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

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)

• You will like this post. Nov 5, 2021 at 7:56
• @TeresaLisbon I was about to make a similar observation: $(n+1)^2-n^2=2n+1$, hence in $[n,n+1[$, $\{x^2\}$ have to "return to the origin" $2n+1$ times. That leaves $2n$ roots, because the last intersection point would be at exactly $n+1$, which is out of the interval. Nov 5, 2021 at 8:03
• @Jean-ClaudeArbaut Thankfully, you can use the resource and your observations to write an answer, so I'll look forward to that! Nov 5, 2021 at 8:06
• You missed the solution $x=1$. Nov 5, 2021 at 8:13
• Nov 5, 2021 at 8:32

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 $$i\in[1,10]$$.

• If $0 \leq n < 2i$, than all the solutions $\frac{n}{2i}$ are under $1$, so they could not be counted in $[1, 10]$. Nov 5, 2021 at 8:09
• @MathStackExchange: see my update. Nov 5, 2021 at 8:11
• Great. Now it's working with the mention that for $i = 10$, to stay in the interval, $f$ should be $0$... Nov 5, 2021 at 8:15
• @MathStackExchange: of course, for $i=10$, $n=0$ is the only possible value. Nov 5, 2021 at 8:31

Wow it's have so many solutions...

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

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 $$k. And this is giving $$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.)