Re: polynomial congruence
- To: mathgroup at smc.vnet.net
- Subject: [mg26145] Re: [mg26138] polynomial congruence
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Thu, 30 Nov 2000 01:03:53 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
As long as p is not very large it is easy to do. For example, suppopse you
want to solve all the congruences
In[1]:=
f[x_, y_] := x^2 + y^2 - 25
In[2]:=
Table[Solve[{f[x, y] == 0, Modulus == 11}[[1]], x], {y, 0, 10}]
Out[3]=
{{{x -> -5}, {x -> 5}}, {{x -> -2 Sqrt[6]},
{x -> 2 Sqrt[6]}}, {{x -> -Sqrt[21]}, {x -> Sqrt[21]}},
{{x -> -4}, {x -> 4}}, {{x -> -3}, {x -> 3}},
{{x -> 0}, {x -> 0}}, {{x -> -I Sqrt[11]},
{x -> I Sqrt[11]}}, {{x -> -2 I Sqrt[6]},
{x -> 2 I Sqrt[6]}}, {{x -> -I Sqrt[39]},
{x -> I Sqrt[39]}}, {{x -> -2 I Sqrt[14]},
{x -> 2 I Sqrt[14]}},
{{x -> -5 I Sqrt[3]}, {x -> 5 I Sqrt[3]}}}
Note that these answers really lie in the algebraic closure of the finite
filed Z/11.
on 11/28/00 3:56 PM, Constantinos Draziotis at roth at math.auth.gr wrote:
>
> Hello,i am a new user of mathematica,i will appreciate very much if you
> can help me with this(it seems simple) problem:i want to solve a
> polynomial congruence modulo prime number i.e f(x,y)=0modulo(p)(prime
> number) with y=0,1,2,3...,n (n:integer).i have to find the classes
> xmodulo(p)
>
> thanks for your time
>
> Costas
>