Let $x,y,u,v$ be positive integers with $x,y$ coprime and $u,v$ coprime
( $xy,uv$ not necessarily coprime). Assume $x+y \ne u+v$.


How small the radical of $xy(x+y)uv(u+v)$ can be infinitely often?


Can we get $O(|(x+y)(u+v)|^{1-C})$ for $C>0$?


These are just two pairs of good $abc$ triples so we can get $C=0$
with pairwise coprimality.



How small the radical of $xy(x+y)uv(u+v)$ can be infinitely often?
Here is a solution where the radical is $O(k^9)$ and $(x+y)(u+v)=O(k^{12})$


The idea is that $x,y,z=a^2,b^2,c^2$ for a Pythagorean triple and $u,v,u+v=A^2,B^2,C^2$ for another with $C=c^2.$ I used the most familiar type of triple (hypotenuse and long leg differ by $1$), there might be others that do better, or special values of $k.$


* $x=(2k+1)^2$
* $y=\left(2k(k+1)\right)^2$
* $u=(y^2-x^2)^2=((2k^2-1)(2k^2+4k+1))^2$
* $v=(2xy)^2=(4k(k+1)(2k+1))^2$


then


* $x+y=(2k^2+2k+1)^2$
* $u+v=(2k^2+2k+1)^4$


Thus $(x+y)(u+v)=O(k^{12})$


But the radical of $xy(x+y)uv(u+v)$ is at most


$$k(k+1)(2k+1)(2k^2-1)(2k^2+2k+1)(2k^2+4k+1)=O(k^9)$$


