$\{x^2\} = \{x\}^2$, how many solutions in interval $[1, 10]$

Question

Find how many solutions there are in the interval $[1, 10]$ to the fractional part equation: $$\left\{x\right\}^2 = \left\{x^2\right\}$$ Where $\{\cdot\}$ is the fractional part function, meaning that: $$\left\{a\right\} = a - \left\lfloor a \right\rfloor$$

Some research about the problem:

I graphed both functions on a Graphing Calculator:

And the problem was looking like it had a tremendous amount of solutions!

Approach

The equation is equivalent to: $$(x - \lfloor x\rfloor)^2 = (x^2 - \lfloor x^2\rfloor)$$

So: $$x^2 - 2x\lfloor x\rfloor + \lfloor x\rfloor^2 = x^2 - \lfloor x^2\rfloor$$

Further investigations lead to: $$2x\lfloor x \rfloor \in \mathbb{Z}$$

In which I hope I could find a clue for solving, using divisibility arguments, however no information appeared obvious to me from this.

(References: the graph was made using GeoGebra Graphing)

Answer 1

Let $x=i+f, 0\le f<1$ (integer plus fractional parts).

The equation turns to

$$\{(i+f)^2\}=f^2$$ which simplifies to $$2if=n$$ for some $n$.

Hence the solutions come with all fractions $$f=\frac{n}{2i}$$ with $0\le n <2i$.

Now count the possible values of $n$ for $i\in[1,10]$.

Answer 2

Wow it's have so many solutions...

$\left\{{x}^{\mathrm{2}} \right\}={x}^{\mathrm{2}} −\left[{x}^{\mathrm{2}} \right]=\left\{{x}\right\}^{\mathrm{2}} \\ $ ${x}^{\mathrm{2}} −\left\{{x}\right\}^{\mathrm{2}} =\left[\left(\left[{x}\right]+\left\{{x}\right\}\right)^{\mathrm{2}} \right] \\ $ $\left[{x}\right]\left(\left[{x}\right]+\mathrm{2}\left\{{x}\right\}\right)=\left[{x}\right]^{\mathrm{2}} +\left[\mathrm{2}\left[{x}\right]\left\{{x}\right\}\right] \\ $ $\mathrm{2}\left[{x}\right]\left\{{x}\right\}=\left[\mathrm{2}\left[{x}\right]\left\{{x}\right\}\right] \\ $ $\left(\mathrm{2}\left[{x}\right]\left\{{x}\right\}\right)\in{Z} \\ $ ${x}\in\left\{\mathrm{1},\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}},\mathrm{2},\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}},\mathrm{2}\frac{\mathrm{2}}{\mathrm{4}},\mathrm{2}\frac{\mathrm{3}}{\mathrm{4}},\right. \\ $ $\:\:\:\:\:\:\:\:\mathrm{3},\mathrm{3}\frac{\mathrm{1}}{\mathrm{6}},\mathrm{3}\frac{\mathrm{2}}{\mathrm{6}},\mathrm{3}\frac{\mathrm{3}}{\mathrm{6}},\mathrm{3}\frac{\mathrm{4}}{\mathrm{6}},\mathrm{3}\frac{\mathrm{5}}{\mathrm{6}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{4},\mathrm{4}\frac{\mathrm{1}}{\mathrm{8}},\mathrm{4}\frac{\mathrm{2}}{\mathrm{8}},\mathrm{4}\frac{\mathrm{3}}{\mathrm{8}},\mathrm{4}\frac{\mathrm{4}}{\mathrm{8}},\mathrm{4}\frac{\mathrm{5}}{\mathrm{8}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{4}\frac{\mathrm{6}}{\mathrm{8}},\mathrm{4}\frac{\mathrm{7}}{\mathrm{8}},\mathrm{5},\mathrm{5}\frac{\mathrm{1}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{2}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{3}}{\mathrm{10}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{5}\frac{\mathrm{4}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{5}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{6}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{7}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{8}}{\mathrm{10}},\mathrm{5}\frac{\mathrm{9}}{\mathrm{10}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{6}\frac{\mathrm{1}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{2}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{3}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{4}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{5}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{6}}{\mathrm{12}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{6}\frac{\mathrm{7}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{8}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{9}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{10}}{\mathrm{12}},\mathrm{6}\frac{\mathrm{11}}{\mathrm{12}},\mathrm{7},\mathrm{7}\frac{\mathrm{1}}{\mathrm{14}} \\ $ $\:\:\:\:\:\:\:\:\mathrm{7}\frac{\mathrm{2}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{3}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{4}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{5}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{6}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{7}}{\mathrm{14}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{7}\frac{\mathrm{8}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{9}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{10}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{11}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{12}}{\mathrm{14}},\mathrm{7}\frac{\mathrm{13}}{\mathrm{14}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{8},\mathrm{8}\frac{\mathrm{1}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{2}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{3}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{4}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{5}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{6}}{\mathrm{16}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{8}\frac{\mathrm{7}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{8}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{9}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{10}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{11}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{12}}{\mathrm{16}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{8}\frac{\mathrm{13}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{14}}{\mathrm{16}},\mathrm{8}\frac{\mathrm{15}}{\mathrm{16}},\mathrm{9},\mathrm{9}\frac{\mathrm{1}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{2}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{3}}{\mathrm{18}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{9}\frac{\mathrm{4}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{5}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{6}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{7}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{8}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{9}}{\mathrm{18}}, \\ $ $\:\:\:\:\:\:\:\:\mathrm{9}\frac{\mathrm{10}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{11}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{12}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{13}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{14}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{14}}{\mathrm{18}}, \\ $ $\left.\:\:\:\:\:\:\:\:\mathrm{9}\frac{\mathrm{15}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{16}}{\mathrm{18}},\mathrm{9}\frac{\mathrm{17}}{\mathrm{18}}, \mathrm{10}\right\} \\ $

Answer 3

Short way.

Write $x=n+\frac{k}{n}$ where $n$ is integer, $0 \leq k < n$, $k$ not necessarily an integer. Then:

$$\{x^2 \}=\{ (n+\frac{k}{n})^2 \}=\{ n^2+2k+(\frac{k}{n})^2 \}=\{2k+(\frac{k}{n})^2 \}$$

аnd

$$\{x \}^2=\{ n+\frac{k}{n} \}^2=\{ \frac{k}{n} \}^2$$

That means that we need to have

$$\{2k+(\frac{k}{n})^2 \}=\{ \frac{k}{n} \}^2$$

and this is possible only if $2k$ is an integer, since $k<n$. And this is giving $2n$ solution for each $n$ giving the total number of solutions:

$$1+\sum_{n=1}^{9}2n=9\cdot10+1=91$$

(Notice that $10$ participates only once.)