Timeline for When is the sum of two quadratic residues modulo a prime again a quadratic residue?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2023 at 18:24 | comment | added | KConrad | The count you are sure about but don't have a citation to goes back at least to Aladov in 1896. See Section 3 in kconrad.math.uconn.edu/blurbs/ugradnumthy/… and look at $N_p(1,1)$ in the table there; the notation $N_p(\varepsilon_1,\ldots,\varepsilon_r)$ is defined at the start of Section 2. | |
| May 18, 2011 at 9:42 | answer | added | GH from MO | timeline score: 3 | |
| May 17, 2011 at 21:12 | vote | accept | Julián Aguirre | ||
| May 17, 2011 at 21:09 | comment | added | Julián Aguirre | @GH Some of the solutions will have $a=b$, which I do not want to count. May be I was not clear about this. | |
| May 17, 2011 at 3:46 | comment | added | GH from MO | Note that your motivating problem can be solved directly. Let $q$ be any of the $(p+1)/2$ quadratic residues, and let's count the solutions of $ab+1=q$. For $q=1$ the equation is equivalent to $a=0$ or $b=0$, so there are $2p-1$ solutions. For $q\neq 1$ we have $a\neq 0$ and $b\neq 0$ which determine each other uniquely, so there are $p-1$ solutions. Altogether the number of (ordered) diophantine pairs equals $2p-1+(p-1)^2/2=(p^2+2p-1)/2$. | |
| May 17, 2011 at 1:05 | answer | added | arithboy | timeline score: 2 | |
| May 17, 2011 at 0:08 | answer | added | Sonia Balagopalan | timeline score: 16 | |
| May 16, 2011 at 22:05 | comment | added | Siksek | This is an easy consequence of the fact that $x^2+y^2=z^2$ is a curve of genus $0$ and so have exactly $p$ projective solutions. | |
| May 16, 2011 at 22:05 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
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| May 16, 2011 at 22:01 | answer | added | Felipe Voloch | timeline score: 5 | |
| May 16, 2011 at 21:53 | history | asked | Julián Aguirre | CC BY-SA 3.0 |