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My question

So there are three parts;

(a) Find a solution to the congruence $x^3 \equiv$ $4$ mod $5$ by trying all possible $x$ values.

(b) Find a solution to the congruence $x^3 \equiv$ $4$ mod $11$ by trying all possible $x$ values.

(c) Find a solution to the congruence $x^3 \equiv$ $4$ mod $55$ by using the Chinese Remainder Theorem

What have I done?

(a)

So if the congruence is in the form:

$$a \equiv b \text{ mod }m$$

then;

$$m | (a-b)$$

So what I have come up with is the set of $x$ values that satisfies that congruence, they are:

$$x = \{...-11,-6,-1,4,9,14,19...\}$$

I'm a bit unsure if the question is asking for one solution...or a generalised form for ALL solutions. Well, generalised in the sense of modular arithmetic I suppose.

(b)

Same deal, except the set of solutions for $x$ is:

$$\{...-28,-17,-6,5,16,27...\}$$

(c)

As for this one, it would be helpful to have my solutions from (a) and (b) in congruence form. But I'm not sure how to change my 'set' of answers to something more general.

$$ \text{Thanks guys.} $$

Rubicon
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    For (a) you are expected to try $0,1,2,3,4$ and conclude that $x=4\bmod5$ is the unique solution. Similarly for (b) you are expected to try $0,1\dots,10$ and conclude that $x=5\bmod11$ is the unique solution. Obviously for (c) you use the CRT. – almagest May 29 '16 at 06:03
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    Your solution for (a) is the congruence class $[4] (\bmod 5)$ (although you're missing $-1$). – Joffan May 29 '16 at 06:08
  • @almagest Oh yep I see. Yeah that's how I got 4 and 5 in the first place...trial and error through the values. Joffan Yeah that was a typo :) Does that change anything because it is a congruence class? – Rubicon May 29 '16 at 06:20

0 Answers0