Timeline for Jacobi symbols for two-square sums of primes
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24 at 10:39 | answer | added | Alexey Ustinov | timeline score: 1 | |
| Mar 24 at 5:54 | comment | added | GH from MO | @GerryMyerson I added a third proof (called "Second proof" in my post) based on quartic reciprocity in Gaussian integers. | |
| Mar 23 at 23:51 | comment | added | GH from MO | @GerryMyerson I added another proof (called "First proof" in my post) that proceeds differently. This proof actually shows that if $p=A^2+B^2$ is a prime with $B\equiv 0\pmod{4}$, then every odd prime divisor of $B$ is a fourth power modulo $p$, and the $2$-power part of $B$ is also a fourth power modulo $p$. | |
| Mar 11 at 13:01 | vote | accept | Roland Bacher | ||
| Mar 11 at 1:09 | history | edited | GH from MO |
edited tags
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| Mar 10 at 22:29 | comment | added | GH from MO | @GerryMyerson Good point. Based on this result and another result of Emma Lehmer (1977), I managed to prove that $A$ and $B$ are fourth powers modulo $p$ when $p\equiv 1\pmod{16}$. | |
| Mar 10 at 10:45 | comment | added | Gerry Myerson | There are explicit formulas for $A$ and $B$. If $p=4k+1$, we can take $A$ to be the absolute least residue of $(2k)!/(2(k!)^2)\bmod p$, and $B$ the absolute least residue of $A(2k)!$. Maybe this helps. See, e.g., math.stackexchange.com/q/45155 | |
| Mar 10 at 0:43 | history | became hot network question | |||
| Mar 9 at 18:11 | answer | added | GH from MO | timeline score: 11 | |
| Mar 9 at 18:06 | comment | added | Roland Bacher | @KConrad Thanks, good point. It is hopefully better now. | |
| Mar 9 at 18:05 | history | edited | Roland Bacher | CC BY-SA 4.0 |
Rephrased the question concisely at the beginning.
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| Mar 9 at 17:30 | comment | added | KConrad | The post is long. Could you summarize the observations as its own focused paragraph: what appears to happen when $p\equiv 1 \bmod 8$, then $p\equiv 1\bmod 16$, and then say what breaks for modulus 32? That way a reader can see the main point more efficiently. | |
| Mar 9 at 16:37 | history | asked | Roland Bacher | CC BY-SA 4.0 |