How to factorize values that include modular arithmetic?

Asked Dec 04 '23 at 03:14

Active Dec 04 '23 at 03:54

Viewed 29 times

I have an equation:

$b*(x*rinv*s*b\mod n) - (x*b)$

In which i need to isolate x but when i do:

$ x*(rinv*s*b \mod n) - b)$

The results are different, Why?

Here are the values:

x = 9425462603919758403509431170377741597727182287022881396228286868666458414738
b = 100906376405775661990562963813068351310177981465009640700224287315820973458674
a = 44286717587595936018074743368265679694432636709170485099126082338272318217947
n =115792089237316195423570985008687907852837564279074904382605163141518161494337

How can i factor this correctly, I am really confused with modulo right now.

edited Dec 04 '23 at 03:54

Bill Dubuque

272,048

asked Dec 04 '23 at 03:14

terry franklin

1

What is your order of operation between the $$ and $\bmod$ in $(xrinvsb\mod n)$? – peterwhy Dec 04 '23 at 03:21
Please fix the unbalanced parenthesis. – peterwhy Dec 04 '23 at 03:23
Are both answers the same modulo n? – Steven Alexis Gregory Dec 04 '23 at 03:32
Generally $\ \begin{align}&\color{#c00}b(xa \bmod n)-bx\ \not\equiv &\ \ x(a \bmod n, -, b)\end{align} \pmod{!n},\ $ but it is true if you delete $,\color{#c00}b,,$ by the congruence sum and product rules in the linked dupe. – Bill Dubuque Dec 04 '23 at 03:50

How to factorize values that include modular arithmetic?

0 Answers0