Algebra - Square of natural numbers

Question

$a$ and $b$ are natural numbers, their product $a\times b$ 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 $x\times y$ 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!

Answer 1

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

Answer 2

Hint:

Use the fundamental theorem of arithmetic.