2
$\begingroup$

What are the following $\text{unknown}_1$ and $\text{unknown}_2$? as I find that $\frac{x^2y-xy^3+2y^2-2}{6x^2y-5x^2-xy^3+7y^2+13} = \text{a constant}$ but use this constant can not find back the rem result.

it is for calculating modulo http://www.mathematik.uni-kl.de/~zca/Reports_on_ca/02/paper_html/node27.html

actually I want to know where do $\text{unknown}_1$ and $\text{unknown}_2$ come from?

formula = $\theta_m(a) = \mathrm{rem}(a, \text{m})$; but software can not calculate when m is a constant, if m is not a constant, what do m represent ?

$P = 6x^2y-5x^2-xy^3+7y^2+13$

$\theta_5(P) = \mathrm{rem}(P, \text{unknown}_1)$;

$\theta_5(P) = \mathrm{rem}(6x^2y-5x^2-xy^3+7y^2+13,\text{unknown}_1) = x^2y-xy^3+2y^2-2$

$\theta_7(P) = \mathrm{rem}(P, \text{unknown}_2)$;

$\theta_7(P) = \mathrm{rem}(6x^2y-5x^2-xy^3+7y^2+13,\text{unknown}_2) = -x^2y+2x^2-xy^3-1$

updated: software using http://www.mmrc.iss.ac.cn/mmp/ i guess maple has similar function

i do not understand a chinese book's writing, not mentioning above unknown, only said coefficient

is the following command to reduce the coefficient module 5 , but it is the result of above which coefficient to redcue? rem(6*x^2*y-5*x^2-x*y^3+7*y^2+13,pmod(6*x^2*y-5*x^2-x*y^3+7*y^2+13,5));

and why use 5 and 7 ? where do these integer come from?

i find a statement divide each coefficient by q and take the remainder but type command according to this statement is not correct rem(6*x^2*y-5*x^2-x*y^3+7*y^2+13, quo(6*x^2*y-5*x^2-x*y^3+7*y^2+13, 5));

  • 0
    I guess the same, but it is written in another style, why m = 5 and m = 7,do it choose coefficient of -5*x^2 and 7*y^2, why negative sign not considered?2011-11-21

1 Answers 1

1

It would appear that unknown1 is 5 and unknown2 is 7. If you reduce the coefficients of $6x^2y-5x^2-xy^3+7y^2+13$ modulo 5, you get $x^2y-xy^3+2y^2-2$, and similarly for the other example.

  • 0
    $6x^2y-5x^2-xy^3+7y^2+13$ is congruent to $x^2y-xy^3+2y^2-2$ modulo 5, because the difference between the two expressions is $5(x^2y-x^2+y^2+3)$, a multiple of 5. All the coefficients are reduced, and it looks like they are reduced to being between $-2$ and $2$.2011-11-21