(a) Determine the positive integer $x$ for which $\frac{1}{4}-\frac{1}{x}=\frac{1}{6}$.

(b) Determine all pairs of positive integers $(a, b)$ for which $a b-b+a-1=4$.

(c) Determine the number of pairs of positive integers $(y, z)$ for which $\frac{1}{y}-\frac{1}{z}=\frac{1}{12}$.

(d) Prove that, for every prime number $p$, there are at least two pairs $(r, s)$ of positive integers for which $\frac{1}{r}-\frac{1}{s}=\frac{1}{p^{2}}$.
(a) Re-arranging and noting that $x \neq 0$, we obtain the following equivalent equations:

$$
\begin{array}{r}
\frac{1}{4}-\frac{1}{x}=\frac{1}{6} \\
\frac{1}{4}-\frac{1}{6}=\frac{1}{x} \\
\frac{3}{12}-\frac{2}{12}=\frac{1}{x} \\
\frac{1}{12}=\frac{1}{x}
\end{array}
$$

and so $x=12$.

(b) Factoring, we obtain the following equivalent equations:

$$
\begin{array}{r}
a b-b+a-1=4 \\
b(a-1)+(a-1)=4 \\
(a-1)(b+1)=4
\end{array}
$$

Since $a$ and $b$ are integers, then $a-1$ and $b+1$ are integers. Therefore, $b+1$ and $a-1$ are a divisor pair of 4 .

Since $b \geq 1$, then $b+1 \geq 2$.

Since $b+1>0$ and $4>0$ and $(a-1)(b+1)=4$, then $a-1>0$.

This means that $b+1$ and $a-1$ must form a positive divisor pair of 4 with $b+1 \geq 2$.

There are two possibilities:

- $b+1=2$ and $a-1=2$; this gives $(a, b)=(3,1)$.
- $b+1=4$ and $a-1=1$; this gives $(a, b)=(2,3)$.

Therefore, the pairs of positive integers that solve the equation are $(3,1)$ and $(2,3)$.
(c) Since $y \neq 0$ and $z \neq 0$, we multiply the given equation by $12 y z$ and manipulate to obtain the following equivalent equations:

$$
\begin{aligned}
\frac{1}{y}-\frac{1}{z} & =\frac{1}{12} \\
12 z-12 y & =y z \\
0 & =y z+12 y-12 z \\
0 & =y(z+12)-12 z \\
-144 & =y(z+12)-12 z-144 \\
-144 & =y(z+12)-12(z+12) \\
-144 & =(y-12)(z+12)
\end{aligned}
$$

Since $y$ and $z$ are integers, then $y-12$ and $z+12$ are integers.

Since $z>0$, then $z+12>0$.

Since $-1440$ and $(y-12)(z+12)=-144$, then $y-12-12$.

Since $(y-12)(z+12)=-144$ and $y-120$, then $s+p^{2}>0$.

Since $-p^{4}0$ and $\left(r-p^{2}\right)\left(s+p^{2}\right)=-p^{4}$, then $r-p^{2}<0$.

Since $r$ is a positive integer, then $1 \leq r<p^{2}$.

Since $\left(r-p^{2}\right)\left(s+p^{2}\right)=-p^{4}$ and $r-p^{2}<0$, then $r-p^{2}$ and $s+p^{2}$ are a divisor pair of $-p^{4}$ with $-p^{2}<r-p^{2}<0$.

We consider the possibilities that $r-p^{2}=-1$ and $r-p^{2}=-p$.

In these cases, $s+p^{2}=p^{4}$ and $s+p^{2}=p^{3}$, respectively.

These give the pairs $(r, s)=\left(p^{2}-1, p^{4}-p^{2}\right)$ and $(r, s)=\left(p^{2}-p, p^{3}-p^{2}\right)$.

Since $p$ is prime, then $p \geq 2$, which means that these pairs are distinct and that each component is a positive integer.

Therefore, there are at least two pairs $(r, s)$ of positive integers for which $\frac{1}{r}-\frac{1}{s}=\frac{1}{p^{2}}$.