TITLE: Proof that if $a/b$ is a perfect square then $a\cdot b$ is also a perfect square
QUESTION [5 upvotes]: It's my first question here, my name is Massimiliano. 
How can I prove that if $a/b$ is a perfect square then $ab$ is also a 
perfect square  without using the unicity of prime decomposition? 
My work: 
If $a/b$ is a perfect square then there exists $c$ such that $c^2 = a/b$ so 
$c^2b^2 = (cb)^2 = ab$, and therefore $ab$ is a perfect square.
But (to the contrary) if $c^2 = ab$ and  I divide by $b^2$ I obtain 
$(c/b)^2 = a/b$. But I don't know if $c/b$ is actually an integer. 
Thank you for help!

REPLY [3 votes]: If $\frac{a}{b}=q^2$, $a=bq^2$, then $a\cdot b = bq^2\cdot b = (bq)^2$.