I have gotten that $(a,m)(a,n) | (a,mn)$, but I'm not able to show that it is the greatest such divisor. The proof I have so far is as follows:

Let $d_1 = (a,m)$ and $d_2 = (a,n)$ so $m = l_1d_1$, $n = l_2d_2$, and $a = j_1d_1 = j_2d_2$. Since $(m,n) = 1$, we have \begin{align*} 1 &= mk_1 + nk_2\\ &= (l_1d_1)k_1 + (l_2d_2)k_2\\ &= (l_1k_1)d_1 + (l_2k_2)d_2, \end{align*} so $(d_1, d_2) = 1$. From above $a = j_1d_1 = j_2d_2$, and by Theorem 2.5, $d_1 \mid j_2$ and $d_2 \mid j_1$, so $j_1j_2 \mid a$.

(Where Theorem 2.5 states that if $(a,b) = 1$ and $a \mid bc$, then $a \mid c$)


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