Solve for positive real x

perash · #1 $Old$ November 5th, 2008, 01:02 PM

Solve for positive real x

$\lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$

HallsofIvy · #2 $Old$ November 6th, 2008, 05:49 AM

Quote:

Originally Posted by perash $View Post$

Solve for positive real x

$\lfloor \sqrt{x}\lfloor \sqrt{x} \rfloor \rfloor +\lfloor \sqrt{x}\rfloor +1=x$

It should be obvious from the definition of "floor" that $\lfloor x\rfloor+ 1> x$ for all positive x so there is no positive x satisfying that equation.

billa · #3 $Old$ November 6th, 2008, 09:59 AM

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.

JaneBennet · #4 $Old$ November 10th, 2008, 05:46 AM

Quote:

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.