14. [40] A number $n$ is called bummed out if there is exactly one ordered pair of positive integers $(x, y)$ such that
$$
\left\lfloor x^{2} / y\right\rfloor+\left\lfloor y^{2} / x\right\rfloor=n .
$$

Find all bummed out numbers.
Answer: $2,6,8,10$
Solution: Suppose $n$ is bummed out. If $(a, b)$ is one solution for $(x, y)$ to the given equation $\left\lfloor x^{2} / y\right\rfloor+\left\lfloor y^{2} / x\right\rfloor=n$, then $(b, a)$ is another, so the unique solution $(a, b)$ better have the property that $a=b$ and $n=2 a \geq 2$. In particular, $n$ is an even positive integer.
Now, if $n=2 a \geq 12$, then setting $x=a-1 \geq 5, y=a+1 \geq 7$, we have
$$
\left\lfloor\frac{x^{2}}{y}\right\rfloor+\left\lfloor\frac{y^{2}}{x}\right\rfloor=\left\lfloor a-3+\frac{4}{a+1}\right\rfloor+\left\lfloor a+3+\frac{4}{a-1}\right\rfloor=2 a=n,
$$
so $n$ cannot be bummed out.
Moreover, $\left\lfloor 1^{2} / 2\right\rfloor+\left\lfloor 2^{2} / 1\right\rfloor=4$, so 4 is not bummed out. The only possibilities left are $2,6,8$, and 10 .
To check these, note that
$$
n=\left\lfloor\frac{x^{2}}{y}\right\rfloor+\left\lfloor\frac{y^{2}}{x}\right\rfloor>-2+\frac{x^{2}}{2 y}+\frac{x^{2}}{2 y}+\frac{y^{2}}{x} \geq-2+\frac{3 x}{\sqrt[3]{4}}
$$
so
$$
x<\frac{\sqrt[3]{4}}{3}(n+2)<.53(n+2)
$$
and similarly for $y$. So we only have to check $x, y \leq\lfloor .53(10+2)\rfloor=6$ :