Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the simultaneous linear congruences - $$ax+by\equiv u\mod m; \\ cx+dy\equiv v\mod m.$$ What are the conditions (if any) on $a,b,c,d,m$ so that the mapping of all pairs $(x,y)$ to $(u,v)$ is one-to-one, $x,y,u,v \epsilon \{0,1,...,m-1\}$. And what is the general result for a linear equation of the form $$ AX \equiv B \mod m.$$

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

Just as in ordinary linear algebra, the question is equivalent to the question of when the matrix $\left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right]$ has an inverse ($\bmod m$). The answer, just as in ordinary linear algebra (suitably interpreted), is that it is necessary and sufficient that the determinant $ad - bc$ be a unit $\bmod m$. (This is a good exercise.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.