Re: polynomial congruence
- To: mathgroup at smc.vnet.net
- Subject: [mg26158] Re: [mg26138] polynomial congruence
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Thu, 30 Nov 2000 01:04:03 -0500 (EST)
- References: <200011280656.BAA02476@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Constantinos Draziotis 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
Not certain I understand the question. I am assuming you have a
polynomial in x and y, a fixed prime modulus, you specify values for y,
and then want to solve for x. If so, then the example below will show a
way to do this.
In[31]:= poly = x^3 + 4*x*y^2 - 7*x^2*y + 2*y^3 + 5*x^2 - x*y -
6*y^2 -3*x + 2*y + 7;
In[32]:= mod = 19;
In[33]:= Table[Roots[(poly/.y->j) == 0, x, Modulus->mod], {j,0,mod-1}]
Out[33]= {False, False, False, False, x == 3, x == 16, x == 4, False,
x == 6 || x == 11 || x == 15, x == 17, x == 5, False, x == 4,
x == 9 || x == 10 || x == 10, x == 9 || x == 13 || x == 14,
x == 4 || x == 9 || x == 11, False, False, x == 2 || x == 12 || x ==
12}
Note that "False" means there is no modular image in Z_mod (integers
modulo mod) that solves the given equation, that is, all solutions lie
in a nontrivial extension field.
Daniel Lichtblau
Wolfram Research
- References:
- polynomial congruence
- From: "Constantinos Draziotis" <roth@math.auth.gr>
- polynomial congruence