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Thread: Solve for integer values of x. - - - (A^B = 1 form of equation)

  1. #1
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    Solve for integer values of x. - - - (A^B = 1 form of equation)

    .


    Here is a problem I made up:

    Let x belong to the set of integers.

    That is, let x belong to the set containing the negative integers, zero, and the positive integers.


    $\displaystyle Let \ \ A \ = \ (x \ + \ (x - 1)\sqrt{x^2 - 1 \ } \ ), \ \ and \ \ let \ \ B \ = \ (-3x^2 - x + 4)$.

    Imaginary numbers will be allowed for the $\displaystyle \ \sqrt{x^2 - 1} \ $ expression.


    Solve $\displaystyle \ A^B = 1, \ \ $ for all integer values of x only.
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  2. #2
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    Re: Solve for integer values of x. - - - (A^B = 1 form of equation)

    Find the values where $A=1$, where $B=0, A\neq 0$, or where $A=\phi, n|B$ where $\phi$ is an $n$-th root of unity:

    $$1=x+(x-1)\sqrt{x^2-1}$$

    If $x\neq 1$:

    $$\sqrt{x^2-1}=-1$$

    Only solution over integers is $x=1$.

    $$-3x^2-x+4=0$$

    $$(3x+4)(-x+1)=0$$

    Only integer is $x=1$ again.

    Finally, we try $A=\phi$:

    $$\phi=x+(x-1)\sqrt{x^2-1}$$

    Again, if $x\neq 1$ we have:

    $$\left( \dfrac{\phi-x}{x-1} \right)^2=x^2-1$$

    $$(x-\phi)^2=(x^2-1)(x-1)^2$$

    Since the RHS is an integer, so too must the left hand side be an integer. This means that $x=-1$ or $x=0$.

    Therefore the solution is $x=-1,0,1$.
    Last edited by SlipEternal; Aug 7th 2018 at 04:00 PM.
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