Timeline for Why are powers of $\exp(\pi\sqrt{163})$ almost integers?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2013 at 18:27 | history | edited | Alicia Garcia-Raboso | CC BY-SA 3.0 |
MathJax
|
| Nov 9, 2009 at 23:54 | comment | added | Kevin Buzzard | In fact the root we want is the one which is about -10^{87} and within 10^{-6} of an integer. The other roots are "random" complex numbers (not near integers, or even near the reals in general) of size about 10^5. | |
| Nov 9, 2009 at 23:51 | comment | added | Kevin Buzzard | I tried this on a computer. The conclusion is that j(5*tau) is a root of an irreducible degree 6 polynomial with coefficient of the order of 10^{100}. | |
| Nov 9, 2009 at 22:30 | comment | added | David E Speyer | I don't think $j(n \tau)$ should be an integer. I think that would indicate that the order $Z[n \tau]$ has trivial class group, and I'm pretty sure that's not true. But I'm rusty on how the elliptic curve/class field dictionary works for nonmaximal orders. | |
| Nov 9, 2009 at 22:27 | comment | added | Kevin Buzzard | I think that the conclusion from this would in fact be that j(n.tau) satisfied a polynomial of low degree but with humungous coefficients. Computer calculation indicate that j(5.tau) is in fact not an integer for tau as above, although it is close to one. | |
| Nov 9, 2009 at 22:08 | history | answered | David Hansen | CC BY-SA 2.5 |