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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

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 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));

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Please make the question self-contained. What are $\text{unknown}_1,\text{unknown}_2$ and the rem result? What software are you using? – Ross Millikan Nov 21 '11 at 3:22
updated software using, Maple have similar function – M-Askman Nov 21 '11 at 3:40
What is the difference between $\theta_m(a)$ and $a \pmod{m}$? – Ross Millikan Nov 21 '11 at 3:47
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? – M-Askman Nov 21 '11 at 4:04
up vote 1 down vote accepted

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.

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how to reduce coefficient modulo 5? which coefficient to reduce ? i updated the command above,but result is not the same – M-Askman Nov 21 '11 at 5:48
$6x^2y-5x^2-xy^3+7y^2+13$ 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 $2$. – Gerry Myerson Nov 21 '11 at 12:33

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