positive fractions, denominator 4, difference equals quotient (4,2) are the only positive integers whose difference is equal to their quotient. Find the sum of two positive fractions, in their lowest terms, whose denominators are 4 that also share this same property.

Let's work with $n$ instead of 4.  We achieve $a=\frac{b^2}{b-n}=b+n+\frac{n^2}{b-n}$, hence $b-n$ must divide $n^2$.  If $a/b$ must be in lowest terms, this is only possible if $b=n+1$: would $b$ be a proper divisor of $n^2$ it would contain a proper divisor $k$ of $n$. Now if $k$ is a proper divisor of $n$ it is a proper divisor of $b=n+k$ and $b/n$ would not be in lowest terms.
Thus all fractions satisfying the conditions are $\frac{n+1}{n}$ and $\frac{(n+1)^2}{n}$.
