ABSTRACT
Let Z denote the integers. Say d divides m, equivalently, that m is a multiple of d, if there exists an integer q such that m = qd. Write d|m if d divides m. It is easy to prove, from the definition, that if d|x and d|y then d|(ax+ by) for any integers x, y, a, b: let x = rd and y = sd, and
ax+ by = a(rd) + b(sd) = d · (ar + bs)
1.1.1 Theorem: Given an integer N and a non-zero integer m there are unique integers q and r, with 0 ≤ r < |m| such that
N = q ·m+ r The integer r is the reduction modulo m of N .
