Timeline for Questions involving primes $p\equiv1\pmod4$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 25, 2019 at 6:29 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
added 2 characters in body
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| Dec 25, 2019 at 5:38 | answer | added | XIP | timeline score: 2 | |
| Dec 18, 2019 at 21:55 | comment | added | reuns | I don't know if it helps but $\Bbb{Z}/(2+2i)^\times = 1,-1,i,-i$ so that $\sum_{z\in 1+(2+2i)\Bbb{Z}[i]} z |z|^{-2s} = \sum_n a_n n^{-s} = \exp(f(2s)+\sum_{p\equiv 1\bmod 4}a_pp^{-s})$ where $a_p = 2 \Re(\pi_p)=(-1)^{b_p}2s_p$, $s_p+it_p \equiv(-1)^{b_p}\bmod (2+2i)$ and the PNT for the Rankin Selberg L-function $\sum_na_n^2 n^{-s}$ gives $\sum_{p \le x}s_p^2\sim C\frac{x^2}{\log x}$. Then we can do the same with $\sum_na_n^{2k} n^{-s-k}$ and approach $\sum_{p\le x}\frac{t_p}{p^{1/2}}=\sum_{p\le x}\sqrt{1-\frac{s_p^2}{p}}$ by $\sum_{k\le K}(-1)^k{1/2\choose k}\sum_{p\le x}\frac{s_p^{2k}}{p^k}$ | |
| Dec 18, 2019 at 11:39 | comment | added | Dmitry Krachun | @reuns note that if you fix one number $\pi$ with $|\pi|=p$ for each $p$ using some weird rule, then there will be no equidistribution to speak of: You can always take points with argument in $(0, \pi/8)\cup (\pi/4, 3\pi/8)$, say. | |
| Dec 18, 2019 at 11:35 | comment | added | Dmitry Krachun | @reuns I am now confused. I think in the question we take all pairs $(x, y)$ in the first quadrant such that $x^2+y^2=p$ and then only look at those suitable modulo 2. | |
| Dec 18, 2019 at 9:47 | comment | added | reuns | @DmitryKrachun We can take $arg(\pi) \in [0,\pi/2]$ if we look at the equidistribution of the $\pi$ (because each comes in pair with $i\overline{\pi}$), but if for each $p\equiv 1 \bmod 4$ we fix a representative $|\pi|^2 = p$ then we'll only have the equidistribution in $[0,\pi /4]$ | |
| Dec 18, 2019 at 9:43 | comment | added | Dmitry Krachun | @Zhi-WeiSun The second questions is way out of reach, I think. My guess is that in the numerator non-negligible contribution comes from primes of the from $k^2+\ell^2$ with $\ell=o(k^\eps)$ for any fixed $\eps>0$. But even existence of such primes is not proven. | |
| Dec 18, 2019 at 9:37 | comment | added | Dmitry Krachun | @reuns Clearly, equidistribution is also true for all primes in $\mathbb{Z}[i]$ in the first quadrant (i.e. quater of a disc). As far as I understand, the theorem I quoted also says that they are equidistributed in the intersection of the quater of a disc with any "arithmetic progression". So here we need to take progression modulo 2. Unfortunately, I am not an expert in algebraic number theory, so I might be missing somthing. | |
| Dec 18, 2019 at 7:50 | comment | added | reuns | @DmitryKrachun I don't see why, the equidistribution is for the unique representative $|\pi|^2 = p$ with $arg(\pi) \in [0,\pi/4]$ (it follows from the PNT for the symmetric power L-functions of $\sum_{z\in \Bbb{Z}[i]} z^4|z|^{-4-2s}$) here the OP is taking a weird representative $arg(\pi) \in [0,\pi/2], 2\ |\ \Im(\pi)$. | |
| Dec 18, 2019 at 2:17 | comment | added | Zhi-Wei Sun | I have just added Question 2. | |
| Dec 18, 2019 at 2:16 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
Add Question 2.
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| Dec 17, 2019 at 21:34 | comment | added | Dmitry Krachun | Your conjecture follows from Theorem 6 in chapter XV in the S. Lang's book "Algebraic Number Theory" (which is probably slightly more general than the result of E.Hecke). See also the discussion from this post: mathoverflow.net/questions/133410/hecke-equidistribution/133447 | |
| Dec 17, 2019 at 17:10 | comment | added | Fedor Petrov | You count Gaussian primes in the quater-circle with abscissa divisible by 2 (or, strictly speaking, not count them but sum up the values of abscissa). I guess it should be a partial case of E. Hecke's theorem, unfortunately I do not know German to read it. | |
| Dec 17, 2019 at 16:42 | comment | added | Zhi-Wei Sun | When we write a prime $p\equiv1\pmod4$ as $x^2+y^2$ with $1\le x\le y$, we can say nothing about the parity of $x$. This is why the conjecture looks challenging. | |
| Dec 17, 2019 at 16:36 | comment | added | Fedor Petrov | Here matwbn.icm.edu.pl/ksiazki/aa/aa79/aa7935.pdf it is claimed that "E. Hecke [H] showed that Gaussian primes are equidistributed over arithmetic progressions within regular planar domains." Your conjecture concerns the Gaussian primes in the domain $\{x,y>0,x^2+y^2<n\}$ and progressions with difference 2. | |
| Dec 17, 2019 at 16:18 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |