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Math Help - integer squares problem

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

    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?
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    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
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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 (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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