I have huge problems to understand the following (simple?) theorem:
Let $a,b,c,d$ naturals such that $a\cdot b = c\cdot d$ and $\operatorname{gcd}(b,d) = 1$. Then it follows that $$ b\mid c \text{ and } d\mid a.$$
I just don't get a proof for it. Can anyone provide me maybe a cause for thought? I really don't know how to start.
Using the theorem that $$ (b, d)=1 \Rightarrow \exists m, n \in \mathbb{Z} \text { s.t. } m b+n d=1 \tag*{(*)} $$ Multiplying $(*)$ by $c$ yields $$ \begin{aligned} & m b c+n c d=c \\ \Rightarrow \quad & m b c+n a b=c \\ \Rightarrow \quad & b(m c+n a)=c \\ \Rightarrow \quad & b \mid c \end{aligned} $$ Multipling $(*)$ by $a$ yields $$ \begin{aligned} & m a b+n a d=a \\ \Rightarrow \quad & m c d+n a d=a \\ \Rightarrow \quad & d(m c+n a)=a \\ \Rightarrow \quad & d \mid a \end{aligned} $$