Interesting Ceiling and Floors

Let $x = 4k(k + 1)$ and $y = 4(k + 1)$ for $k > 1$ . Then $x + y = 4(k + 1)^2$ and:

$\left \lfloor \dfrac{x + y}{x^2} \right \rfloor = \left \lfloor \dfrac{4(k + 1)^2}{16k^2(k + 1)^2} \right \rfloor = \left \lfloor \dfrac{1}{4k^2} \right \rfloor = 0$

$\left \lceil \dfrac{y^2}{x + y} \right \rceil = \left \lceil \dfrac{16(k + 1)^2}{4(k + 1)^2} \right \rceil = 4$

$\left \lfloor \dfrac{(x + y)^2}{x^2 + y^2} \right \rfloor = \left \lfloor \dfrac{16(k + 1)^4}{16k^2(k + 1)^2 + 16(k + 1)^2} \right \rfloor = \left \lfloor \dfrac{(k + 1)^2}{k^2 + 1} \right \rfloor = \left \lfloor \dfrac{k^2 + 2k + 1}{k^2 + 1} \right \rfloor = \left \lfloor 1 + \dfrac{2k}{k^2 + 1} \right \rfloor = 1$

Therefore, for any $k > 1$ ,

$\left \lfloor \dfrac{x + y}{x^2} \right \rfloor + \left \lceil \dfrac{y^2}{x + y} \right \rceil = \left \lfloor \dfrac{(x + y)^2}{x^2 + y^2} \right \rfloor + 3 \Longrightarrow 0 + 4 = 1 + 3$

which is a true statement, so there are infinitely many solutions .

$x\to \infty, \: y=\text{int}( 2\sqrt{x})$ also works. First term is 0, second is 4 and right side is 4 for $x\neq y$ as others have explained.

The text below the graph just reads that for x>6 is empty, so I didn't include it. Let me know if You have some suggestions and constructive criticism. I'm open, Thanks :D

The right hand side expression can be written as $\lfloor \frac{2xy}{x^2+y^2} \rfloor + 4$ .

Because $(x-y)^2 \ge 0$ we see that $x^2+y^2 \ge 2xy$ , and the floor evaluates to 1 when x=y and vanishes otherwise.

Sidenote: for $x=y$ the equation becomes $\lfloor \frac{2}{x} \rfloor+ \lceil \frac{x}{2} \rceil=5$ with solutions $(x,y) \in \{ (9,9)(10,10)\}$ .

Now Suppose $x \ne y$

we want to solve $\lfloor \frac{x+y}{x^2}\rfloor + \lceil \frac{y^2}{x+y} \rceil\ = 4$

To keep the 2nd term $\lceil \frac{y^2}{x+y} \rceil$ from getting larger than 4 we need $x\ge \frac14 y^2 - y$

Choosing $y>8$ implies that $x>y$ and the first term $\lfloor \frac{x+y}{x^2}\rfloor$ vanishes. The only remaining condition is $3 \lt \frac{y^2}{x+y} \le 4$ or $\frac{y^2}{4}-y \le x \lt \frac{y^2}{3}-y$

The size of the interval is $\frac {y^2}{12}$ . So to any $y>8$ we can find at least 5 values of x.