[(m, n) = 1 ∧ d|mn] ⇒ ∃d1, d2 ∈ Z Such That d1d2 = d, d1|m, d2|n And (d1, d2) = 1

Let there be integers \(m\) and \(n\) such that \((m, n)=1\). According to the fundamental theorem of arithmetic, both \(m\) and \(n\) can be written as a product of primes:

$$\begin{align} m &= p_1^{m_1} p_2^{m_2} p_3^{m_3} \ldots p_k^{m_k} \\ n &= q_1^{n_1} q_2^{n_2} q_3^{n_3} \ldots q_h^{n_h}\end{align}$$

\(p_i\) is a prime factor of \(m\) and \(q_j\) is a prime factor of \(n\). Since \((m,n)=1\), then \(p_i \ne q_j\) for all \(i\) and \(j\). For example:

$$ \begin{gather} 4116 = 2^2 * 3 * 7^3 \\ 196625 = 5^3 * 11^2 * 13 \\ (4116, 196625) = 1 \quad \text{ (no common prime divisor)}\end{gather} $$

Let \(d\) be a divisor of \(mn\). This means all prime factors of \(d\) are either prime factors of \(m\) or prime factors of \(n\). This means \(d = p_1^{a_1} p_2^{a_2} \ldots p_k^{a_k} q_1^{b_1} q_2^{b_2} \ldots q_h^{b_h}\) where \(a_i \le m_i\) and \(b_i \le n_i\). This means we can represent \(d\) as \((p_1^{a_1} p_2^{a_2}\ldots) (q_1^{b_1} q_3^{b_3}\ldots)=d_1d_2\), where \(d_1\) has all the \(p_i\) factors and \(d_2\) has all the \(q_j\) factors:

$$ \begin{gather} d_1 = p_1^{a_1} p_2^{a_2}\ldots p_k^{a_k} \\ d_2 = q_1^{b_1} q_2^{b_2} \ldots q_h^{b_h} \\ (d=d_1d_2) \end{gather}$$

Since \(d_1\) has \(p_i\) factors, then \(d_1|m\), and since \(d_2\) has \(q_i\) factors, then \(d_2|n\). Also, since \(p_i \ne q_j\) for all \(i\) and \(j\), then \(d_1\) and \(d_2\) have no common prime factors. Therefore \((d_1, d_2)=1\).

For example, let \(m=4116\) and let \(n=196625\). The prime factors are:

$$ \begin{align} m &= 2^2 \cdot 3 \cdot 7^3 \\ n &= 5^3 \cdot 11^2 \cdot 13 \end{align} $$

Let \(d=2*5^2*7^2*11=26950\). We know \(d|mn\) as all it's prime factors exists in either \(m\) or \(n\). This means we can do this:

$$ \begin{gather} d=(2*7^2)(5^2*11) = (d_1)(d_2) \\ d_1 = 2*7^2 = 98 \\ d_2 = 5^2*11 = 275\end{gather} $$

Since \(d_1\) has factors of \(m\), then \(d_1|m\), and since \(d_2\) has factors of \(n\), then \(d_2|n\). Also, since \(d_1\) and \(d_2\) have no common prime factors, then \((d_1, d_2)=1\).

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