Quadruple Fraction Sum

We start by showing that solutions must have $a = 3$ . If $a > 3$ , then the left side of the equation is at most $2/4 + 1/5 + 1/6 - 1/d = 13/15 - 1/d < 1$ , so there are no solutions for $a > 3$ . If $a < 3$ , then $2/a \geq 1$ , so for a solution we require $1/b + 1/c - 1/d \leq 0$ , or $1/b + 1/c \leq 1/d$ . But since $b < c < d$ , it holds that $1/b + 1/c > 1/d$ , so there are no solutions for $a < 3$ . Hence, any solution must have $a = 3$ , so now by substitution, we need only look for solutions to $1/b + 1/c - 1/d = 1/3$ with $3 < b < c < d$ .

We show that solutions must have $b = 4$ or $b = 5$ . If $b < 4$ , then $1/b \geq 1/3$ , and so for a solution we require $1/c - 1/d \leq 0$ . But since $c < d$ , $1/c - 1/d > 0$ , so there are no solutions for $b < 4$ . If $b > 5$ , then $1/b \leq 1/6$ , so for a solution we require $1/c - 1/d \geq 1/6$ . However, since $c > b > 5$ , we have $1/c \leq 1/6$ , and so $1/c - 1/d \leq 1/6 - 1/d < 1/6$ , so there are no solutions for $b > 5$ . Hence, any solution must have $b = 4$ or $b = 5$ .

We first examine potential solutions with $b = 5$ . Then, our solutions must satisfy $1/c - 1/d = 2/15$ where $5 = b < c < d$ . We show that solutions must have $c = 6$ or $c = 7$ . If $c > 7$ , then $1/c \leq 1/8$ , so $-1/d = 2/15 - 1/c \geq 2/15 - 1/8 = 2/15 - 2/16 > 0$ , a contradiction since this implies $-1/d$ is positive. Hence, solutions require $c \leq 7$ . Combined with $c > 5$ , our only possibilities are $c = 6$ and $c = 7$ . We easily verify whether these two cases yield solutions by solving for $d$ and checking if $d$ is an integer. Both yield solutions, with $c = 6$ corresponding to $d = 30$ , and $c = 7$ corresponding to $d = 105$ , so this case yields two solutions total.

Now, consider solutions with $b = 4$ . Then, we wish to solve $1/c - 1/d = 1/12$ . We show that solutions require $c \leq 11$ , since if $c > 11$ , then $1/c \leq 1/12$ , so $1/c - 1/d < 1/12$ , and there are no solutions. Combined with $c > 4$ due to $b = 4$ , we have 7 cases to check. Five of them lead to solutions: $c = 6$ corresponds to $d = 12$ , $c = 8$ corresponds to $d = 24$ , $c = 9$ corresponds to $d = 36$ , $c = 10$ corresponds to $d = 60$ , and $c = 11$ corresponds to $d = 132$ . Thus, this case yields five solutions total.

This completes our case analysis, and so we conclude that there are exactly 7 solutions: $(3,5,6,30)$ , $(3,5,7,105)$ , $(3,4,6,12)$ , $(3,4,8,24)$ , $(3,4,9,36)$ , $(3,4,10,60)$ , and $(3,4,11,132)$ .