Can't Factor This Fractional Part

The trick here is that the square is inside of the fractional part so that we cannot easily factor the given expression. However, using the substitution found in the details and assumptions, we can get

$\left\{x^2\right\} - 2\left\{x\right\} + 1 = 0$ $\rightarrow x^2 - \left\lfloor x^2 \right\rfloor - 2x + 2\left\lfloor x \right\rfloor + 1 = 0$ $\rightarrow x^2-2x+1 = \left\lfloor x^2 \right\rfloor - 2\left\lfloor x \right\rfloor$ $\rightarrow (x-1)^2 = \left\lfloor x^2 \right\rfloor - 2\left\lfloor x \right\rfloor$

Since the right hand side of the above expression is an integer, the left hand side must also be an integer. Then $x-1 = \sqrt{n}$ , for some nonnegative integer $n$ , so that we are guaranteed that the LHS is an integer. Since $x$ increases with $n$ , we can test successively larger values of $n$ until a solution is found. We only have to go to $n = 3$ , and since this is the smallest solution, the answer is $\lfloor 1000\{ 1 + \sqrt{3} \} \rfloor = 732$