euclid's magic

Using Euclidean Algorithm 5k+3=1 (3k+2)+2k+1, where 2k+1 is the remainder 3k+2=1 (2k+1)+k+1, 2k+1=1 (k+1)+k, k+1=1 (k)+1, k=k*(1)+0. The process stops when remainder equals zero so GCD = 1

If A>B then, GCD(A,B) = GCD(B,R) , where R is the remainder of A/B .

GCD(5k+3,3k+2)
= GCD(3k+2, 2k+1) = GCD(2k+1,k+1) = GCD(k+1, k) = GCD(k,1) = 1

If $k$ can be any integer, let $k = 0$ . It is easy thus to see that $\text{gcd} (2, 3) = 1.$