Exercise 4. Find all real numbers $a$ such that $a+\frac{2}{3}$ and $\frac{1}{a}-\frac{3}{4}$ are integers.
Solution to Exercise 4 Let's denote these two integers by $m$ and $n$. Since $a=m-\frac{2}{3}$ and $\frac{1}{a}=n+\frac{3}{4}$, we have $3a=3m-2$ and $\frac{4}{a}=4n+3$. Therefore, by multiplying these equations, we get $12=(3m-2)(4n+3)$. Since $4n+3$ is odd and it divides 12, it must be one of the following numbers: $-3, -1, 1, 3$. The case $4n+3=-3$ is impossible. If $4n+3=-1$ then $n=-1$ and $3m-2=-12$, which is impossible. The case $4n+3=1$ is impossible. The only remaining case is $4n+3=3$ which gives $n=0$ and $3m-2=4$ hence $m=2$, and then $a=\frac{4}{3}$