Timeline for Integral polynomials dividing N!
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 22, 2019 at 14:30 | history | edited | José Hdz. Stgo. | CC BY-SA 4.0 |
edited title
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| Nov 12, 2016 at 5:51 | comment | added | Jyrki Lahtonen | See also Joni Teräväinen's answer at Math.SE. | |
| Nov 12, 2016 at 1:30 | vote | accept | S. Pek | ||
| Nov 11, 2016 at 21:15 | answer | added | Lucia | timeline score: 16 | |
| Nov 11, 2016 at 18:57 | comment | added | Joe Silverman | @LSpice Since the statement is clearly true for non-zero constant polynomials, I guess the OP was using "of finite degree" to mean "degree not equal to $-\infty$", where by convention the zero polynomial is assigned degree $-\infty$. (This is the right value if you want to define the $x$-adic absolute value to be $|f(x)|=e^{-\deg(f)}$.) But none-the-less, I think it's confusing to say "finite degree" instead of just saying that the polynomial is not the zero polynomial. | |
| Nov 11, 2016 at 16:44 | answer | added | José Hdz. Stgo. | timeline score: 10 | |
| Nov 11, 2016 at 14:45 | comment | added | Charles | @LSpice I think in that usage "finite" means $0<x<+\infty$; it makes at least some sense in the context of infinite products $\prod_ia_i=\exp\left(\sum_i\log a_i\right)$ where a factor of 0 corresponds to a summand of $-\infty$. | |
| Nov 11, 2016 at 14:06 | comment | added | LSpice | @JoeSilverman, there is a weird colloquial useage where sometimes 'finite' is used in place of 'positive' (non-mathematicians will often try to emphasise that an unlikely event is not impossible by saying it has a 'finite probability'), and that might be what was meant here. (Of course the conjecture is true by luck for constant polynomials, except $P = 0$.) | |
| Nov 11, 2016 at 13:56 | comment | added | Joe Silverman | I fixed the formatting and changed the wording a little. In the future, it would be better if you format your question using LaTeX. Two other comments. First, you said that "P(N) is a polynomial of finite degree". A polynomial, by definition, has finite degree. Second, it's best to distinguish the polynomial from its value. | |
| Nov 11, 2016 at 13:54 | history | edited | Joe Silverman | CC BY-SA 3.0 |
fixed formatting
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| Nov 11, 2016 at 13:39 | comment | added | Jason Starr | There are conjectures, e.g., by G. Martin, about the density of $N$ such that $P(N)$ is $N^{1/d}$-smooth. That might be enough to conclude that $P(N)$ divides $N!$. | |
| Nov 11, 2016 at 13:29 | review | Close votes | |||
| Nov 11, 2016 at 16:23 | |||||
| Nov 11, 2016 at 12:55 | comment | added | S. Pek | Yes I believe it might be. But could be gravely mistaken by missing the obvious. | |
| Nov 11, 2016 at 12:54 | comment | added | Jason Starr | Are you asking whether this holds for every single-variable polynomial with integer coefficients? | |
| Nov 11, 2016 at 12:41 | review | First posts | |||
| Nov 11, 2016 at 13:01 | |||||
| Nov 11, 2016 at 12:38 | history | asked | S. Pek | CC BY-SA 3.0 |