misread the bound

https://en.wikipedia.org/wiki/Abc_conjecture

The problem statement is wrong (it requires $a$ and $b$ to be both strictly greater than $1$ !)

$(2^{5^n},3^{5^n})$

My idea is to set $rad(a)$ and $rad(b)$ to be some constants and just see what happens. In particular, let $a=2^x$ and $b=3^y$. Then we are asked: $$\frac{a+b}{rad(a+b)}>6*2024^{2024}$$Let's just make sure that $a+b \vdots p^k$, where $p^{k-1}>6*2024^{2024}$. It is sufficient. Note that if $2^x+3^y \vdots p$, then by LTE Lemma $2^{xp^{k-1}}+3^{yp^{k-1}} \vdots p^k$. Thus we only need to find a prime $p$ and integers $x,y$ such that $2^x+3^y \vdots p$. As an example, $x=y=1$ and $p=5$ work.