How to find all possible integers for a given congruence?

Asked Apr 03 '20 at 19:14

Active Apr 03 '20 at 22:41

Viewed 48 times

I was given a statement:

$$kx^3 \equiv n \pmod m = 3 \pmod 9$$

Trying to solve it

A user gave me some tips in the comment section under this question. Trying to use it to solve my problem:

$$kx^3 \equiv 3 \pmod 9 \Longleftrightarrow d = gcd(x,9) | 3$$ $$k \equiv \frac{n}{x^3} \mod \frac{m}{d} = \frac{3}{x^3} \mod \frac{9}{3}$$

So the solution is that x has to be 1 or else there will never come a whole integer out. Solving this for $x=1$: $$k = 0.$$

Is that correct?

edited Jun 12 '20 at 10:38

Community

asked Apr 03 '20 at 19:14

Do you know how to solve $\ b x \equiv n\pmod{m}?\ $ Above is a special case $,b = c^3\ \ \ $ – Bill Dubuque Apr 03 '20 at 19:35
Sadly, I do not know how to solve it. I am totally lost. – Apr 03 '20 at 19:44
What is your context? Normally one would not be posed such exercises before one learned the basic techniques of solving linear congruences. Do you know about modular inverses? – Bill Dubuque Apr 03 '20 at 19:46
@BillDubuque well, I know about the Euclidian Algorithm. And yes, I know about modular inverses. – Apr 03 '20 at 19:58
There are many prior answers here explaining how to solve a general linear congruence, e.g. see here and here. Your congruence is a special case (when the lead coef is a cube). – Bill Dubuque Apr 03 '20 at 20:04
@BillDubuque thank you, I have tried it to some extent. But I am very unsure of what I have done is right. I have edited the post to fit that into the question. – Apr 03 '20 at 20:29
By general theory $,a, k \equiv n\pmod{!m},$ is solvable $\iff d=\gcd(a,m)\mid n.,$ If so then the solution is $, k \equiv (n/d)/(a/d) \pmod{!m/d}.\ $ Yours is special case $,a = c^3\ \ \ $ – Bill Dubuque Apr 03 '20 at 20:39
Let us continue this discussion in chat. – Apr 03 '20 at 21:41