Timeline for Distribution of moduli of quadratic residues
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| S Jan 12, 2014 at 21:03 | history | suggested | tc1729 | CC BY-SA 3.0 |
Added LaTeX
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| Jan 12, 2014 at 20:59 | review | Suggested edits | |||
| S Jan 12, 2014 at 21:03 | |||||
| Jul 15, 2013 at 21:41 | vote | accept | Lenny Fukshansky | ||
| Jul 15, 2013 at 14:19 | answer | added | David E Speyer | timeline score: 8 | |
| Jul 12, 2013 at 14:13 | comment | added | Emil Jeřábek | @FelipeVoloch: The OP asked for it to be proportional to the length of $I$, not to $|S(D,x)|$. Anyway, I’d be very cautious with such calculations. For one thing, that the average order of $k$ is $\log\log x$ does not mean that the average order of $2^{-k}$ is $(\log x)^{-\log 2}$. In fact, for constant $k$, there are $\sim\frac{x(\log\log)^{k-1}}{(k-1)!\log x}$ numbers in $[1,x]$ with $k$ prime factors, which would rather suggest that $|S(D,x)|$ is of order $x/\sqrt{\log x}$. It’s still unclear to me whether this is the right exponent, and whether $|S(D,x)|\sqrt{\log x}/x$ actually converges. | |
| Jul 12, 2013 at 11:43 | comment | added | Felipe Voloch | @EmilJeřábek Your initial argument suggests that $|S(D,x)|$ is asymptotic to $x/(\log x)^{\log 2}$, no? If that holds with a reasonable error term (which is not clear at all) then $|S(D,bx)| - |S(D,ax)|$ will be asymptotic to $(a-b)|S(D,x)|$. | |
| Jul 12, 2013 at 11:27 | comment | added | Emil Jeřábek | Here’s a guess: $D$ is a square modulo a random prime (or prime power) with probability 1/2, hence it is a square modulo a random number with $k$ prime factors with probability $2^{-k}$. Since a random integer below $n$ has $k\approx\log\log n$ prime factors on average, which is in particular nonconstant, I’d expect that $\lim_{x\to\infty}\frac{|S(D,x)\cap[ax,bx]|}{|[ax,bx]|}=0$. | |
| Jul 12, 2013 at 8:13 | comment | added | Douglas Zare | If you were to restrict to $q$ prime then for $D\gt 1$, quadratic reciprocity, the Chinese remainder theorem, and the equidistribution of primes in residue classes (Vallée-Poussin) give you equidistribution. For example, suppose $D=6$. $2$ is a quadratic residue mod $q \gt 2$ iff $q = \pm 1 \mod 8$, and $3$ is a quadratic residue iff $q = \pm 1 \mod 12$, so $6$ is a quadratic residue when $q = 1, 5, 19, 23 \mod 24$, and it's a quadratic nonresidue when $q = 7,11,13,17 \mod 24$. | |
| Jul 12, 2013 at 7:27 | review | First posts | |||
| Jul 12, 2013 at 7:28 | |||||
| Jul 12, 2013 at 7:08 | history | asked | Lenny Fukshansky | CC BY-SA 3.0 |