1.3: Problem 18: Prove that \((a^2,b^2) = c^2\) if \((a,b) = c\).
Proof
Suppose \((a,b) = c\). Then, we can write \(a = cx\) and \(b = cy\) for some \(x,y \in \mathbb{Z}\) and \((x,y) = 1\). Observe that
$$
\begin{align*}
(a^2, b^2) = (c^2 \cdot x^2, c^2 \cdot y^2) = c^2 \cdot (x^2, y^2).
\end{align*}
$$
We claim that \((x^2,y^2)=1\). Suppose for the sake of contradiction that it wasn’t and that \((x^2, y^2)=d\). Then \(d \mid x^2\) and \(d \mid y^2\). Consider the prime factorization of \(d = p_1p_2\cdots\). Take \(p_1\). \(p_1\) is prime and divides both \(x^2\) and \(y^2\). But since \(p_1\) is prime and it divides a product \(x^2 = x \cdot x\), then it must divide \(x\). Thus, \(p\) will divide both \(x\) and \(y\). This is a contradiction since \((x,y)=1\). Therefore, we must have \((x^2,y^2) = 1\) and thus, \((a^2,b^2) = c^2\). \(\ \blacksquare\)