Solving An Equation with the Floor Function

Expand, simplify and rearrange, we get $\large x = \frac { \lfloor x \rfloor ^2 + \lfloor x^2 \rfloor }{2 \lfloor x \rfloor }$

For $\lfloor x \rfloor = 1$ , then $1 \leq x < 2$ , and $1 \leq x^2 < 4$ , values of $\lfloor x^2 \rfloor$ are $1,2,3$ , then $x = \frac {2}{2}, \frac {3}{2}$ , a total of $2$ values

Similarly, For $\lfloor x \rfloor = 2$ , then $2 \leq x < 3$ , and $4 \leq x^2 < 9$ , values of $\lfloor x^2 \rfloor$ are $4,5,6,7,8$ , then $x = \frac {8}{4}, \frac {9}{4}, \frac {10}{4}, \frac {11}{4}$ , a total of $4$ values

Repeat: for $\lfloor x \rfloor = 3$ , then $9 \leq x^2 < 16$ , values of $\lfloor x^2 \rfloor$ are $9, 10,11,12,13,14,15$ , then $x = \frac {18}{6}, \frac {19}{6}, \frac {20}{6}, \frac {21}{6}, \frac {22}{6}, \frac {23}{6}$ , a total of $6$ values

Apply the same argument, for $\lfloor x \rfloor = 4,5,6,7,8,9$ , total of values are $8,10,12,14,16,18$ respectively

And lastly, $x=10$ is the largest solution.

Add them all up yields: $2+4+6+8+10+12+14+16+18+1 = \boxed{91}$

Write $x = n+y$ , $n$ an integer, $0 \le y < 1$ . Then the left side is $(n+y)^2 - \lfloor (n+y)^2 \rfloor$ and the right side is $y^2$ . Equating and canceling gives $n^2 + 2ny - \lfloor n^2 + 2ny + y^2 \rfloor = 0$ , but we can take the $n^2$ out of the floor since it's an integer. Hence $2ny - \lfloor 2ny + y^2 \rfloor = 0,$ which happens if and only if $2ny$ is an integer (since $0 \le y^2 < 1$ ).

For each $n$ between $1$ and $9$ , there are $2n$ such $y$ , namely $\frac{0}{2n}, \frac1{2n}, \ldots, \frac{2n-1}{2n}$ . And we can't forget $n = 10, y = 0$ . So the answer is $(2 + 4 + \cdots + 16 + 18) + 1 = \fbox{91}$ .

Main idea : Manipulate the equation to obtain $x$ as a rational of a certain form, then break the remaining equation with this new expression of $x$ .

Complete proof :

We manipulate the equation as follows:

$x^2 - \lfloor x^2 \rfloor = (x - \lfloor x \rfloor)^2$

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

$\lfloor x^2 \rfloor + \lfloor x \rfloor^2 = 2x \lfloor x \rfloor$

Note that the left hand side is an integer, so $x = \dfrac{k}{2 \lfloor x \rfloor}$ for some integer $k$ to make the right hand side to be an integer too. Let $x = a + y$ where $a$ is an integer and $y \in [0,1)$ , so $x = \dfrac{k}{2a}$ for some integer $k$ , so $x = a + \dfrac{b}{2a}$ for some integers $a \ge 1$ , $b \in [0,2a)$ . Note that $\lfloor x \rfloor = a$ .

Plugging in, we have:

$\left\lfloor \left( a + \dfrac{b}{2a} \right)^2 \right\rfloor + a^2 = 2a \left( a + \dfrac{b}{2a} \right)$

$\left\lfloor \left( a + \dfrac{b}{2a} \right)^2 \right\rfloor + a^2 = 2a^2 + b$

$\left\lfloor a^2 + b + \left( \dfrac{b}{2a} \right)^2 \right\rfloor = a^2 + b$

Since $0 \le \dfrac{b}{2a} < 1$ , it follows that $0 \le \left( \dfrac{b}{2a} \right)^2 < 1$ , and so the last equation is always true for any positive integer $a$ and integer $b \in [0, 2a)$ .

Now, note that $a$ ranges from $1$ to $9$ ; the last value $x = 10$ is handled later. For each of these $a$ , we have $2a$ values of $b$ and hence $2a$ values of $x$ satisfying the equation, so in total, for $a = 1,2,\ldots,9$ , we have $2+4+\ldots+18 = 90$ values of $x$ in $[1,10)$ . The last value $x = 10$ also satisfies the equation, bringing the total to $\boxed{91}$ .

Motivation : To be honest I didn't expect those fractional values. At first I was thinking about $x = \sqrt{n}$ or so, trying to mess things up with that, before finally struck by the idea of opening up the right part. The remaining of my train of thoughts follows the above.

The question has the equivalent form of solving for all $x\in [1,10]$ such that $\{x^2\}=\{x\}^2$ where $\{\cdot \}$ denotes the fractional part of $x$ . Now, $\{x^2\}=\{\{x\}^2+2\{x\}\lfloor x\rfloor+\lfloor x\rfloor ^2\}$ . This implies that $2\{x\}\lfloor x\rfloor$ must be a positive integer which in turn implies that $\{x\}$ must be rational. So, for a given $\lfloor x\rfloor=k$ this can happen in $2k$ ways if $1\le k\le 9$ . If $k=10$ this can happen in only one way. So, total number of such numbers is $2\sum_{k=1}^9 k +1=91$

Note that for $a\leq x < a+1$ , there are $2a$ values of $x$ such that the equation holds ( $x = \frac{\lfloor x \rfloor ^2 + \lfloor x^2 \rfloor}{2 \lfloor x \rfloor}$ - the result follows from the definition of the floor function). Hence, on $[1,10)$ , there are $2 \cdot \frac{9 \cdot 10}{2} = 90$ solutions, and of course all integers solve the equation, so there are a total of $\boxed{91}$ solutions.

the equation can be written as : { $x^{2}$ } = {x}^2 {a} denotes fractional part of a.
let q={x}. and a = floor of x.
{ $x^{2}$ } = { $q^2$ + 2aq}}.
$q^{2}$ = { $q^{2}$ +2aq}.
now let a=1.
$q^{2}$ = { $q^{2}$ +2q}.
floor of $q^{2}$ +2q can be 1 or 2.
for each floor their exists one possible value of q.
similarly for a=2 floor of $q^{2}$ +4q can be 1 , 2 , 3 or 4.
for each floor their exists one possible value of q.
similarly for a=1,2,3,4,5,6,7,8,9 there are 2,4,6,8,10,12,14,16,18 solutions respectively.
their exists one more solution for x=10.
thus there are a total of 2+4+6+8+10+12+14+16+18+1=91 values of x

After using $x= [x] + \theta$

We have $\theta^{2} +2[x]\theta] = 2[x]\theta$ .

Now since LHS is an integer RHS is also an integer.

Thus all x satisfying are 10 integers in range given.

But when $[x]= \xi$

We have $2\xi -1$ values of $\theta$ .

Hence total such values are 81.

Thus answer= $\boxed{91}$ .