Re: Solving x^2+n*y^2=a without factoring positive $n$?

Georgi Guninski on Fri, 29 Nov 2019 17:29:53 +0100

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Re: Solving x^2+n*y^2=a without factoring positive $n$?

To: pari-dev@pari.math.u-bordeaux.fr
Subject: Re: Solving x^2+n*y^2=a without factoring positive $n$?
From: Georgi Guninski <gguninski@gmail.com>
Date: Fri, 29 Nov 2019 18:29:36 +0200
References: <CAGUWgD-GHjDcR=-uiwAEW4ezZ64O1s5X-x=zGKt0ni=MpAMSGA@mail.gmail.com> <20191128163502.4q43rvadseq7zcy3@yellowpig> <CAGUWgD-66YrU2c7of+9wKLsX4BOPJq_+r8ft7QtbnY76d0RG=w@mail.gmail.com>

> Sorry, I mean quadratic in log(n).

(I am having some problems with gmail's filters).

Are you sure you can bound the complexity only with n
without a?

Fix n and set X=2^2^2^n, Y=X+1, a=X^2+n Y^2

I think you can't express X with only log(n)^2 operations.

Follow-Ups:
- Re: Solving x^2+n*y^2=a without factoring positive $n$?
  - From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>

References:
- Solving x^2+n*y^2=a without factoring positive $n$?
  - From: Georgi Guninski <gguninski@gmail.com>
- Re: Solving x^2+n*y^2=a without factoring positive $n$?
  - From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Re: Solving x^2+n*y^2=a without factoring positive $n$?
  - From: Georgi Guninski <gguninski@gmail.com>

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