Htykuut

$a$ and $b$ are natural numbers, their product $atimes b$ is full/complete sqare, prove that then $a$ and $b$ are full squares. $gcd (a,b)=1$ .

Natural number is full square if you can write it in form $n^2$.
I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help. Then i also tried to use the fact that when you divide $xtimes y$ with $z$, and $(x,z)=1$ then $z$ divides $y$, but I don't know how to use it in this case.
Can anyone help me and give me instructions what to do? Thanks a lot!

hint:

every number can be written as a product of prime number to a certain power.

$n = prod_{p in mathbb{P}} p^{alpha_p} $

using this plus the definition of full square and your work you can do it:

$a = prod_{p in mathbb{P}} p^{alpha_p} $ and
$b = prod_{q in mathbb{P}} q^{alpha_q} $

we have also $ab = prod_{n in mathbb{P}} q^{2*alpha_n} $

using your hint, prove that every $alpha_p$ and $alpha_q$ is even

Hint:

Use the fundamental theorem of arithmetic.