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$a$ 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!

Bill Dubuque
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Haus
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    "I tried dividing with $3$, because full square when divided with three has remainder $0$ or $1$, but that doesn't help." That's not so strange. There are lots of non-squares too which give $0$ or $1$ as a remainder. Dividing by $3$ can only prove (in some cases) that a number is not a square, by giving a remainder of $2$. It can never prove that something is a square. – Arthur Dec 04 '18 at 10:27
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    It is confusing to put an essential condition like $\gcd(a,b)=1$ after the statement of the problem. I suggest "$a$ and $b$ are natural numbers, with $\gcd(a,b)=1$..." (Also, in English, a full/complete square is called a perfect square.) – TonyK Dec 04 '18 at 15:02

2 Answers2

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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

Frayal
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Hint:

Use the fundamental theorem of arithmetic.

5xum
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