Timeline for When are infinitely many points in the orbit of a polynomial integers?
Current License: CC BY-SA 2.5
9 events
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| Oct 28, 2010 at 16:50 | history | edited | Nick S | CC BY-SA 2.5 |
deleted 27 characters in body
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| Oct 28, 2010 at 5:09 | comment | added | Gerry Myerson | @David, since the problem is about iterates at 0, your counterexample does not apply. Easily fixed; take $P(x)=(x^2+4)/2$. | |
| Oct 28, 2010 at 5:06 | comment | added | Gerry Myerson | I don't see how Lagrange applies to this problem. Suppose $P^{(n)}(0)$ is an integer if and only if $n$ is a power of 2. Then there's no guarantee that there are $m+1$ integer, or even rational, values of $x$ such that $P(x)$ is an integer. To get rational coefficients, you need both argument and value to be rational. | |
| Oct 28, 2010 at 4:19 | history | edited | Tom Church | CC BY-SA 2.5 |
corrected tex so it compiles
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| Oct 28, 2010 at 3:14 | comment | added | David E Speyer | For a different kind of counterexample, consider $P(x) = (x^2-1)/2+1$. Starting with any odd integer $>1$ gives an infinite orbit of odd integers, but $P$ does not take integers to integers. | |
| Oct 28, 2010 at 3:08 | comment | added | Yuval Filmus | @Nick For the question about rationals, you can try $P(x) = \sqrt{2}x + \sqrt{2}-1$. Then $P(P(x)) = 2x+1$, and so the orbit will contain $2^n-1$ for all $n$. Your method works only if there is an $n$ such that infinitely many times $P^{(t+n)}(0),P^{(t)}(0)$ are both rationals. Conceivably, this can fail to happen. | |
| Oct 28, 2010 at 3:03 | vote | accept | Yuval Filmus | ||
| Oct 28, 2010 at 2:34 | history | edited | Nick S | CC BY-SA 2.5 |
added 79 characters in body; added 12 characters in body; added 221 characters in body
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| Oct 28, 2010 at 2:28 | history | answered | Nick S | CC BY-SA 2.5 |