Prove that if $d \mid (n_1 \times n_2)$ where $d > 1$ and $\gcd (n_1, n_2) = 1$, then only one of the following holds: either $d \mid n_1$ or $d \mid n_2$.
We'll assume, for the purpose of contradiction, that both $d \mid n_1$ and $d \mid n_2$ hold, where $d>1$.
That is, $d$ is a divisor of $n_1$ and $n_2$.
The gcd of $n_1$ and $n_2$ is therefore at least $d$. Since $d>1$, then the gcd is also strictly greater than one.
This contradicts the given assumption that $\gcd (n_1, n_2)=1$.
Therefore, by contradiction, both $d \mid n_1$ and $d \mid n_2$ cannot hold.
This leaves two possibilities:
- either $d \mid n_1$ or $d \mid n_2$ but not both.
- neither $d \mid n_1$ or $d \mid n_2$ hold.
If neither hold, then this contradicts the assumption $d \mid (n_1 \times n_2)$, leaving only the desired conclusion that only one of $d \mid n_1$ and $d \mid n_2$ holds.