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.
- Is this correct?
- 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.
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.