Fractional part has gone crazy

As $x\in (\sqrt{2},\sqrt{3}), \dfrac{1}{x}\in \left (\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{2}}\right ), x^2\in (2,3)$

$\left\lfloor\dfrac{1}{x}\right\rfloor=0,\lfloor x^2\rfloor=2$

$\left \{ \dfrac{1}{x}\right \}=\dfrac{1}{x}, \{x^2\}=x^2-2$

$x^2-2=\dfrac{1}{x}\\\Rightarrow x^3-2x-1=0\\\Rightarrow (x+1)(x^2-x-1)=0$

Since $x\ne -1$ , $x^2=x+1$ $x^4-\dfrac{3}{x}=x(2x+1)-3×(x^2-2)\\=-x^2+x+6=-x-1+x+6=\boxed{5}$

Let the number satisfying the conditions be $a$

Let $f(y) = y^2 - \frac{1}{y}$ . for $y \in (\sqrt{2},\sqrt{3})$

Then $f'(y) = 2y + \frac{1}{y^2} > 0$ . Therefore $f(y)$ is increasing.

Therefore $f(\sqrt{3}) \geq f(y) \geq f(\sqrt{2})$

$\Rightarrow 3 > 3 - \frac{1}{\sqrt{3}} \geq f(y) \geq 2 - \frac{1}{\sqrt{2}} > 1$

$\Rightarrow 3 > f(y) > 1$ .

Now, we know that $f(a)$ is an integer, since the fractional parts cancel out. But the only interget in the range of $f(y)$ is $2$ . Therefore $f(a) = 2$

$\Rightarrow a^2 - \frac{1}{a} = 2$

$\Rightarrow a^3 - 1 = 2a$

$\Rightarrow (a+1)(a^2 - a - 1) = 0$

Since $a \not = -1$ , $a^2 = a + 1 \Rightarrow a - 1 = \frac{1}{a}$

Therefore, $a^4 -3\frac{1}{a}$

$= (a+1)^2 -3(a-1)$

$= a^2 + 2a + 1 - 3a + 3$

$= a+1 + 2a + 1 - 3a + 3 = \boxed{5}$