Timeline for Regarding the digit expansion of $\sqrt 7$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 10 at 12:48 | comment | added | Timothy Chow | @AlekseiKulikov See Integer points of an elliptic curve. | |
| Aug 10 at 7:56 | comment | added | Aleksei Kulikov | @so-calledfriendDon Yes indeed, you are right! I was just not aware that computing integral points on elliptic curves is easier. | |
| Aug 9 at 22:22 | comment | added | so-called friend Don | Maybe I miss something, but don't you only need finitely many integral points on the elliptic curve $x^2-7y^3=2$? That follows from a theorem of Siegel. Note that if $x^2-7y^3=2$, then $(7x)^2 - (7y)^3 = 98$, so we get a point on the Mordell curve $U^2 = V^3 + 98$. Data of of Bennett and Ghadermarzi at personal.math.ubc.ca/~bennett/BeGa-data.html shows that the only integer points $(U,V)$ there are $(\pm 21, 7)$. Tracing this back corresponds to the "trivial" solution $3^2-2^1 = 7^1$. For theoretical justification see personal.math.ubc.ca/~bennett/BeGh-LMSJCM-2015.pdf | |
| Aug 9 at 20:15 | comment | added | Stanley Yao Xiao | @AlekseiKulikov you are right that we do not have a general effective Faltings' theorem in the height aspect. However, for genus 2 and hyperelliptic curves in general there are algorithms that determine the set of rational points. Most of these methods are a clever application of the Chabauty-Coleman method. | |
| Aug 9 at 19:53 | comment | added | Aleksei Kulikov | @StanleyYaoXiao Faltings's theorem is what I had in mind, except I did not know how to compute the genus. As for " It's probably not hard to determine the set of rational points; this is within the purview of computational arithmetic geometry. " -- it is one thing to say such things and the other is to carry out the computation, in my opinion; as far as I know in general Faltings's theorem is ineffective currently. | |
| Aug 9 at 16:13 | comment | added | Stanley Yao Xiao | The curve defined by the equation $x^2 - 7y^6 = 2$ is hyperelliptic of genus $2$, which means that it has finitely many rational points by Faltings' theorem. It's probably not hard to determine the set of rational points; this is within the purview of computational arithmetic geometry. | |
| Aug 9 at 15:19 | history | edited | Aleksei Kulikov | CC BY-SA 4.0 |
added 1166 characters in body
|
| Aug 9 at 14:42 | comment | added | user534817 | Thank you very much. In fact, the value $𝑛$ is not allowed. Sorry for this tipo. The interval should be $[n+1,2n]$. | |
| Aug 9 at 12:06 | history | answered | Aleksei Kulikov | CC BY-SA 4.0 |