<s>[INST] How small the radical of $xy(x+y)uv(u+v)$ can be infinitely often? 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.
 [/INST] 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
$$k(k+1)(2k+1)(2k^2-1)(2k^2+2k+1)(2k^2+4k+1)=O(k^9)$$
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