Prove if given L|M and gcd(M, D) = 1, then gdc (D, L) = 1 is true

Question

I want to prove if given $L|M$ and $gcd(M, D) = 1,$ then $gdc (D, L) = 1$ is true (As stated in the title)

This seems very easy, yet I am having trouble on where to really start?

I feel like Bezout's Identity should be used here, however I don't know how to really apply it.

Any suggestions, or beginning steps would be gladly appreciated

Answer 1

Let $M = nL$.

Since $GCD(M,D) = 1$, there exist integers $a, b$ such that

$$\begin{align*} aM+bD &= 1\\ anL + bD &= 1 \end{align*}$$

So $1$ is a multiple of $GCD(D, L)$.

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From transitivity of divisibilities $GCD(D,L) \mid L$ and $L\mid M$, then $GCD(D,L) \mid M$.

Then $GCD(D,L)$ is a common divisor of both $M$ and $D$, and so $GCD(D,L)\mid GCD(M,D)$.

From given that $GCD(M,D) = 1$, then $GCD(D,L)$ is also $1$.