Congruences (2/17) If m (the modulus) is positive and if a and b are integers, then we say a is congruent to b mod m, writing a  b (mod m), provided that m | (b – a). Again, this says that a and b have the same remainder r upon division by m, where 0 ≤ r < m, and of course a  r (mod m) and b  r (mod m). Congruences “respect” addition, subtraction and multiplication, i.e., if a 1  b 1 (mod m) and a 2  b 2 (mod m), then (1) a 1  a 2  b 1  b 2 (mod m) (proof?), and (2) a 1 a 2  b 1 b 2 (mod m) (proof??)

However, Watch Out for Division Division (i.e. cancellation) may not work! Example (2)(6)  (2)(8) (mod 4), but 6  8 (mod 4) Guess what’s wrong here. In regular integer (i.e., Diophantine) equations, you can cancel provided what? So what will be the corresponding condition in congruences? Proof??

Solving Linear Congruences Suppose we have a congruence which contains a variable x which is linear in x. Then by adding, subtracting and multiplying, we can simplify this congruence to the form a x  c (mod m) and we seek x. How do we find it? By the solutions to a congruence of any kind, we shall mean solutions which are not congruent to each other (mod m), and in general we list them in their “smallest” form, i.e., so that they all satisfy 0 ≤ x 0 < m. Hence a nice thing about congruences is that there are at most m solutions, i.e., we can do “guess and check” on a finite set!

Some Examples Simply using guess and check, find all solutions of the following linear congruences a x  c (mod m) : 3x  5 (mod 6) 3x  9 (mod 6) 5x  4 (mod 6) 4x  12 (mod 8) 12x  15 (mod 100) (Be more clever. Think even and odd.) What patterns can we discern as to the number of solutions which occur? Hint: It has to do with the GCD(a, m) and whether that number divides c.

Assignment for Wednesday Hand-in #2 is due. Read Chapter 8 to the top of page 60.