The solution set for [x] {x} = 1 (where {x} and [x] are, respectively, fractional part function and greatest integer function) is

$\left[x\right]\left\{x\right\}=1$

Obviously, x cannot be negative.

Further, $\left[x\right]\ne 0.$

$\begin{array}{l}\mathrm{For} \left[x\right]=1,\left\{x\right\}=1, \mathrm{whichisnotpossible}.\\ \mathrm{Hence}, x\notin \left[0,2\right).\end{array}$

$\begin{array}{l}\mathrm{For} x\in \left[2,3\right),\left[x\right]=2 \mathrm{or} \left\{x\right\}=\frac{1}{2} \mathrm{or} x=\frac{5}{2}=2+\frac{1}{2}\\ \mathrm{For} x\in \left[3,4\right),\left[x\right]=3 \mathrm{or} \left\{x\right\}=\frac{1}{3} \mathrm{or} x=3+\frac{1}{3}.\\ \mathrm{For} x\in \left[4,5\right),\left[x\right]=4 \mathrm{or} \left\{x\right\}=\frac{1}{4} \mathrm{or} x=4+\frac{1}{4}\end{array}$

Thus, solution is $x=m+\frac{1}{m}, \mathrm{where} m\in N-\left\{1\right\}.$