5. Determine all pairs $(x, y)$ of integers with $x \leq y$ for which $\frac{1}{x}+\frac{1}{y}=\frac{1}{4}$.
5. Throughout this solution, we use the fact that if $a$ and $b$ are positive, then $\frac{1}{a}$ is less than $\frac{1}{b}$ exactly when $a$ is greater than $b$. (That is, $\frac{1}{a}b$.)
Since $\frac{1}{x}+\frac{1}{y}$ is positive, then $x$ and $y$ cannot both be negative (otherwise $\frac{1}{x}+\frac{1}{y}$ would be negative).
Since $x \leq y$, this means that either both $x$ and $y$ are positive or $x$ is negative and $y$ is positive.
Suppose that $x$ and $y$ are both positive. This means that $\frac{1}{x}$ and $\frac{1}{y}$ are both positive.
Since $\frac{1}{x}+\frac{1}{y}=\frac{1}{4}$, then $\frac{1}{x}$ and $\frac{1}{y}$ are both less than $\frac{1}{4}$.
This means that $x>4$ and $y>4$.
Since $x \leq y$, then $\frac{1}{x} \geq \frac{1}{y}$ which means that $\frac{1}{4}=\frac{1}{x}+\frac{1}{y} \leq \frac{1}{x}+\frac{1}{x}$ or $\frac{1}{4} \leq \frac{2}{x}$ which gives $\frac{2}{8} \leq \frac{2}{x}$ and so $x \leq 8$. (In other words, since $\frac{1}{x}$ is larger, then it must be at least one-half of $\frac{1}{4}$ so $\frac{1}{x}$ is at least $\frac{1}{8}$ and so $x$ is at most 8 .)
Since $x>4$ and $x \leq 8$, then $x$ could equal $5,6,7$, or 8 .
If $x=5$, then $\frac{1}{y}=\frac{1}{4}-\frac{1}{5}=\frac{5}{20}-\frac{4}{20}=\frac{1}{20}$ and so $y=20$.
If $x=6$, then $\frac{1}{y}=\frac{1}{4}-\frac{1}{6}=\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$ and so $y=12$.
If $x=7$, then $\frac{1}{y}=\frac{1}{4}-\frac{1}{7}=\frac{7}{28}-\frac{4}{28}=\frac{3}{28}$ and so $y=\frac{28}{3}$ which is not an integer.
If $x=8$, then $\frac{1}{y}=\frac{1}{4}-\frac{1}{8}=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}$ and so $y=8$.
Therefore, when $x$ and $y$ are both positive, the solutions are $(x, y)=(5,20),(6,12),(8,8)$.
Suppose that $x$ is negative and $y$ is positive. This means that $\frac{1}{x}$ is negative and $\frac{1}{y}$ is positive. Since $\frac{1}{x}+\frac{1}{y}=\frac{1}{4}$, then $\frac{1}{y}$ is greater than $\frac{1}{4}$.
This means that $y<4$.
Since $y$ is positive and $y<4$, then $y$ could equal 1,2 or 3 .

If $y=1$, then $\frac{1}{x}=\frac{1}{4}-\frac{1}{1}=\frac{1}{4}-\frac{4}{4}=-\frac{3}{4}$ and so $x=-\frac{4}{3}$, which is not an integer.
If $y=2$, then $\frac{1}{x}=\frac{1}{4}-\frac{1}{2}=\frac{1}{4}-\frac{2}{4}=-\frac{1}{4}$ and so $x=-4$.
If $y=3$, then $\frac{1}{x}=\frac{1}{4}-\frac{1}{3}=\frac{3}{12}-\frac{4}{12}=-\frac{1}{12}$ and so $x=-12$.
Therefore, when $x$ is negative and $y$ is positive, the solutions are $(x, y)=(-4,2),(-12,3)$.
In summary, the solutions are $(x, y)=(-4,2),(-12,3),(5,20),(6,12),(8,8)$.
ANSWER: $(x, y)=(-4,2),(-12,3),(5,20),(6,12),(8,8)$