Thread: Solve for positive real x
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Old November 10th, 2008, 05:46 AM
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Originally Posted by perash View Post
Solve for positive real x

\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 let’s 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 x=n^2+2n,\ n\in\mathbb{Z}^+.

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

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

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

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

This proves that all 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 x=n^2+2n-k, where 1\leqslant k\leqslant2n. Suppose there is such a solution.

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

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

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

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

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

i.e. \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 1\leqslant k\leqslant2n \Rightarrow k-\frac{n^2+2n}2\leqslant\frac{2n-n^2}2 and \frac{n^2+2n}2-k\leqslant\frac{n^2+2n-2}2.

If k-\frac{n^2+2n}2\geqslant0, then \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 \frac{n^2+2n}2-k\geqslant0, then \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 \fbox{1}.

This completes the proof.
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