Integral part and fractional part

"The decimal part stays the same" means that we want $f(x)=x^2-x$ to be an integer. (In the example given in the problem, that integer is $17.14005-4.14005=13.$ ) Since $f(x)=x^2-x$ on $[4,5]$ is an increasing, continuous function, with range $[f(4),f(5)]=[12,20]$ , the 7 integers 13, 14,...19 are in the range.

It is given that the numbers and their square have the same fractional part i.e.

$\large\displaystyle \left\{x\right\} = \left\{x^2\right\}$

(Here $\left\{x\right\}$ = fractional part of x )

Now $x = \lfloor x \rfloor + \left\{x\right\}$

$x^2 = \lfloor x^2 \rfloor + \left\{x^2\right\} = \lfloor x^2 \rfloor + \left\{x\right\}$

Therefore $x^2 - x = \lfloor x^2 \rfloor + \left\{x\right\} - \lfloor x \rfloor + \left\{x\right\} = \lfloor x^2 \rfloor - \lfloor x \rfloor = integer$

Therefore for obtaining the no. of such integers between $4$ and $5$ we need to count the no. of integer solutions between $f\left(4\right)$ and $f\left(5\right)$ where $f\left(x\right) = x^2 - x$ which is 7.

Also the numbers are of the form

$\large\displaystyle 4 + \frac{-7+\sqrt{49 + 4n}}{2}$ where $n\in\left[1,7\right]$

and can be obtained by solving the quadratics

$a^2 + a = a + 1$

$a^2 + a = a + 2$

$a^2 + a = a + 3$

$a^2 + a = a + 4$

$a^2 + a = a + 5$

$a^2 + a = a + 6$

$a^2 + a = a + 7$

Here a represents the fractional part of $x$ .

You could use either function: $x^2-x$ is an integer iff $x^2-x+1$ is ;)