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.
 A: I prove your first sentence below:
If $x, y\in\mathbb Z$ satisfy $x-y=x/y,$ then we have :
$$x(y-1)=y^2.$$
Thus, if a prime $p$ divides $y-1,$ then $p$ divides $y$ as well, a contradiction as $y$ and $y-1$ are coprime. Hence no prime divides $y-1,$ i.e. $y=2.$ Therefore $x(2-1)=4.$  
As to the second sentence, if $(x-y)/4=x/y,$ then we have:
$$x(y-4)=y^2.$$
Since any common divisor of $y$ and $y-4$ must divide $4,$ denoting $\gcd(y,y-4)=g,$ we consider three cases:
I. $g=1.$
We have already shown that, in this case, $y-4=1,$ i.e. $y=5$ and $x=25.$
II. $g=2.$
In this case $y-4$ cannot have any odd prime divisor, and $4$ cannot divide $y-4$ either, thus $y-4=2,$ i.e. $y=6$ and $x=18.$
III. $g=4.$
Again no common odd prime divisor of $y$ and $y-4$ can occur, while $4$ divides $y-4.$ Hence $y-4=4,$ i.e. $y=8,$ and $x=16.$
So you know all the positive fractions with the required property, and it remains to calculate the sum of two of them, which I left to you.
Hope this helps.
Edit:
The fractions are supposed to be in the lowest terms, thus the cases II. and III. do not count at all. Hence the only pair of fractions with the required property is $(\frac{x}{4},\frac{y}{4})=(\frac{25}{4},\frac{5}{4}),$ in which case the sum is $\frac{15}{2}.$
A: If $\frac a4, \frac b4$ are the required fractions with $a, b \in \mathbb N$, then $$\frac a4 - \frac b4 = \frac ab$$
So $$\frac{4a}b = a-b$$
The right hand side is an integer, so we must have $b \mid 4a$. But by assumption, the fractions are in their lowest terms, so $b \not \mid 4$. So $b \mid a$. Let $a = kb$ with $k \in \mathbb N$.
Then $$4k = b(k-1)\\$$
By assumption, $(b,4) = 1$, so since $(k, k-1) = 1$, we have $k \mid b$.
Taking $k = 5$, we get $b = 5$, $a = 25$ with sum $30$ as one solution to your problem.
A: $$ m- n = \frac{m}{n} $$ So $$ m=nk,\ nk-n=k,\ (n-1)(k-1)=1 $$
That is $$n=2,\ k=2,\ m=4 $$ 
A: 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 $\frac{(n+1)^2}{n}$.
