Abstract Alg. Proof w/ Mod Congruence and Relative Primes

Colleen G
Messages
6
Reaction score
0

Homework Statement


If ≡(mod), ≡(mod),and gcd(,)=1,provethat ≡ (mod ).

Homework Equations


If ≡(mod)→n|ab-cd
≡(mod)→n|b-d
gcd(,)=1→ relatively prime. So bx+ny=1

Need to show n|a-c→a-c=nw

The Attempt at a Solution


If n|ab-cd, then nk=ab-cd
If n|b-d, then nl=b-d
If n|ab-cd AND n|b-d, then n|p(ab-cd)+q(b-d). So pab-pcd+qb-qd.
pad-pcd+qb-qd
=pab+qb-pcd-qd
=b(pa+q) +d(-pc-q)

I'm stuck! Don't know if this is going anywhere.
 
Physics news on Phys.org
Colleen G said:
If ≡(mod), ≡(mod),and gcd(,)=1,provethat ≡ (mod ).
You need to edit your post so the variables show up. ( https://www.physicsforums.com/help/latexhelp/ )
Do you mean something like ab \equiv cd \ (mod \ n) ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top