Find the smallest positive real $x$ such that $\lfloor{x^2}\rfloor - x\lfloor{x}\rfloor = 6$

Question

Problem

Find the smallest positive real $x$ such that $\lfloor{x^2}\rfloor - x\lfloor{x}\rfloor = 6$.

What I've done

I set $m = \lfloor x \rfloor$ and $n = \{x\}$. Then I proceeded as below:

$\lfloor (m+n)^2 \rfloor -(m+n)m = 6$

Since $m$ is a non-negative integer, thus $\lfloor (m+n)^2 \rfloor = \lfloor m^2 \rfloor + \lfloor 2mn + n^2 \rfloor = m^2+ \lfloor 2mn + n^2 \rfloor$.

Then I got $\lfloor 2mn+n^2 \rfloor -mn = 6$

Where I'm stuck

I don't know how to go on.

Answer 1

First note that if $m=0$ then the equation becomes $0=6,$ which is absurd, so $m>0$.

Then, note $$k = mn = \lfloor 2mn+n^2 \rfloor - 6$$ is an integer and $k/m = n < 1,$ so $k=mn < m$.

Then the last equation just states $k=6$ and since $6=k<m$ the smallest $m$ can be is $7$.

We conclude $$x = m + n = m + \frac{k}{m} = 7 + \frac67$$

Answer 2

First we can find $x \not\in \mathcal{I}$, otherwise $\lfloor x^2 \rfloor - x\lfloor x\rfloor = 0$

Now write $x$ as $x \equiv l + \frac{n}{m}$, with $l,m,n \in \mathcal{N}$ and $m>n>0$

Another point we can quickly observe is that $l=m$, otherwise $\lfloor x^2 \rfloor - x\lfloor x\rfloor \not \in \mathcal{I}$

Thus $$\left\lfloor\left (m+\frac{n}{m}\right)^2 \right\rfloor - (m^2+n) = 0$$ You can quickly find $n=6$ Thus the smallest $m$ is $m=7$

So the answer is $7\frac{6}{7}$