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Browning in this paper proves that if $f \in \mathbb{Z}[x]$ is an irreducible polynomial of degree $d \geq 3$ and $k$ is an integer $\geq 3d/4 + 1/4$, then we have $$\\#\{n \in \mathbb{Z} \cap [1, X] : f(n) \text{ is $k$-free}\}\sim C_{1}(k, f)X$$ as $X \rightarrow \infty$ where $$C_{1}(k, f) = \prod_{p}\left(1 - \frac{\varrho(p^{k})}{p^{k}}\right)$$ and $\varrho_{f}(n) = \\#\{a \pmod{n}: f(a) \equiv 0 \pmod{n}\}$. Is there anywhere that gives a similar asymptotic for the case when $f$ is reducible?

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The same asymptotic is conjectured to hold for arbitrary $f$, reducible or not. Following an idea of Granville, it is possible to prove that it follows from the ABC conjecture. (See section 12.2 of "Heights in diophantine geometry" by Bombieri and Gubler). However, even the existence of infinitely many square-free values of $n^4 + 1$ is not known unconditionally. On the other hand, using an elementary sieve argument, it is possible to prove the conjecture for the case when $f$ splits into factors of degree at most 2. (And, I believe, even of degree $\leq 3$, although this is more difficult.) –  Vesselin Dimitrov Feb 2 '13 at 14:40
    
@lzk712 tried to fix the tex by escaping backslashes. –  joro Feb 2 '13 at 15:36
    
What is $d$? The degree of $f$? –  Gerry Myerson Feb 2 '13 at 23:01
    
@Gerry Myerson: Yes $d$ is the degree of $f$. Thanks!; @joro: Thanks for the formatting fix.; @Vesselin Dimitrov: Do you know where one can find the case when $f$ splits into factors of degree at most 2 worked out? Everyone seems to just say by "sieve methods". –  lzk712 Feb 3 '13 at 1:28
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Take a look at the posting of Ravi B on this link: artofproblemsolving.com/Forum/… . For k-frees, the argument works assuming $f$ splits into factors of degree at most k. As for the Browning result you quote, my guess is that the proof would likewise work assuming $f$ splits into irreducibles whose degrees satisfy the mentioned inequality. –  Vesselin Dimitrov Feb 3 '13 at 1:52
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