Timeline for Orders of the conjugates of an algebraic prime number in its residue field
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2014 at 0:47 | answer | added | David E Speyer | timeline score: 1 | |
| May 8, 2014 at 18:31 | comment | added | David E Speyer | Whoa, I take it back! I just checked the first 93 primes which are $1 \bmod 8$, and $a$ is always square! (To fill in a few details above, $\left( \frac{a-bi}{a+bi} \right) = \left( \frac{2a}{a+bi} \right) = \left( \frac{2a}{p} \right)$, but $\left( \frac{2}{p} \right) \equiv 1$ since $p \equiv 1 \bmod 8$.) | |
| May 8, 2014 at 18:24 | comment | added | David E Speyer | So, to state a concrete version, if we look at primes that are $p \equiv 1 \bmod 8$ and we write $p = a^2+b^2$, we would like to know whether $\left( \frac{a}{p} \right)$ is $1$ and $-1$ equally often. (For primes that are $5 \bmod 8$, switching $a$ and $b$ switches the residue class, so we don't get a good question.) It seems hard to imagine the answer could be no, but I also see no attack. | |
| May 7, 2014 at 11:15 | answer | added | GNiklasch | timeline score: 3 | |
| S May 7, 2014 at 10:23 | history | suggested | GNiklasch | CC BY-SA 3.0 |
Notational embellishments; order of quantifiers in the "For instance…" sentence (to avoid a trivial "no")
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| May 7, 2014 at 10:17 | review | Suggested edits | |||
| S May 7, 2014 at 10:23 | |||||
| May 7, 2014 at 10:13 | history | edited | Albertas | CC BY-SA 3.0 |
Corrected spelling.
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| May 7, 2014 at 9:48 | history | asked | Albertas | CC BY-SA 3.0 |