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Math Help - X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

  1. #1
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    X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

    Okay, so I'm stuck on this for an hour now, so I'd really appreciate some help, please:

    X, Y, Z are coprime integers. We also know that 1/x + 1/y = 1/z. Prove that (X+Y) is a square number.

    -----
    This should be entry level university math, but I'm stuck with it. I hope I posted to the right topic. Thanks for the help.
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  2. #2
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    Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

    Are x, y, z pairwise coprime or coprime together?

    In mathematics, the case of variable names matters. For example, X and x are different variables.
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    Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

    I apologize. Please consider x=X and so on. My mistake.

    They are coprime together: The problem says that their greatest common divisor is 1.

    Thank you.
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    Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

    OK, here is a sketch. Suppose that p, q, r are prime, i, j₁, j₂ and k are positive integers, and x=p^iq^{j_1}, y=q^{j_2}r^k. In the general case, the variables here should be vectors. Without loss of generality, let j_1\le j_2. Then

    z=\frac{xy}{x+y} =\frac{p^iq^{j_1+j_2}r^k}{q^{j_1}(p^i+q^{j_2-j_1}r^k)} =\frac{p^iq^{j_2}r^k}{p^i+q^{j_2-j_1}r^k}.

    Let s=p^i+q^{j_2-j_1}r^k}. Then the only primes that could divide s are p, q and r. If p | s, then p\mid q^{j_2-j_1}r^k, a contradiction. Similarly, r | s is false. Thus s=q^j for some j. Moreover, j=j_2; otherwise q ∈ GCD(x, y, z). Finally, if j_2>j_1, then q | s implies q\mid p^i, a contradiction. Therefore, j_1=j_2=j, s=q^j and x+y=q^{2j}.

    See if you can shorten this and if it can be extended to the general case.
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  5. #5
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    Re: X, Y, Z are coprime integers. 1/x + 1/y = 1/z. Prove that (X+Y) is a square numbe

    Quote Originally Posted by emakarov View Post
    OK, here is a sketch. Suppose that p, q, r are prime, i, j₁, j₂ and k are positive integers, and x=p^iq^{j_1}, y=q^{j_2}r^k. In the general case, the variables here should be vectors. Without loss of generality, let j_1\le j_2. Then

    z=\frac{xy}{x+y} =\frac{p^iq^{j_1+j_2}r^k}{q^{j_1}(p^i+q^{j_2-j_1}r^k)} =\frac{p^iq^{j_2}r^k}{p^i+q^{j_2-j_1}r^k}.

    Let s=p^i+q^{j_2-j_1}r^k}. Then the only primes that could divide s are p, q and r. If p | s, then p\mid q^{j_2-j_1}r^k, a contradiction. Similarly, r | s is false. Thus s=q^j for some j. Moreover, j=j_2; otherwise q ∈ GCD(x, y, z). Finally, if j_2>j_1, then q | s implies q\mid p^i, a contradiction. Therefore, j_1=j_2=j, s=q^j and x+y=q^{2j}.

    See if you can shorten this and if it can be extended to the general case.
    To extend your ideas to the general case, may I suggest that we write x = da and y = db where d = gcd(x,y).
    From this we get dab=(a+b)z and using your method we show that d = a + b.
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