       Re: PolynomialMod

• To: mathgroup at smc.vnet.net
• Subject: [mg121349] Re: PolynomialMod
• From: Artur <grafix at csl.pl>
• Date: Mon, 12 Sep 2011 04:19:34 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```Maybe I was used wrong example

Example 1.

f = 1 + 3 x + 5 x2  + 5 x3  + 5 x4  + 3 x5  + x6;

If we have smaller order equation for example
x^5-x-1=0
that x^5=x+1
multiply both sides by x
x^6=x^2+x
if we substitute in f  x^6->x^2+x and x^5->x+1 we have

f = 1 + 3 x + 5 x2  + 5 x3  + 5 x4  + 3 (x+1) +(x^2+x);
now after reduction f=4 + 7 x + 6 x^2 + 5 x^3 + 5 x^4

That same result we will obtain uses

PolynomialMod[1 + 3 x + 5 x^2 + 5 x^3 + 5 x^4 + 3 x^5 + x^6,  x^5 - x - 1]

of course f=0 is divisible by my p from previous example but we can do similar cycle of substitutions as in Example 1 but if f<>0 isn't divisible

in my example let symbolically

x^3-b x-c==0

x^3=b x+c
x^4=b x^2+c x
x^5=b x^3+c x^2=b (b x+c)+c x^2=b^2 x+b c+c x^2
x^6=b^2 x^2+b c x+c x^3=b^2 x^2+b c x+c (b x+c)=b^2 x^2+2 b c x+c^2
and apply these equations to starting function f

f=1 + 3 x + 5 x2  + 5 (b x+c) + 5 (b x^2+c x) + 3 (b^2 x+b c+c x^2) + (b^2 x^2+2 b c x+c^2)
after reduction we have finally result (Collect by x)

f=(1 + 5 c +  3 b c + c^2) + (3 + 5 b + 3 b^2 + 5 c + 2 b c) x + (5 + 5 b + b^2 + 3 c) x^2

generally g=0 and f is some function (not necessary zero)

Artur Jasinski

W dniu 2011-09-11 16:48, Andrzej Kozlowski pisze:
> f = 1 + 3 x + 5 x2  + 5 x3  + 5 x4  + 3 x5  + x6;
>
> p = -1 + (-(3/2) - (I Sqrt)/2) x + (3/2 - (I Sqrt)/2) x2  + x3;

```

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