Timeline for Three consecutive quadratic residues problem
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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Oct 14, 2012 at 15:47 | answer | added | Franz Lemmermeyer | timeline score: 7 | |
Jan 17, 2012 at 11:46 | vote | accept | Seblis | ||
Jan 17, 2012 at 10:28 | comment | added | Franz Lemmermeyer | The Lehmer's worked on problems like these; searching for "runs of residues" in Math Scinet should produce relevant literature. Also, the Vorlesungen über Zahlentheorie by Hasse contain a chapter where the connection between such problems and elliptic curves over finite fields is worked out in detail. | |
Jan 17, 2012 at 7:39 | comment | added | Asaf | It might be worth mentioning that the theorem is indeed true for infinitely many primes, by a result of Chowla (and other improvements) that says that there are infinitely many primes with the property that the first non-quad. residue is about omega(log(p)). | |
Jan 17, 2012 at 7:19 | answer | added | Noam D. Elkies | timeline score: 20 | |
Jan 17, 2012 at 6:06 | comment | added | Noam D. Elkies | Um, make that $N \geq 10$: either $5$, $9$, or $10$ works, because $5 \cdot 8 \cdot 10 = 20^2$ so one of $5$, $8$, $10$ is a quadratic residue and can be paired with $4$, $9$, or $9$ respectively. | |
Jan 17, 2012 at 4:04 | comment | added | Noam D. Elkies | FWIW once $N\geq25$ the sequence $3,4,5,\ldots,N$ does contain for each $p$ some $n$ such that $n$ and $n-1$ are both squares mod $p$; indeed one of $4$, $9$, and $25$ works, because each is a square, as is one of $3$, $8$, and $24 = 3 \cdot 8$. It's not obvious to me whether one should expect such a construction to work also for $n,n-1,n-2$. | |
Jan 16, 2012 at 20:43 | comment | added | Will Jagy | It is unfortunate that you phrase this as though it is a homework question. Also that you have not given enough detail on dependencies, with no idea why we should regard this as part of a research project. That is, why do you think this is true, what work have you put into this so far? | |
Jan 16, 2012 at 20:25 | history | asked | Seblis | CC BY-SA 3.0 |