2016/104) Find all real numbers a such that 3 $< a < 4$ and $a(a - 3\{a\})$ is an integer. where $\{x\}$ is fractional part

let a  = 3 + x
so we have (3+x) (3+x-3x) = (3+x)(3-2x) = n an integer
$9 - 3x -  2x^2 = $ integer
$2x^2 + 3x -k = 0$
the values can be 1 or 2 or 3 or 4
giving the value of x = $\frac{\sqrt{9+8k}-3}{4}$
putting k = 1 to 4 we get a = $3 + \frac{\sqrt{17}-3}{4}$,$ 3 + \frac{1}{2}$,$3 + \frac{\sqrt{33}-3}{4}$, $3 + \frac{\sqrt{41}-3}{4}$