most recent 30 from http://mathoverflow.net 2013-05-20T05:21:46Z http://mathoverflow.net/feeds/question/116089 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116089/polynomial-version-of-the-conjecture-about-power-free-values-of-polynomials joro 2012-12-11T15:02:18Z 2012-12-12T08:28:48Z

<p><a href="http://www.math.unicaen.fr/~nitaj/abc.html#Consequences" rel="nofollow">The conjecture about Power free-values of polynomials</a> 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 $n$ $$ |n|^{\deg{(F)}-1-\epsilon} \le C_{\epsilon,F} \operatorname{rad}(F(n))$$</p> <p>The conjecture implies this polynomial version. For $f(x) , g(x) \in \mathbb{Z}[x]$ and $f(x)$ squarefree,</p> <p>$$\deg (\operatorname{rad}(f(g(x)))) > \deg(g(x)) (\deg(f(x))-1) \qquad (1)$$</p> <p>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 $f(x)=x^2-1$.</p> <blockquote> <p>Is (1) proved for polynomials?</p> </blockquote>

http://mathoverflow.net/questions/116089/polynomial-version-of-the-conjecture-about-power-free-values-of-polynomials/116097#116097 Peter Mueller 2012-12-11T15:34:44Z 2012-12-12T08:28:48Z

<p>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.</p>