how many solution

Note that the LHS $\lfloor x + x^2 \rfloor$ is an integer, the RHS must also be an integer. Therefore $x$ can only be of the form $\frac n3$ , where $n$ is an integer. Since $x^2 + x = \left(x+\frac 12 \right)^2 - \frac 14$ , $\implies x^2 + x \ge - \frac 14$ and $\lfloor x + x^2 \rfloor \ge - 1$ . Therefore the smallest possible RHS $n = -1$ .

$\begin{array} {lll} n=-1 & \implies \lfloor x + x^2 \rfloor = \left \lfloor -\frac 13 - \frac 19 \right \rfloor = - 1 & \blue{= \text {LHS, acceptable}} \\ n = 0 & \implies \lfloor x + x^2 \rfloor = \left \lfloor 0 + 0 \right \rfloor = 0 & \blue{= \text {LHS, acceptable}} \\ n = 1 & \implies \lfloor x + x^2 \rfloor = \left \lfloor \frac 13 + \frac 19 \right \rfloor = 0 & \red{\ne \text {LHS, unacceptable}} \\ n = 2 & \implies \lfloor x + x^2 \rfloor = \left \lfloor \frac 23 + \frac 49 \right \rfloor = 1 & \red{\ne \text {LHS, unacceptable}} \\ n = 3 & \implies \lfloor x + x^2 \rfloor = \left \lfloor 1 + 1 \right \rfloor = 2 & \red{\ne \text {LHS, unacceptable}} \\ n = 4 & \implies \lfloor x + x^2 \rfloor = \left \lfloor \frac 43 + \frac {16}9 \right \rfloor = 3 & \red{\ne \text {LHS, unacceptable}} \\ n = 5 & \implies \lfloor x + x^2 \rfloor = \left \lfloor \frac 53 + \frac {25}9 \right \rfloor = 4 & \red{\ne \text {LHS, unacceptable}} \\ n = 6 & \implies \lfloor x + x^2 \rfloor = \left \lfloor 2 + 4 \right \rfloor = 6 & \blue{= \text {LHS, acceptable}} \\ n= 7 & \implies \lfloor x + x^2 \rfloor = \left \lfloor \frac 73 - \frac {49}9 \right \rfloor = 7 & \blue{= \text {LHS, acceptable}} \end{array}$

For $n \ge 8$ , $\text {LHS > RHS}$ and there is no solution. Therefore the sum of solutions is $-\frac 13 + 0 + 2 + \frac 73 = \boxed 4$ .

• $3x$ is integer
• if $x>3$ then $x+x^2=x(x+1)>3(x+1)$ so no solution there.
• if $x<-1$ then LHS is non-negative but RHS negative Only need to check x=k/3 where k is integer between -3 and 9

Solutions are $-\frac{1}{3}, 0, 2, \frac{7}{3}$