Least $x$ Such That $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$

Question

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.

Answer 1

$$x = I+ F \qquad (I \in \mathbb Z, \quad 0 \le F < 1).$$

$$\lfloor 2IF+F^2\rfloor =10 \tag{A.}$$

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 $0 \le \epsilon < 1$.

Answer 2

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 $5 \leq x < \sqrt{35}.$

Answer 3

$$\lfloor{x}^{\mathrm{2}} \rfloor−\lfloor{x}\rfloor^{\mathrm{2}} =\mathrm{10} \\ $$ $$\ast \\ $$ $${x}^{\mathrm{2}} −\lfloor{x}\rfloor^{\mathrm{2}} =\mathrm{10}+\left[\mathrm{0};\mathrm{1}\right) \\ $$ $${x}^{\mathrm{2}} −\lfloor{x}\rfloor^{\mathrm{2}} =\left[\mathrm{10};\mathrm{11}\right) \\ $$ $$\ast\ast \\ $$ $${x}^{\mathrm{2}} =\left(\lfloor{x}\rfloor+\left[\mathrm{0};\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$ $${x}^{\mathrm{2}} −\lfloor{x}\rfloor^{\mathrm{2}} =\mathrm{2}\lfloor{x}\rfloor\left[\mathrm{0};\mathrm{1}\right)+\left[\mathrm{0};\mathrm{1}\right)^{\mathrm{2}} \\ $$ $${x}^{\mathrm{2}} −\lfloor{x}\rfloor^{\mathrm{2}} =\lfloor{x}\rfloor\left[\mathrm{0};\mathrm{2}\right)+\left[\mathrm{0};\mathrm{1}\right) \\ $$ $$\ast\ast\ast \\ $$ $$\lfloor{x}\rfloor\left[\mathrm{0};\mathrm{2}\right)+\left[\mathrm{0};\mathrm{1}\right)=\left[\mathrm{10};\mathrm{11}\right) \\ $$ $$\lfloor{x}\rfloor\left[\mathrm{0};\mathrm{2}\right)=\left(\mathrm{9};\mathrm{11}\right) \\ $$ $$\lfloor{x}\rfloor=\frac{\left(\mathrm{9};\mathrm{11}\right)}{\left[\mathrm{0};\mathrm{2}\right)}=\left[\mathrm{5};\infty\right) \\ $$ $$\ast\ast\ast\ast \\ $$ $${x}^{\mathrm{2}} =\left[\mathrm{10};\mathrm{11}\right)+\lfloor{x}\rfloor^{\mathrm{2}} \\ $$ $${x}_{{MIN}} =\sqrt{\mathrm{10}+\mathrm{5}^{\mathrm{2}} }=\sqrt{\mathrm{35}} \\ $$