Prove that if $gcd(a, b) = c$ then $c^2|ab$. I recently just started this topic in class and have been going over some examples without much success. I understand the concept behind if a|b and b|c then a|c but when it comes to more complex ones, I become a  tad confused..
*Thanks for the help everyone!!
 A: Even I can do this one!
If $\gcd(a,b) = c$ then $c | a$ and $c | b$, so $c$ squared divides $ab$.
A: Hint:


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*If $\frac{a}{x} \in \mathbb{N}$ and $\frac{b}{x} \in \mathbb{N}$ then then $\frac{ab}{xy} = \frac{a}{x}\cdot\frac{b}{y} \in \mathbb{N}$.


I hope this helps $\ddot\smile$
A: Hint:
If $\gcd(a, b) = c$, then $c|a$ and $c|b$. Hence, $a = k_{1}c$ for some $k_{1} \in \mathbb{Z}$ and $b = k_{2}c$ for some $k_{2} \in \mathbb{Z}$. 
Can you take it from here? (Try writing $ab$ as some multiple of $c^{2}$ given the information you have!)
A: Hint: write down with words what the definition of gcd is. Then make sure you understand it. Then you will see that $gcd(a,b)=c \Rightarrow c^2|ab$ follows immediately from the definition.
A: Easy!  The key is to realize that if $\gcd(a, b) = c$, then we have both $c \vert a$ and $c \vert b$; greatest common divisor is, after all, a divisor.  Thus there exist integers $k_1$, $k_2$ such that $a = k_1c$ and $b = k_2c$;  then $ab = k_1k_2c^2$, showing that in fact $c^2 \vert ab$.  Nice one!  Cheers and QED, Bob Lewis
