Related to [this question](https://mathoverflow.net/questions/202683/counterexample-to-proposition-of-granville-related-to-abc-conjecture).

For polynomial $f$, let $rad(f)$ denote the radical of $f$,
the product of irreducible factors.

Suppose that $G(x,y) \in \mathbb{C}[x,y]$ is homogeneous
without any repeated factors.  


Let $r,s \in \mathbb{C}[u_1,\ldots,u_n] $ be coprime
and at least one of them depends on at least two variables.
Let $f \in \mathbb{C}[u_1,\ldots,u_n],R,S \in \mathbb{C}[t]$.

A "bad" identitiy is identity of the form $r=R(f),s=S(f)$.

> Q1 Is it true that $\deg{(rad(G(r,s)))}\ge \max\{\deg(r),\deg(s)\}(\deg(G)-2) + 2$, except for "bad" identities?

--- 

`Pasten`'s nice comment about linear substitution
showed bad identities can't be excluded, here is the construction.

From the linked question, let $G(x,y)=x^4+xy^3$.

Certain univariate $r,s$ are counterexample.

The radical is divisible by $t+2$. Set $t=(u+v)^2-2$.

Explicitly:


    r=8*u^6 + 48*u^5*v + 120*u^4*v^2 + 160*u^3*v^3 + 120*u^2*v^4 + 48*u*v^5 + 8*v^6 - 48*u^4 - 192*u^3*v - 288*u^2*v^2 - 192*u*v^3 - 48*v^4 + 96*u^2 + 192*u*v + 96*v^2
    s=u^8 + 8*u^7*v + 28*u^6*v^2 + 56*u^5*v^3 + 70*u^4*v^4 + 56*u^3*v^5 + 28*u^2*v^6 + 8*u*v^7 + v^8 - 8*u^6 - 48*u^5*v - 120*u^4*v^2 - 160*u^3*v^3 - 120*u^2*v^4 - 48*u*v^5 - 8*v^6 + 24*u^4 + 96*u^3*v + 144*u^2*v^2 + 96*u*v^3 + 24*v^4 - 96*u^2 - 192*u*v - 96*v^2 + 144


The degree of the radical is $17$, which replaces $+2$ with $+1$.





I _suspect_ $abc$ implies this, so $abc$ for multivariate polynomials
might solve this.