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Thread: Solve for positive real x

  1. #1
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    Solve for positive real x

    Solve for positive real x

    $\displaystyle \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
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  2. #2
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    Quote Originally Posted by perash View Post
    Solve for positive real x

    $\displaystyle \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
    It should be obvious from the definition of "floor" that $\displaystyle \lfloor x\rfloor+ 1> x$ for all positive x so there is no positive x satisfying that equation.
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  3. #3
    Member billa's Avatar
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    try x=3, and x=10000
    I think they work

    floor(x) +1 is less than x but that is not the case in this problem because we are taking the sq root of x.

    I couldnt solve this, but it doesnt work for any perfect squares and x has to be a whole number.
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by perash View Post
    Solve for positive real x

    $\displaystyle \lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$
    There may be a quick proof using fixed-point theorems, but letís consider the following algebraic method.

    Since the LHS is an integer, all solutions are (positive) integers. I claim that all solutions are of the form $\displaystyle x=n^2+2n,\ n\in\mathbb{Z}^+.$

    Note that $\displaystyle \lfloor\sqrt{n^2+2n}\rfloor=n.$ This is because $\displaystyle n<\sqrt{n^2+2n}<n+1$ for all $\displaystyle n\in\mathbb{Z}^+.$

    And $\displaystyle \lfloor n\sqrt{n^2+2n}\rfloor=n^2+n-1$ because $\displaystyle \forall n\in\mathbb{Z}^+,$

    $\displaystyle n^2+n-1\ <\ n\sqrt{n^2+2n}\ <\ n^2+n$

    i.e. $\displaystyle n^4+2n^3-n^2-2n+1\ <\ n^4+2n^3\ <\ n^4+2n^3+n^2$

    This proves that all $\displaystyle x=n^2+2n,\ n\in\mathbb{Z}^+,$ are solutions. Now we must show that there are no other solutions.

    Any other solution would have to be of the form $\displaystyle x=n^2+2n-k,$ where $\displaystyle 1\leqslant k\leqslant2n.$ Suppose there is such a solution.

    Again we have $\displaystyle \lfloor\sqrt{n^2+2n-k}\rfloor=n$ since $\displaystyle n\leqslant\sqrt{n^2+2n-k}<n+1$ for all $\displaystyle 1\leqslant k\leqslant2n.$

    Then, in order for the solution to hold, we would need to have $\displaystyle \lfloor n\sqrt{n^2+2n-k}\rfloor=n^2+n-k-1.$ This would mean that we would need

    $\displaystyle n\sqrt{n^2+2n-k}\ <\ n^2+n-k$

    i.e. $\displaystyle n^4+2n^3-kn^2\ <\ n^4+2n^3+n^2-2kn^2-2kn+k^2$

    i.e. $\displaystyle k^2-(n^2+2n)k\ >\ -n^2$

    i.e. $\displaystyle \left[k-\frac{n^2+2n}2\right]^2\ >\ \frac{\left(n^2+2n\right)^2}4-n^2=\frac{n^4+4n^3}4\quad\ldots\fbox{1}$

    Now $\displaystyle 1\leqslant k\leqslant2n$ $\displaystyle \Rightarrow$ $\displaystyle k-\frac{n^2+2n}2\leqslant\frac{2n-n^2}2$ and $\displaystyle \frac{n^2+2n}2-k\leqslant\frac{n^2+2n-2}2.$

    If $\displaystyle k-\frac{n^2+2n}2\geqslant0,$ then $\displaystyle \left[k-\frac{n^2+2n}2\right]^2\leqslant\frac{\left(2n-n^2\right)^2}4=\frac{n^4-4n^3+4n^2}4<\frac{n^4+4n^3}4.$

    If $\displaystyle \frac{n^2+2n}2-k\geqslant0,$ then $\displaystyle \left[\frac{n^2+2n}2-k\right]^2\leqslant\frac{\left(n^2+2n-2\right)^2}4=\frac{n^4+4n^3-8n+4}4<\frac{n^4+4n^3}4.$

    In either case, we have a clear contradiction of $\displaystyle \fbox{1}.$

    This completes the proof.
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