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Math Help - Solve the equation

  1. #1
    Senior Member
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    Solve the equation

    Solve the equation for integers:
    2^{\frac {x-y}y}-\frac 32y=1.
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  2. #2
    Junior Member
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    Mar 2008
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    by guessing.....................lol'
    x=4,y=2
    want to get an integer, so y should be multiple of 2, so i like 2.....
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  3. #3
    Super Member

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    Hello, james_bond!

    Solve the equation for integers: . 2^{\frac {x-y}y}-\frac 32y\;=\;1
    We have: . 2^{\frac{x-y}{y}} \;=\;\frac{3}{2}y + 1\;\;{\color{blue}[1]}

    Since the right side is rational, the exponent must be an integer.
    . . Hence, the right side is a power-of-2.

    Then: . \frac{3}{2}y + 1 \:=\:2^k for some positive integer k.
    . . And: . y \:=\:\frac{2}{3}\left(2^k-1\right)

    So 2^k-1 must be divisible by 3.
    . . This happens when k is even: . k \,=\,2n

    . . Hence, we have: . \boxed{y \;=\;\frac{2}{3}\left(2^{2n}-1\right)}\;\;{\color{blue}[2]}


    Then [1] becomes: . 2^{\frac{x-y}{y}} \;=\;2^{2n} \quad\Rightarrow\quad \frac{x-y}{y} \:=\:2n

    . . and we have: . x \:=\:y(2n+1)\;\;{\color{blue}[3]}

    Substitute [2] into [3]: . \boxed{x \;=\;\frac{2}{3}\left(2^{2n}-1\right)(2n+1)}


    Solution: . \begin{Bmatrix}x &=&\frac{2}{3}\left(2^{2n}-1\right)(2n+1) \\ \\[-3mm] y &=& \frac{2}{3}\left(2^{2n}-1\right) \end{Bmatrix}\quad\text{ for }n \in I^+


    The first few solutions are: . (6,2),\;(50,10),\;(294,42),\;(1530,170),\;\hdots

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