Least $x$ Such That $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$ A friend recently texted me the following: 

Compute the least $x$ such that $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$.

Is there a way to do what my friend is asking analytically? I graphed it on Desmos, and got $x\approx 5.91608$. I realized that this should be (and indeed appears to be) $\sqrt{35}$, as then we have $35-25=10$. To show this, I tried my typical way of solving floor problems, which is to break up $x=I+F$, where $I=\lfloor x\rfloor$ and $F=x-\lfloor x\rfloor$. After simplifications, I got $\lfloor F(2I+F)\rfloor =10$. But then upon any further manipulation, I get back the original equation. Is there some further manipulation I am not aware of that lets me solve this analytically? Or am I on the wrong track entirely? Should I perhaps try to prove that $\sqrt{35}$ is the least value of $x$ satisfying the above condition? Sorry if I’m missing something obvious.
 A: I hid most of the steps in "spoiler" sections so you can mouse over them to see part of the answer without showing everything.
We can rule out $x < 0,$ since

 if $x < 0$ then $\lfloor x\rfloor \leq x < 0$ and therefore $\lfloor x\rfloor^2 \geq x^2 \geq \lfloor x^2\rfloor.$

For $x > 0,$ we have $x < \lfloor x+1\rfloor,$
from which $x^2 < \lfloor x+1\rfloor^2$
and therefore $\lfloor x^2\rfloor < \lfloor x+1\rfloor^2.$
It follows that  

 $$\lfloor x^2\rfloor - \lfloor x\rfloor^2 < \lfloor x+1\rfloor^2 - \lfloor x\rfloor^2 = 2\lfloor x\rfloor + 1.$$

So

 you can have $\lfloor x^2\rfloor - \lfloor x\rfloor^2 \geq 10$ only if $2\lfloor x\rfloor + 1 > 10,$ which implies $\lfloor x\rfloor > 4.5,$ which implies $\lfloor x\rfloor\geq 5.$

So now you just need to prove that

 $\lfloor x^2\rfloor - \lfloor x\rfloor^2 < 10$ for $5 \leq x < \sqrt{35}.$

A: $$x = I+ F \qquad (I \in \mathbb Z, \quad 0 \le F < 1).$$
$$\lfloor 2IF+F^2\rfloor =10 \tag{A.}$$
A quick check shows that $I$ needs to be positive.
Let's first consider what happens when 
$$F^2 + 2IF = 10 \tag{B.}$$
By the quadratic equation, 
$$F = \dfrac{-2I+\sqrt{4I^2+40}}{2} 
    = -I+\sqrt{I^2+10}$$
This clearly implies $F \ge 0$. We also need
\begin{align}
   F &< 1 \\
   -I+\sqrt{I^2+10} &< 1 \\
   \sqrt{I^2 + 10} &< I + 1 \\ 
   I^2 + 10 &< I^2 + 2I + 1 \\
   10 &< 2I + 1 \\
   I &\ge 5
\end{align}
Which leads, unsurprisingly, t0 $x = \sqrt{I^2+10}$ with the restriction $I \ge 5$.
For $I = 5$, we get $F = -5 + \sqrt{35}$. So $x = \sqrt{35}$.
$\lfloor x^2 \rfloor -\lfloor x\rfloor ^2 = 35 - 25 = 10$ 
In general, there will be a solution when 
$$10 \le \lfloor 2IF+F^2\rfloor < 11$$
So our more general equation must have the form
$$ F^2 + 2IF = 10 + \epsilon$$
where $0 \le \epsilon < 1$. Then we would get
$$F = -I+\sqrt{I^2+10 + \epsilon}$$
and $$x = \sqrt{I^2+10 + \epsilon}$$
where $0 \le \epsilon < 1$.
A: $$\left[{x}^{\mathrm{2}} \right]−\left[{x}\right]^{\mathrm{2}} =\mathrm{10} \\ $$
$$\left[\left(\left[{x}\right]+\left\{{x}\right\}\right)^{\mathrm{2}} \right]−\left[{x}\right]^{\mathrm{2}} =\mathrm{10} \\ $$
$$\left[{x}\right]^{\mathrm{2}} +\left[\mathrm{2}\left[{x}\right]\left\{{x}\right\}\right]−\left[{x}\right]^{\mathrm{2}} =\mathrm{10} \\ $$
$$\mathrm{2}\left[{x}\right]\left\{{x}\right\}\in\left[\mathrm{10},\mathrm{11}\right) \\ $$
$$\left\{{x}\right\}=\left[\frac{\mathrm{10}}{\mathrm{2}\left[{x}\right]},\frac{\mathrm{11}}{\mathrm{2}\left[{x}\right]}\right) \\ $$
$${x}\left({MIN}\right)\Rightarrow\left[{x}\right]=\mathrm{6} \\ $$
$${x}\left({MIN}\right)=\mathrm{6}+\frac{\mathrm{10}}{\mathrm{2}\cdot\mathrm{6}}=\mathrm{6}\frac{\mathrm{5}}{\mathrm{6}}. \\ $$
