If $a,b \in \Bbb N$ for which LCM$(a,b)=16\cdot(a,b)$, then $a|b$ or $b|a$

Is always correct statement that if natural numbers $a,b \in \Bbb N$ for which LCM$(a,b)=16\cdot(a,b)$, then $a|b$ or $b|a$?

I used formula that LCM$(a,b)=\frac{a\cdot b}{(a,b)}$

$\frac{a\cdot b}{(a,b)}=16\cdot(a,b) \implies a\cdot b= (4\cdot(a,b))^2$

Is it somehow useful? What should I do next?

Solution 1:

You're nearly there!

Let $ a = k_a (a,b) , b = k_b (a,b)$ where $(k_a, k_b ) = 1$.

Hint: What is the value of $ k_a k_b$ according to your equation?

Hint: Can we conclude that one of them must be 1?
If yes, then we either have $ a \mid b$ or $b\mid a$. If no, then we have a counterexample.

Solution 2:

Let $d= (a,b)$. We know that $LCM (a,b) ={ab\over d}$. Then as you noted we have $$ab =16d^2$$ We can also write $a=dx$ and $b=dy$ where $x,y$ are coprime. So $$xy=16$$ Now it is not difficult to finish.