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Thread: One equation, three unknowns

  1. #1
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    One equation, three unknowns

    Does the equation

    \dfrac{a}{b} \ + \ \dfrac{b}{c} \ + \ \dfrac{c}{a} \ = \ 1

    have any solutions for a, b, c, which belong to the set of negative and positive integers?
    Last edited by greg1313; Dec 23rd 2016 at 06:35 AM.
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    Re: One equation, three unknowns

    It helps if you tell us what you have tried or thought about. Otherwise, you learn little from the exercise.

    $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} = 1 \implies abc * \left ( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \right ) = abc * 1 \implies$

    $a^2c + ab^2 + bc^2 = abc.$

    $CASE\ I:\ ASSUME\ a, b, c \in \mathbb Z\ and\ a, b, c > 0.$

    $a^2c + ab^2 + bc^2 = abc \implies a^2c < abc \implies \dfrac{a^{\cancel {2}}\cancel c}{\cancel a \cancel c} < \dfrac{\cancel a b \cancel c}{\cancel a \cancel c} \implies a < b.$

    $Also\ a^2c + ab^2 + bc^2 = abc \implies b^2c < abc \implies \dfrac{b^{\cancel {2}}\cancel c}{\cancel b \cancel c} < \dfrac{a \cancel b \cancel c}{\cancel b \cancel c} \implies b < a.$

    $Contradiction \implies assumption\ that\ a, b, c \in \mathbb Z\ and\ a, b, c > 0\ is\ FALSE.$

    Now what?
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    Re: One equation, three unknowns

    Quote Originally Posted by JeffM View Post
    Now what?
    a, b, c belong to the set of the negative and the positive integers


    Case II:

    a < b < c,

    a, b < 0 and c > 0
    Last edited by greg1313; Dec 23rd 2016 at 10:29 PM.
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    Re: One equation, three unknowns

    Quote Originally Posted by greg1313 View Post
    a, b, c belong to the set of the negative and the positive integers


    Case II:

    a < b < c,

    a, b < 0 and c > 0
    First, if you are going to do this by cases, your second case needs to be a, b, and c all negative.

    Second, in your original statement of the problem you did not stipulate that a < b < c. Is that a stipulation or not?

    Third if it is NOT a stipulation, you can without loss of generality set up case III as

    $a < 0 < b \le c$ and case IV as $a \le b < 0 < c.$

    Now all that needs to be done is to figure out which if any of those cases is feasible.
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    Re: One equation, three unknowns

    Interesting li'l headacher, Greg.

    One thing I'm sure of:
    no solution for a,b,c = -999 to 999 skip 0

    A quadratic in terms of c gives following restrictions:
    (let u = a^2 - ab, v = 4ab^3)
    u^2 - v => 0
    SQRT(u^2 - v) = integer
    -u +- SQRT(u^2 - v) = even

    Quadratics in terms of a and b give similar restrictions, of course.
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    Re: One equation, three unknowns

    Quote Originally Posted by JeffM View Post
    First, if you are going to do this by cases, your second case needs to be a, b, and c all negative.
    No, I wouldn't make that my second case, because I know all of the negative signs would cancel out, and that
    would simplify it back to Case I.


    Second, in your original statement of the problem you did not stipulate that a < b < c. Is that a stipulation or not?
    No, that is not a stipulation of the original problem.

    Third if it is NOT a stipulation, you can without loss of generality set up case III as

    $a < 0 < b \le c$ and case IV as $a \le b < 0 < c.$

    Now all that needs to be done is to figure out which if any of those cases is feasible.
    Case IV may wind up being feasible for the integers (less 0), because there is a solution in real numbers I found:

    a =  -2

    b =  2 - 2\sqrt{2} \ \approx \ -0.8

    c =  2


    \dfrac{-2}{2 - 2\sqrt{2}} \ + \ \dfrac{2 - 2\sqrt{2}}{2} \ + \ \dfrac{2}{-2} \ = \ 1
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