Here, $a \in R$ and equation is

$-3\{x-[x]\}^{2}+2\{x-[x]\}+a^{2}=0$

$\text { Let } \quad t=x-[x]$

$\therefore-3 t^{2}+2 t+a^{2}=0$

$\Rightarrow t=\frac{1 \pm \sqrt{1+3 a^{2}}}{3}$

$\because \quad t=x-[x]=\{x\} \quad[\text { fractional part }]$

$\therefore \quad 0 \leq t \leq 1$

$\Rightarrow 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3} \leq 1$

Taking positive sign

$\because \quad[\{x\}>0]$

$\therefore 0 \leq \frac{1 \pm \sqrt{1+3 a^{2}}}{3}<1$

$\Rightarrow \sqrt{1+3 a^{2}}<2$

$\Rightarrow \quad 1+3 a^{2}<4$

$\Rightarrow a^{2}-1<0$

$\Rightarrow \quad(a+1)(a-1)<0$

For no integral solution of a we consider the interval $(-1,0) \cup(0,1)$