Proof of Chinese Remainder Theory

PsychonautQQ
Messages
781
Reaction score
10

Homework Statement



Theorm: Let m and n be relatively prime integers. If s and t are arbitrary integers there exists a solution x in Z to the simultaneous congruences:
x~s (mod m) and x~t (mod n)

Part of proof that confuses me: Since gcd(m,n) = 1, the Euclidean algorithm gives p and q in Z such that 1 = mp + nq. Take x = (mp)t + (nq)s.

How do they get the "Take x = (mp)t + (nq)s" part?

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
The Chinese remainder theorem says that a solution exists (or, the part you're trying to prove at least). What he's done is told you the form of the solution to the congruences, and will later show that this is always a solution by virtue of its form, thus proving the theorem.
 
Last edited:
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...
Back
Top