Floor Function in a Quadratic

Given that $x^2 - 8\lfloor x \rfloor + 7 = 0$ , $\implies x^2 = 8\lfloor x \rfloor - 7$ .

Since $x^2 = 8\lfloor x \rfloor - 7$ ,

$\begin{aligned} \implies x^2 & \le 8x - 7 \\ x^2 - 8x + 7 & \le 0 \\ (x-1)(x-7) & \le 0 \\ \implies 1 \le x & \le 7 \end{aligned}$

Therefore there are only seven possible values $\lfloor x \rfloor$ , $1$ to $7$ , and we need to check which of these values satisfy the equation $x^2 = 8\lfloor x \rfloor - 7$ or $\left \lfloor \sqrt{8\lfloor x \rfloor -7} \right \rfloor = \lfloor x \rfloor$ .

$\begin{array} {lll} \lfloor x \rfloor = 1 & \implies \lfloor \sqrt{8(1)-7} \rfloor = \lfloor \sqrt 1 \rfloor & \blue{= \lfloor x \rfloor} \\ \lfloor x \rfloor = 2 & \implies \lfloor \sqrt{8(2)-7} \rfloor = \lfloor \sqrt 9 \rfloor & \red{\ne \lfloor x \rfloor} \\ \lfloor x \rfloor = 3 & \implies \lfloor \sqrt{8(3)-7} \rfloor = \lfloor \sqrt {17} \rfloor & \red{\ne \lfloor x \rfloor} \\ \lfloor x \rfloor = 4 & \implies \lfloor \sqrt{8(4)-7} \rfloor = \lfloor \sqrt {25} \rfloor & \red{\ne \lfloor x \rfloor} \\ \lfloor x \rfloor = 5 & \implies \lfloor \sqrt{8(5)-7} \rfloor = \lfloor \sqrt {33} \rfloor & \blue{= \lfloor x \rfloor} \\ \lfloor x \rfloor = 6 & \implies \lfloor \sqrt{8(6)-7} \rfloor = \lfloor \sqrt {41} \rfloor & \blue{= \lfloor x \rfloor} \\ \lfloor x \rfloor = 7 & \implies \lfloor \sqrt{8(7)-7} \rfloor = \lfloor \sqrt {49} \rfloor & \blue{= \lfloor x \rfloor} \end{array}$

Therefore the sum of squares of solutions is $1+33+41+49=\boxed{124}$ .

If we assume x is an integer (so that x = int(x)), we find that an integer solution will satisfy x^2 - 8 x + 7 = 0 so that x = 1 or x = 7. Further, an analysis of the graphs of f(x) = x^2 + 7 and g(x) = 8 int(x) shows that there are no solutions greater than 7 (a quadratic will eventually dominate a step function for large enough x).

To find the non-integer solutions, we note that x^2 + 7 = 8*int(x). Since x^2+7 is always positive, this eliminates any negative solutions. Further, since int(x) is always an integer, this means that x^2+7 is an integer multiple of 8. This means x^2 must be one more than a multiple of 8. But, since we have established that 0 < x and x is at most 7, this means that x^2 must be between 0 and 49. The only numbers that are between 0 and 49 and are also one more than a multiple of 8 are 1, 17, 25, 33, 41, or 49. Checking each of these against the original equation, only x^2 = 1, 33, 41, or 49 actually work. Hence, the sum of the squares of the solutions are 1 + 33 + 41 + 49 = 124.