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Math Help - A much needed proof

  1. #1
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    A much needed proof

    Given integer n>1, prove that we can find integer k>1 and a1, a2, ......ak >1 such that
    a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)

    Does this need some complex theorem to start with? Much thanks!
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  2. #2
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    Quote Originally Posted by gentsl View Post
    Given integer n>1, prove that we can find integer k>1 and a1, a2, ......ak >1 such that
    a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)

    Does this need some complex theorem to start with? Much thanks!
    Say a_k = x then on LHS we have k*x on RHS we have n*k/x.
    So we require that x^2 = n so chose x = sqrt(n) for all a_1,...,a_k
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  3. #3
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    This means n = +/- sqrt a_k therefore k >1 must apply ? smilarly for a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)?
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  4. #4
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    Quote Originally Posted by gentsl View Post
    This means n = +/- sqrt a_k therefore k >1 must apply ? smilarly for a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)?
    No it means that if you set a_1=a_2= .. =a_k=\sqrt{n} then the left hand side equals k\sqrt{n} and the right hand side is equal to n \left(\frac{k}{\sqrt{n}}\right)=k\sqrt{n} , so the condition of the problem is satisfied by these values (at least it is when n is a perfect square, otherwise the a's are not integers and we still need a solution).

    RonL
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