Timeline for Does the equation $241+2^{2s+1}=m^2$ have a solution?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 28, 2015 at 21:56 | history | suggested | tomasz | CC BY-SA 3.0 |
texified title
|
| Sep 28, 2015 at 21:29 | review | Suggested edits | |||
| S Sep 28, 2015 at 21:56 | |||||
| Sep 28, 2015 at 7:28 | vote | accept | few_reps | ||
| Sep 28, 2015 at 7:27 | comment | added | few_reps | @MikeBennett : Thanks, indeed, Ridout's theorem (together with the PNT for primes in an arithmetic progression) makes the job for Q2. | |
| Sep 28, 2015 at 3:12 | comment | added | so-called friend Don | If anyone is wondering, the argument Mike Bennett refers to is worked out in math.dartmouth.edu/~carlp/2tokmmminus1v8.pdf . See Theorem 1. | |
| Sep 27, 2015 at 23:36 | comment | added | Mike Bennett | By Ridout's theorem ( a $p$-adic version of Roth's theorem), the number of integers (prime or otherwise) of the shape $|m^2-2^{2s+1}|$ up to $x$ is $\ll x^{1/2+\epsilon}$, so one gets zero density among the primes. | |
| Sep 27, 2015 at 23:09 | comment | added | Noam D. Elkies | Q2: the density is surely zero; I guess that a proof is within reach but possibly not easy. Stefan Kohl's argument (together with Dirichlet's theorem) shows that at any rate the upper density is less than $1$. | |
| Sep 27, 2015 at 22:38 | answer | added | Will Jagy | timeline score: 5 | |
| Sep 27, 2015 at 22:32 | answer | added | Stefan Kohl♦ | timeline score: 38 | |
| Sep 27, 2015 at 20:35 | history | asked | few_reps | CC BY-SA 3.0 |