integer squares problem

**Deadstar** · May 1st 2010, 03:20 AM

Suppose that $x$ , $y$ and $z$ are positive integers satisfying $\gcd(x, y) = 1$ and $xy = z^2$ . Show that there are integers $u$ and $v$ such that $x = u^2$ and $y = v^2$ .

I have some ideas for this. Trivially we can take $u = 1$ and $v = z$ . I see nothing that says all numbers have to be different. (Although I suppose we could do this if $z$ is prime or a square leaving the below for when $z$ is composite and not a square)

Alternatively, write $z$ as a product of primes...

$z = p_1^{k_1}p_2^{k_2}\dots p_n^{n_1}$ .

Then $z^2 = (p_1^{k_1}p_2^{k_2}\dots p_n^{n_1})^2$ then we divide this up as we please such as...

$z^2 = (p_1^{k_1})^2 (p_2^{k_2}\dots p_n^{n_1})^2$ since $(p_2^{k_2}\dots p_n^{n_1})^2 = (y)$ wont be divisible by $(p_1^{k_1})^2 = (x)$ and hence our $\gcd(x,y) = 1$ condition holds.

How does this all look?

**tonio** · May 1st 2010, 03:39 AM

Originally Posted by Deadstar

Suppose that $x$ , $y$ and $z$ are positive integers satisfying $\gcd(x, y) = 1$ and $xy = z^2$ . Show that there are integers $u$ and $v$ such that $x = u^2$ and $y = v^2$ .

I have some ideas for this. Trivially we can take $u = 1$ and $v = z$ . I see nothing that says all numbers have to be different. (Although I suppose we could do this if $z$ is prime or a square leaving the below for when $z$ is composite and not a square)

Alternatively, write $z$ as a product of primes...

$z = p_1^{k_1}p_2^{k_2}\dots p_n^{n_1}$ .

Then $z^2 = (p_1^{k_1}p_2^{k_2}\dots p_n^{n_1})^2$ then we divide this up as we please such as...

$z^2 = (p_1^{k_1})^2 (p_2^{k_2}\dots p_n^{n_1})^2$ since $(p_2^{k_2}\dots p_n^{n_1})^2 = (y)$ wont be divisible by $(p_1^{k_1})^2 = (x)$ and hence our $\gcd(x,y) = 1$ condition holds.

How does this all look?

Not so neat but I think you've got the gist of the idea: but why would you decide that $(p_i^{k_1})^2=(x)$ ?? Just write $x,y,z$ as a product of primes, and check that every prime of $x$ and also of $y$ must have an even power...!

Tonio

**Deadstar** · May 1st 2010, 03:41 AM

Originally Posted by tonio

Not so neat but I think you've got the gist of the idea: but why would you decide that $(p_i^{k_1})^2=(x)$ ?? Just write $x,y,z$ as a product of primes, and check that every prime of $x$ and also of $y$ must have an even power...!

Tonio

Yeah it was just an example. General idea though was to divide up into primes.

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