied to both $2$-adic and $3$-adic valuations). Commented Jun 14, 2023 at 13:33
• @Jyrki no I'm not. Commented Jun 14, 2023 at 13:34
• Bringing up the $p$-adic valuation is overkill for the following reason. The representation of a rational number as a fraction $r/s$ in the reduced form, i.e. $\gcd(r,s)=1$, is unique up to sign. The denominators of the reduced forms of these numbers are $2^{a_i}3^{b_i}$ with $i=1,2$ respectively. There may be a special case when there are zeros among the exponents. Commented Jun 14, 2023 at 13:41

Note $$a,b,c,d$$ the given integers. It is equivalent to prove $$\frac{1}{2^a}+\frac{1}{3^b}=\frac{1}{2^c}+\frac{1}{3^d}\Rightarrow (a,b)=(c,d)$$ We have $$\frac{1}{2^a}+\frac{1}{3^b}=\frac{1}{2^c}+\frac{1}{3^d}\iff\frac{1}{2^a}-\frac{1}{2^c}=\frac{1}{3^d}-\frac{1}{3^b}$$ This equality is not possible, except when both sides are zero because if not then the $$LHS$$ is a finite decimal while the $$RHS$$ is periodic (not finite) decimal. We are done.