Talk:Linear congruence theorem

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Sorry to make something that seems easy a bit more confusing, but the steps taken to solve the system of congruences (the steps are known as the Chinese remainder theorem, by the way) only work if the bases (numbers being used as mod's) are relatively prime. In the example mod 4 and mod 6 are both used. Both these have a HCF of 2, not one, so they are not relatively prime. They can be decomposed into relatively prime congruences, but I'm not sure how to do that (I was searching for a way to do it when I stumbled across this page) — Preceding unsigned comment added by 150.203.233.5 (talk • contribs) 03:35, 13 March 2003


 * This works in the given example because the equations can be divided by the GCD which yields a system where all moduli are coprime: 2x ≡ 2 (mod 6) <=> x ≡ 1 (mod 3) and 2x ≡ 4 (mod 8) <=> x ≡ 2 (mod 4), leaving equations (mod 3), (mod 7) and (mod 4), with pairwise coprime "bases". Another criterion concerning solvability of a system (when the "bases" are not coprime) is given in Chinese remainder theorem. &mdash; MFH:Talk 00:18, 8 August 2011 (UTC)

x must be an integer

Given the fact that a,b,n are integers do not assume x is an integer,such assumption need to be mentioned.

$$3x \equiv 2 \pmod {6}\ $$ d = gcd(3,6) = 3 and 3 does not divide 2,there is no solution. If x=2/3, then $$2 \equiv 2 \pmod {6}\ $$ Lolnameless (talk) 04:22, 18 December 2011 (UTC)lolnameless (talk) 04:22, 18 December 2011 (UTC)