1.3: Problem 19: Let \(a\) and \(b\) be positive integers such that \((a,b) = 1\) and \(ab\) is a perfect square. Prove that \(a\) and \(b\) are perfect squares. Prove that the result generalizes to \(k\)th powers.
Proof
Let the prime factorization of \(a\) be
$$
\begin{align*}
a = p_1^{n_1} \cdot p_2^{n_2} \ \cdots \ p_k^{n_k}
\end{align*}
$$
and let the prime factorization of \(b\) be
$$
\begin{align*}
b = q_1^{m_1} \cdot q_2^{m_2} \ \cdots \ q_r^{m_r}
\end{align*}
$$
We know that \(a\) and \(b\) are coprime, so \(a\) and \(b\) share no prime factors and so \(p_i \neq q_j\) for all \(i,j\). Consider the prime factorization of \(ab\)
$$
\begin{align*}
ab = (p_1^{n_1} \cdot p_2^{n_2} \ \cdots \ p_k^{n_k}) \cdot (q_1^{m_1} \cdot q_2^{m_2} \ \cdots \ q_r^{m_r})
\end{align*}
$$
We know \(ab\) is a perfect square, so any exponent of any prime factor must be even. But \(a\) and \(b\) don’t share any prime factors. So every prime factor in \(a\) and \(b\) must have an even exponent. So \(a\) and \(b\) are perfect squares themselves as required. \(\ \blacksquare\)