Not calculus this time

Relevant wiki: Floor and Ceiling Functions - Problem Solving

$\begin{aligned} \lfloor 4x \rfloor + \lceil x \rceil & = x^2 + 4 \end{aligned}$

Since the LHS of the equation is integral, the RHS must also be integral, which means that $x^2 = n \implies x = \sqrt{n}$ , where $n$ is a non-negative integer.

We can assume that the solutions should occur around the solutions of continuous equation (that is equation without the floor and ceiling functions) as follows:

$\begin{aligned} 4x + x & = x^2 + 4 \\ x^2 - 5x + 4 & = 0 \\ (x-1)(x-4) & = 0 \\ \implies x & = 1, \ 4 \end{aligned}$

Checking for solutions around $1 = \sqrt 1$ ,

• When $x=\sqrt 0: \ \lfloor 0 \rfloor + \lceil 0 \rceil \color{#D61F06}{<} 0 + 4 \color{#D61F06}{\text{ not a solution}}$
• When $x<0$ , LHS $\color{#D61F06}{<}$ RHS, $\color{#D61F06}{\text{so no solution}}$
• When $x=\sqrt 2: \ \lfloor 4 \times 1.414 \rfloor + \lceil 1.414 \rceil \color{#D61F06}{>} 2 + 4 \color{#D61F06}{\text{ not a solution}}$ ; and this means that there is no solution until around $x = 4 = \sqrt{16}$

Checking for solutions around $1 = \sqrt 1$ ,

• When $x=\sqrt{15}: \ \lfloor 15.492 \rfloor + \lceil 3.873 \rceil \color{#3D99F6}{=} 15 + 4 \color{#3D99F6}{\text{ a solution}}$
• When $x=\sqrt{14}: \ \lfloor 14.967 \rfloor + \lceil 3.742 \rceil \color{#3D99F6}{=} 14 + 4 \color{#3D99F6}{\text{ a solution}}$
• When $x=\sqrt{13}: \ \lfloor 14.422 \rfloor + \lceil 3.605 \rceil \color{#D61F06}{>} 13 + 4 \color{#D61F06}{\text{ not a solution}}$ ; and this means that there is no solution for $x = \sqrt 2$ to $\sqrt{13}$
• When $x=\sqrt{17}: \ \lfloor 16.492 \rfloor + \lceil 4.123 \rceil \color{#3D99F6}{=} 17 + 4 \color{#3D99F6}{\text{ a solution}}$
• When $x=\sqrt{18}: \ \lfloor 16.971 \rfloor + \lceil 4.243 \rceil \color{#D61F06}{<} 18 + 4$ , not a solution; and this means that there is no solution for $x \ge \sqrt {18}$

Therefore the solutions are $\sqrt 1$ , $\sqrt {14}$ , $\sqrt {15}$ , $\sqrt {16}$ , and $\sqrt {17}$ . And the sum of squares of solution is $1+14+15+16+17 = \boxed{63}$ .

Note the left hand side is necessarily integral. So that means x² is integral.

By inspection, x=1 and x=4 are solutions.

However, the condition that x² is an integer does not imply x is an integer.

Looking at the equation x² + 4 = 5x tells us that the integral intersection points are 1 and 4.

When x is outside the domain (0,5), the right hand side is necessarily greater than the left hand side, so no solutions exist, which can be shown using basic inequalities involving the floor and ceiling functions.

It then remains to analytically exhaust all the possible solutions that lie in the interval (0,5)

Such solutions are square roots of integers.

After some exhaustive bashing, the other 3 solutions are √14, √15 and √17

Hence the answer to the question is 1+16+14+15+17=63

Relevant wiki: Floor and Ceiling Functions - Problem Solving

Since $a-1 < \lfloor a \rfloor \leq a$ and $a \leq \lceil a \rceil < a+1$ , the solutions must satisfy $5x + c = x^2 + 4$ with $-1 < c < 1$ . Completing the square, $(x - 2\tfrac12)^2 = 2\tfrac14 - c,$ so that $x = 2\tfrac12 \pm \sqrt{2\tfrac14 - c}$ and $x^2 \in \langle 0.486,\,1.910 \rangle \cup \langle 13.090,\,18.514 \rangle.$ Moreover, since the lhs of the original equation is an integer, $x^2$ must also be an integer. This limits us to $x^2 \in \{ 1, 14, 15, 16, 17, 18 \}.$ Of these possible values, only $x^2 = 18$ does not work out, leaving us with the sum of squares $\boxed{1 + 14 + 15 + 16 + 17} = 63.$