1. Primes and Squares again

Q: Show that if a, b are positive integers such at (a,b)=1 and ab is a square, then a and b are also squares.

Suppose a and b are not squares, then a=h*k and b=p*q where h,k,p,q are all unique primes. (Is this the right approach to this proof?)

2. Hi,

Hint: A number is a perfect square iff all exponents in its prime decomposition are even.

Use this on ab. Since (a,b)=1, it is clear how prime decompositions of a and of b must look like.

3. Originally Posted by dlbsd
Q: Show that if a, b are positive integers such at (a,b)=1 and ab is a square, then a and b are also squares.

Suppose a and b are not squares, then a=h*k and b=p*q where h,k,p,q are all unique primes. (Is this the right approach to this proof?)

Let $\displaystyle ab=x^2$
Let $\displaystyle p$ be a prime, such that $\displaystyle p|a$
Now $\displaystyle p$ can't divide $\displaystyle b$ as $\displaystyle (a,b)=1$
So,$\displaystyle p|x$
$\displaystyle \Rightarrow p^2|x^2$
$\displaystyle \Rightarrow p^2|ab$
$\displaystyle \Rightarrow p^2|a$

Thus power of prime in a(and by similar logic in b) are all even. Thus either is a perfect sqaure as well.

This is wrong ! Please see below from post by BingK

4. Originally Posted by aman_cc
Thus power of prime in a(and by similar logic in b) are all even. Thus either is a perfect sqaure as well.
Could you explain this to me?
So $\displaystyle p^2|a$ which means there is an integer h such that $\displaystyle a=p^2 * h$

So the power of the prime is 2 but what about h?

5. Originally Posted by dlbsd
Could you explain this to me?
So $\displaystyle p^2|a$ which means there is an integer h such that $\displaystyle a=p^2 * h$

So the power of the prime is 2 but what about h?

This is wrong, please see post below from BingK

The proof is valid for any p. So in your specific question -
case 1: h is prime - then be same logic you can show $\displaystyle h^2|a$. But as that is not the case hence h can't be prime.
so h is composite - then also if q is a prime such that $\displaystyle q|h$ then you can show that $\displaystyle q^2|a$. Infact h itself is a perfect sqaure - as it should be

The idea of the proof is that for "any" prime ,$\displaystyle p$, which divides $\displaystyle a$, $\displaystyle p^2$ also divides a. So in the prime- factorization of a all the powers will be even.
Logic for a can be replicated for b. Thus the statement will be true for ab as well. If all powers of prime are even then the number is a perfect square. Using this fact we conclude ab is a perfect square.

Hope it helps.

6. I think there's a bit of a problem with your logic that if $\displaystyle p^2|a$ then all the powers in the prime factorization of $\displaystyle a$ will be even. If you think about it $\displaystyle p^2|p^3$, and $\displaystyle 3$ is not even ...

But if you use Taluivren's hint, you can take the prime decomposition of $\displaystyle ab$, and all the powers have to be even. Since $\displaystyle (a,b)=1$, we can rearrange the prime decomposition of $\displaystyle ab$ so it is written as the prime decomposition of $\displaystyle a$ multiplied by the prime decomposition of $\displaystyle b$. Then, we look at the respective prime decompositions and notice that the powers are all even. Again, by the hint, we have that $\displaystyle a$ and $\displaystyle b$ are also perfect squares.

7. Originally Posted by Bingk
I think there's a bit of a problem with your logic that if $\displaystyle p^2|a$ then all the powers in the prime factorization of $\displaystyle a$ will be even. If you think about it $\displaystyle p^2|p^3$, and $\displaystyle 3$ is not even ...

But if you use Taluivren's hint, you can take the prime decomposition of $\displaystyle ab$, and all the powers have to be even. Since $\displaystyle (a,b)=1$, we can rearrange the prime decomposition of $\displaystyle ab$ so it is written as the prime decomposition of $\displaystyle a$ multiplied by the prime decomposition of $\displaystyle b$. Then, we look at the respective prime decompositions and notice that the powers are all even. Again, by the hint, we have that $\displaystyle a$ and $\displaystyle b$ are also perfect squares.
@Bingk - I get what you are saying. Yes - it is incorrect. Thanks for pointing that out.