A weak form of the ABC conjecture states that $\operatorname{rad}(ABC)^2-C$ is always positive. Here the radical of a positive integer $n$, denoted $\operatorname{rad}(n)$, is the product of the distinct prime factors of $n$. For instance $\operatorname{rad}(20)=2.5=10$. Note that for coprime numbers $A,B$ we have $\operatorname{rad}(AB)=\operatorname{rad}(A).\operatorname{rad}(B)$.
Can we show that $\operatorname{rad}(ABC)^2-C$ is not a square?
If it would be a square, we would have $\operatorname{rad}(ABC)^2-C=t^2$ for some positive integer $t$. As the left side is divisible by $\operatorname{rad}(C)$, it follows that $t^2$, and hence $t$, is divisible by $\operatorname{rad}(C)$. Rewriting yields $\operatorname{rad}(AB)^2-\frac{C}{\operatorname{rad}(C)^2}=(\frac{t}{\operatorname{rad}(C)})^2$. This implies that $C$ is a powerful number.
I checked the statement for $C \le 1000000$ using Sage.
I tried modular arithmatic to proof the statement, but it seems not a successful approach. Modulo 24 for instance gives a simple table based on the divisibility by 2 and 3:
$$\begin{array}{|c||c|c|} \hline rad(AB)^2 & \text{uneven} & \text{even} \\ \hline \hline \text{factor 3} & 9 & 12 \\ \hline \text{no factor 3} & 1 & 4 \\ \hline \end{array}$$
Unfortunally, this split seems not very helpful.