Define $\mathcal P(n)=\{\{q_1\},\dots,\{q_m\}\}$ where $n=\prod_{i=1}^mq_i$ and all $q_i\in\mathbb P$. Then one can define
$\displaystyle \operatorname{rad}(n)=\prod_{p\in\bigcup\mathcal P(n)}p$
and also a variant of the radical
$\displaystyle \operatorname{radx}(n)=\prod_{p\in\bigoplus\mathcal P(n)}p,\,$ where $\oplus$ stands for symmetric difference.
$\bigoplus\mathcal P(n)=\{q_1\}\oplus\dots\oplus \{q_m\}$
Both $\operatorname{rad}$ and $\operatorname{radx}$ are multiplicative (but not completely multiplicative) and for all $n$ $\operatorname{radx}(n)|\operatorname{rad}(n)$.
Conjecture:
There exist an $\varepsilon>0$ such that there exist infinitely many $(a,b,c)\in\mathbb Z_+^3$, $a,b,c$ mutually co-prime, such that $a+b=c$ and $c>\operatorname{radx}(abc)^{1+\varepsilon}$.
That is, I want a proof that there is no corresponding abc-theorem for $\operatorname{radx}$.