Timeline for Power free values of reducible polynomials
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 19, 2014 at 11:28 | answer | added | Stanley Yao Xiao | timeline score: 1 | |
| Aug 19, 2014 at 11:11 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
fixed the latex
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| Feb 3, 2013 at 1:57 | comment | added | Vesselin Dimitrov | Alternatively, take a look at the proof of Thm. 12.2.15 in Bombieri-Gubler. The conjecture is shown to be equivalent to the statement that the set of $n$ for which there is a prime $p \geq n$ with $p^2 \mid f(n)$ has density 0. For $\deg{f} \leq 2$, this is obvious. When $f$ splits into such polynomials, you get a finite union of sets of density 0. | |
| Feb 3, 2013 at 1:52 | comment | added | Vesselin Dimitrov | 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. | |
| Feb 3, 2013 at 1:28 | comment | added | lzk712 | @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". | |
| Feb 3, 2013 at 1:26 | history | edited | lzk712 | CC BY-SA 3.0 |
Fixed typo
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| Feb 2, 2013 at 23:01 | comment | added | Gerry Myerson | What is $d$? The degree of $f$? | |
| Feb 2, 2013 at 15:36 | comment | added | joro | @lzk712 tried to fix the tex by escaping backslashes. | |
| Feb 2, 2013 at 15:35 | history | edited | joro | CC BY-SA 3.0 |
fixed latex
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| Feb 2, 2013 at 14:40 | comment | added | Vesselin Dimitrov | 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.) | |
| Feb 2, 2013 at 14:21 | history | asked | lzk712 | CC BY-SA 3.0 |