143*. Find the condition under which the difference of two irreducible fractions is equal to their product.
Let $\frac{a}{b}$ and $\frac{c}{d}$ be two irreducible fractions, such that $\frac{a}{b}-\frac{c}{d}=\frac{a}{b} \cdot \frac{c}{d}$, from which $a d-b c=a c$, or $a d=c(a+b)$, or $b c=a(d-c)$. Given that these fractions are irreducible, we conclude that $c$ is a divisor of $a$ and $a$ is a divisor of $c$, i.e., $a=c$ and $d=a+b$. Therefore, the fractions must have the form: $\frac{a}{b}, \frac{a}{a+b}$.