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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.

Bill Dubuque
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mathquester
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  • Each integer can be written in an unique way as a product of prime factors. Write $a.b$ in such product of prime factors. – Lourrran Jul 17 '23 at 10:32
  • So let $a = \prod\limits p^{\alpha_p}, b = \prod\limits q^{\beta_q}, c = \prod\limits r^{\gamma_r}, d = \prod\limits s^{\delta_s}$. I know that there is no common prime factors in $b$ and $d$. – mathquester Jul 17 '23 at 10:45

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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 $$ \begin{aligned} & m a b+n a d=a \\ \Rightarrow \quad & m c d+n a d=a \\ \Rightarrow \quad & d(m c+n a)=a \\ \Rightarrow \quad & d \mid a \end{aligned} $$

Lai
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