# number theory – If d|nm and \$gcd(n, m)= 1\$ then exist d1, d2 such that \$d=d1d2\$ and \$d1|n\$, \$d2|m\$ (without Fund. Theorem of Arit)

We want to prove that if $$d|nm$$ and $$gcd(n,m)=1$$ then $$d=d1d2$$ where $$d1|n$$ and $$d2|m$$ and $$gcd(d1,d2)=1$$

We already proved it using Fundamental Theorem of Aritmethic. But we wonder if there is a way to prove it using only gcd basic theorems.

Our hints

If $$d1|n$$ and $$d2|m$$, then $$d1d2|nm$$

The assumption $$d|nm$$ is a consequence of $$d|m$$ $$(a|b => a|bc)$$

If $$d|nm$$ then $$d|gcd(d,n) gcd(d,m)$$ (Properties)

$$gcd(d_1,d_2)| gcd(n,m)$$ obv