Put these steps in order to form a proof that if a and m are relatively prime integers and m 1, then an inverse of a modulo m exists.

Hence, s is an inverse of a modulo m, by the definition of the inverse modulo m. It follows that so = 1 (mod m), since tm = 0 (mod m). We see that so + tm = 1 (mod m), since so + tm = 1. By Bezout's theorem and the fact that god(a,m) = 1, there are integers s and t such that so + tm = 1. Assume that o and m are relatively prime integers and m 1