Bill Allombert on Tue, 21 Nov 2023 15:10:33 +0100
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Re: sqrt(x,n) for non-prime n?
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: sqrt(x,n) for non-prime n?
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Date: Tue, 21 Nov 2023 15:10:26 +0100
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- References: <568168ea458df688f419ca9967be8edf@stamm-wilbrandt.de> <f1846cf6-3de0-4bdb-ae3a-da80cf8611d9@normalesup.org> <2e004d9474c71af3dfc4dc61b8f398af@stamm-wilbrandt.de> <c13c448308acb6271940db1fa52d97e7@stamm-wilbrandt.de> <90939d8f-440a-41ee-91f5-bdcc3abddad2@normalesup.org>
On Tue, Nov 21, 2023 at 02:44:11PM +0100, Aurel Page wrote:
> On 21/11/2023 13:03, hermann@stamm-wilbrandt.de wrote:
> > So implementing Kunerth's method is only needed if factoring modulus
> > would take too much time.
> If you can compute square roots of arbitrary numbers modulo n, then you can
> factor n. Indeed, take some x modulo n, compute a square root y of x^2 mod
> n: then gcd(x-y,n) is a nontrivial factor of n with good probability.
No, you need two different squareroots that are not ± the others.
For example 2^64 is a root of -1 modulo (2^128+1) but it is not sufficient for
factoring.
But yes, there must be some unstated hypothesis for Kunerth's method to work.
Cheeers,
Bill