
Integers, Prime Factorization, and More on Primes
... Proof. (1) By the Euclidean algorithm, there exist integers m, n such that ma + nb = 1. Multiplying c to both sides we have mac + nbc = c. Since a | bc, i.e., bc = qa for some integer q, then c = mac + nqa = (mc + nq)a, which means that a is a divisor of c. (2) If p - a, then gcd(p, a) = 1. Thus by ...
... Proof. (1) By the Euclidean algorithm, there exist integers m, n such that ma + nb = 1. Multiplying c to both sides we have mac + nbc = c. Since a | bc, i.e., bc = qa for some integer q, then c = mac + nqa = (mc + nq)a, which means that a is a divisor of c. (2) If p - a, then gcd(p, a) = 1. Thus by ...
Name: Math 2412 Activity 2(Due by Feb. 28) Find the quadratic
... values, then what can you conclude about p x and q x ? Consider the previous problem. ...
... values, then what can you conclude about p x and q x ? Consider the previous problem. ...