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Math Help - Difficult square number problem

  1. #1
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    Difficult square number problem

    A pair of positive integers (x, y) is called 'square' if both x + y and xy are perfect square numbers. E.g. (5, 20) is 'square' since 5 + 20 = 25 and 5 x 20 = 100, both which are perfect squares.

    Prove no square pair exists in which one of its numbers is 3.

    I'm not sure how to do this, but I think it involves mods in the solution.

    Please help, thanks, BG
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  2. #2
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    Hello BG5965
    Quote Originally Posted by BG5965 View Post
    A pair of positive integers (x, y) is called 'square' if both x + y and xy are perfect square numbers. E.g. (5, 20) is 'square' since 5 + 20 = 25 and 5 x 20 = 100, both which are perfect squares.

    Prove no square pair exists in which one of its numbers is 3.

    I'm not sure how to do this, but I think it involves mods in the solution.

    Please help, thanks, BG
    There may be a shorter route to the contradiction, but the following works OK.

    Suppose that (3,x) is a square pair. Then, for some p, q:

    3+x = p^2 and 3x=q^2

    \Rightarrow q is a multiple of 3; say q = 3r

    Then 3x = q^2 = 9r^2

    \Rightarrow x = 3r^2

    \Rightarrow 3 + x = 3+3r^2 = p^2

    \Rightarrow 3(1+r^2) = p^2

    \Rightarrow p is a multiple of 3; say p = 3s

    \Rightarrow p^2 = 9s^2 = 3(1+r^2)

    \Rightarrow 1+r^2 = 3s^2

    \Rightarrow r^2 \equiv 2 \mod 3

    But this is impossible, since 0^2\equiv 0 and 1^2 \equiv 2^2 \equiv 1 \mod 3.

    Hence the square pair (3, x) does not exist.

    Grandad
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  3. #3
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    Please don't interpret this as a useless post - but I'm pretty sure that my teacher mentioned mod 4, unless I was mistaken?

    Anyway, Grandad, your proof makes sense, but I'm not sure if it's the approach we're supposed to use. Can anyone else use mod 4?
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