Consider positive coprime integers $A$ and $B$ with $A+B=C$. The triple $(A,B,C)$ is called an ABC-triple if the radical of the product $ABC$ is smaller then $C$. The radical of a positive integer $n$, denoted $\operatorname{rad}(n)$, is the product of the distinct prime factors of $n$. For instance $\operatorname{rad}(20)=2.5=10$. The smallest ABC-triple is $(1,8,9)$ as $\operatorname{rad}(1.8.9)=6<9$.
There is a database with all ABC-triples with $C<10^{18}$. I noticed that the product $A*B*C$ is unique, but can we prove it?
If the hypothesis is true, we can use it to split the Dirichlet series
$$\sum_{n=1}^{\infty}\frac{\operatorname{rad}(n)^t}{n^s}=\prod_{p,prime}{(1+\frac{p^t}{p^s-1})}$$
into two disjunct sets with $n=A*B*C$ and $n \ne A*B*C$.
Note that the ABC-triples $(128,3645,3773)$ and $(648,3125,3773)$ have the same radical as $\operatorname{rad}(ABC)=2.3.5.7.11$. Hence a split of the summand using $\operatorname{rad}(ABC)$ does not work.
I can only conclude the following. Assume that we would have $A_1*B_1*C_1=A_2*B_2*C_2$ for two different ABC-triples, then $A_1 \ne A_2$, $B_1 \ne B_2$ and $C_1 \ne C_2$.
Reasoning. Because if $C_1=C_2$ then we can assume without loss of generality that $A_1 < A_2$. This implies $A_2B_2>A_1B_1$ hence $A_2B_2C_2>A_1B_1C_1$ which is which is contrary to the assumption. We conclude that $C_1 \ne C_2$.
Now suppose $A_1=A_2$. We can assume without loss of generality that $C_1 < C_2$ which implies that $B_1=C_1-A_1 < C_2-A_2=B_2$. This leads to $A_1*B_1<A_2*B_2$ and $A_1*B_1*C_1<A_2*B_2*C_2$ which is also contrary to the assumption. Hence $A_1 \ne A_2$. By symmetry in $A$ and $B$ we conclude $B_1 \ne B_2$.
