Let a and b be integers, not both zero. Assume that gcd(a, b) = 1. Set g = gcd(a2, b2). I
need to prove that g = 1.
By Theorem 2.1.3 there exist integers x and y such that ax + by = 1. Now do some algebra
1 = 1^3 = (ax + by)^3 = a^3x^3 + 3a^2x^2by + 3axb^2y^2 + b^3y^3 = a^2(ax^3 + 3x^2by) + b^2(3axb^2y^2 + by^3).
Set u = ax^3 +3x^2by and v = 3axb^2y^2 +by^3.
Thus a^2u+b^2v = 1.
Since g = gcd(a^2, b^2), there exists, t ∈ Z such that a^2 = gs and b^2 = gt. Hence
1 = a^2u + b^2v = gsu + gtv = g(su + tv),
that is g|1. Since g > 0 we conclude g = 1.
Now assume that d = gcd(a, b) > 1. Then there exist j, k ∈ Z such that a = dj and b = dk.
By Proposition 2.2.5 it follows that gcd(j, k) = 1. By the first part of this proof it follows that
gcd(j^2, k^2) = 1.
Since a^2 = d^2j^2 and b^2 = d^2k^2 and since j^2 and k^2 are relatively prime, Lemma 1 implies that
gcd(a^2, b^2) = gcd(d^2j^2, d^2k^2) = d2.