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Thread: equation on the set of integers:

  1. #1
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    equation on the set of integers:

    Solve the following equation on the set of integers:

    $\displaystyle \sin(\frac{\pi}{3}(x-\sqrt{x^2-3x-12}))=0$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by perash View Post
    Solve the following equation on the set of integers:

    $\displaystyle \sin(\frac{\pi}{3}(x-\sqrt{x^2-3x-12}))=0$
    Which is asking you to find all the solutions of:

    $\displaystyle x-\sqrt{x^2-3x-12}\equiv 0 \mod 3$

    RonL
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  3. #3
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    Hello, perash!

    I haven't solve it yet . . . It's quite involved . . .


    Solve the following equation on the set of integers:

    $\displaystyle \sin\left[\frac{\pi}{3}\left(x-\sqrt{x^2-3x-12}\right)\right]\:=\:0$
    The Captain is absolutely correct . . .


    We have: .$\displaystyle \sin\left[\pi\left(\frac{x - \sqrt{x^2-3x-12}}{3}\right)\right] \:=\:0$

    We know that: .$\displaystyle \sin(\pi k) \:=\:0$ . . . for any integer $\displaystyle k.$

    Hence: .$\displaystyle \frac{x-\sqrt{x^2-3x-12}}{3}$ must be an integer.

    . . We can let: .$\displaystyle \frac{x-\sqrt{x^2-3x-12}}{3} \:=\:k$


    There are more restrictions:

    . . $\displaystyle x^2-3x-12$ must be nonnegative: .$\displaystyle x^2-3x-12 \:\geq\:0\quad\Rightarrow\quad x \leq -3,\;x \geq 6$

    . . And, of course, $\displaystyle x^2-3x-12$ must be a square.

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  4. #4
    Senior Member JaneBennet's Avatar
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    So far I’ve found three possible values for x: −13, 7, 16.
    Last edited by JaneBennet; Dec 22nd 2007 at 08:58 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    $\displaystyle x-\sqrt{x^2-3x-12}\equiv 0 \mod 3$
    Okay, my stab at this is confusing me.

    $\displaystyle x - \sqrt{x^2 - 3x - 12} \equiv 0~\text{ mod 3}$

    $\displaystyle x - \sqrt{x^2} \equiv 0~\text{ mod 3}$

    $\displaystyle x - |x| \equiv 0~\text{ mod 3}$

    Isn't this an identity in mod 3? But the original equation is not true for all x within the domain.

    -Dan
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  6. #6
    Senior Member JaneBennet's Avatar
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    You can’t remove multiples of 3 within the square root. $\displaystyle \sqrt{4-3\times1}$ ≢ $\displaystyle \sqrt{4}\pmod{3}$.
    Last edited by JaneBennet; Dec 22nd 2007 at 11:28 AM. Reason: Wrong example
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JaneBennet View Post
    You can’t remove multiples of 3 within the square root. $\displaystyle \sqrt{4+3\times7}$ ≢ $\displaystyle \sqrt{4}\pmod{3}$.
    Hmmm... I had never noticed that before. Thank you.

    (Upon reflection I should have known that. I studied that section of Algebra at one point.)


    By the way, if you like, you can do the "not equivalent" in LaTeX: $\displaystyle \not \equiv$ by typing \not \equiv

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JaneBennet View Post
    So far Iíve found three possible values for x: −13, 7, 16.
    16 doesn't work.

    I have "reduced" the problem to solving the Diophantine equation:
    $\displaystyle n^2 - 4m^2 = 57$
    (I can explain how I got this, but I'd like the original poster to do some work on their own.)

    So
    $\displaystyle (n + 2m)(n - 2m) = 57$

    There are going to be some obvious limits on m and n here because n + 2m has to be one of 1, 3, 19, 57.

    So we are looking the solution to one of two systems of equations:
    $\displaystyle n + 2m = 57$
    and
    $\displaystyle n - 2m = 1$

    or
    $\displaystyle n + 2m = 19$
    and
    $\displaystyle n - 2m = 3$

    The first has a solution of m = 14 and n = 29, giving an x value of -13. The second has a solution of m = 4 and n = 11, giving an x value of 7.

    So the only two solutions are x = -13 and x = 7.

    -Dan
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