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Thread: integer squares problem

  1. #1
    Super Member Deadstar's Avatar
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    integer squares problem

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


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

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

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

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

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

    How does this all look?
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Suppose that $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$ are positive integers satisfying $\displaystyle \gcd(x, y) = 1$ and $\displaystyle xy = z^2$. Show that there are integers $\displaystyle u$ and $\displaystyle v$ such that $\displaystyle x = u^2$ and $\displaystyle y = v^2$.


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

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

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

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

    $\displaystyle z^2 = (p_1^{k_1})^2 (p_2^{k_2}\dots p_n^{n_1})^2$ since $\displaystyle (p_2^{k_2}\dots p_n^{n_1})^2 = (y)$ wont be divisible by $\displaystyle (p_1^{k_1})^2 = (x)$ and hence our $\displaystyle \gcd(x,y) = 1$ 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 $\displaystyle (p_i^{k_1})^2=(x)$ ?? Just write $\displaystyle x,y,z$ as a product of primes, and check that every prime of $\displaystyle x$ and also of $\displaystyle y$ must have an even power...!

    Tonio
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by tonio View Post
    Not so neat but I think you've got the gist of the idea: but why would you decide that $\displaystyle (p_i^{k_1})^2=(x)$ ?? Just write $\displaystyle x,y,z$ as a product of primes, and check that every prime of $\displaystyle x$ and also of $\displaystyle y$ 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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