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The conjecture about Power free-values of polynomials is: Let $F(X)$ be a polynomial with integer coefficients and no repeated roots. For any $\epsilon > 0$, there exists a constant $C_{\epsilon,F}$ such that for any integer $n$ $$ |n|^{\deg{(F)}-1-\epsilon} \le C_{\epsilon,F} \operatorname{rad}(F(n))$$

The conjecture implies this polynomial version. For $f(x) , g(x) \in \mathbb{Z}[x]$ and $f(x)$ squarefree,

$$\deg (\operatorname{rad}(f(g(x)))) > \deg(g(x)) (\deg(f(x))-1) \qquad (1)$$

The bound is tight because for Chebyshev polynomials $T_n,U_n$, $T_n(x)^2 - 1 = (x^2-1) U_{n-1}^2(x)$ with $f(x)=x^2-1$.

Is (1) proved for polynomials?

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The very web page you link to tells you that this conjecture would follow from abc, and abc is known over polynomials (where it is called the Mason-Stothers theorem.) –  JSE Dec 11 '12 at 21:20
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Haven't seen that anywhere, but it's easy to prove: Write $f(g(X))=\prod(X-\beta_i)^{e_i}$ for distinct complex $\beta_i$, and $e_i\ge1$. Taking the derivative gives $f'(g(X))g'(X)=h(X)\prod(X-\beta_i)^{e_i-1}$ for some polynomial $h(X)$. Now $f(X)$ and $f'(X)$ are relatively prime, hence so are $f(g(X))$ and $f'(g(X))$. Thus $\prod(X-\beta_i)^{e_i-1}$ divides $g'(X)$, implying $$\text{deg}(f(g(X))-\text{deg(rad}(f(g(X)))\le\text{deg}(g'(X))=\text{deg}(g(X))-1,$$ and the claim follows.

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"hence so (realtively prime) are $f(g(X))$ and $f'(g(X))$". Isn't $f(x)=x, g(x)=x^2$ a counterexample? $f(g(x))=x^2$ which is certainly not coprime to the derivative. There is no restriction on $g(x)$, it need not be squarefree. There is an infinite family of counterexamples to this. –  joro Dec 12 '12 at 7:24
    
I didn't say that $f(g(X))$ and its derivative are relatively prime. Rather I say and use that $f(g(X))$ and $f'(g(X))$ are relatively prime, which is trivial: Let $\alpha$ be a common complex root of $f(g(X))$ and $f'(g(X))$. Then $g(\alpha)$ is a common root of $f(X)$ and $f'(X)$, contrary to $f(X)$ and $f'(X)$ being relatively prime. (Remark: I fixed a typo in the answer regarding the derivative of $f(g(X))$ which is $f'(g(X))g'(X)$.) –  Peter Mueller Dec 12 '12 at 8:29
    
I misunderstood, sorry. btw, (1) fails for $g(x)$ constant giving 0=0. –  joro Dec 12 '12 at 8:44
    
Is there still something unclear? (I'm asking because you didn't accept the answer.) –  Peter Mueller Dec 12 '12 at 11:48
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My proof uses the fact if a polynomial $w(X)$ divides $g'(X)$, then $\text{deg}(w(X))\le\text{deg}(g'(X))$, a trivial fact which holds unless $g'(X)=0$, in which case $g(X)$ is a constant. But in this degenerate case, your inequality does not hold, so of course my argument cannot work then either. –  Peter Mueller Dec 12 '12 at 12:26
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