gcd(a,b)=1 => gcd(a^2,b)=1

Question

how do I show that coprime numbers are still coprime, if I square one of them.

And that without using prime factorization.

Answer 1

I think you’re trying to avoid the whole p1^a1 prime factorization argument, but I still think using the factors of a and b is the way to go.

Suppose gcd(a^2,b) > 1. Then b shares a prime factor with a^2. But a^2 shares all its prime factors with a, so b and a share a prime factor, contradicting the assumption that gcd(a,b) = 1.

Answer 2

can use Bezout's identity

gcd(a,b) = 1

=> exists x,y such that ax+by = 1

=> a^(2)x^(2) + by(ax+1) = ax(ax+by) + by = ax+by = 1

=> gcd(a^(2),b)=1