Congratulations to Wei Zhang, age 18 from Merchiston Castle School for the superb solution given below. Andrei Lazanu, age 14, School No. 205, Bucharest, Romania submitted a similar solution. \par Suppose there is a fraction $1/n$ , where $n$ belongs to $\{1,2,3,4,5,6\}$.

$$\eqalign{ 1/1: 1x &=1\ {\rm so}\ x=1 \cr 1/2: 2x &=1\ {\rm so}\ x=4 \cr 1/3: 3x &=1\ {\rm so}\ x=5 \cr 1/4: 4x &=1\ {\rm so}\ x=2 \cr 1/5: 5x &=1\ {\rm so}\ x=3 \cr 1/6: 6x &=1\ {\rm so}\ x=6 }$$

As I showed above, $1/n$ is in the set $\{1,2,3,4,5,6\}$.

Now we have $a/n = a \times 1/n$ , where $a$ belongs to $\{0,1,2,3,4,5,6\}$, this is a number from the set $\{0,1,2,3,4,5,6\}$ times a number from the set of $\{1,2,3,4,5,6\}$. Therefore we will get a set of numbers for $a/n$ , which is $\{0,1,2,3,4,5,6\}$. Solving the equation

$${1\over x} + {1\over y}= {1\over{x+y}}$$

is equivalent to solving $x^2+xy+y^2=0$.

when

x=1

y²+y+1 = 0

y²+y = 6

y=2

y=4

when

x=2

y²+2y+4 = 0

y²+2y = 3

y=1

y=4

when

x=3

y²+3y+2 = 0

y²+3y = 5

y=6

y=5

when

x=4

y²+4y+2 = 0

y²+4y = 5

y=1

y=2

when

x=5

y²+5y+1 = 0

y²+5y = 3

y=3

y=6

when

x=6

y²+6y+1 = 0

y²+6y = 6

y=3

y=5

$$ \begin{tabular}{c c c c c c c c} \mbox{when} & x=1 & y^2+y+1 = 0 & \mbox{so} & y^2+y = 6 & y=2 & \mbox{or} & y=4 \\ \mbox{when} & x=2 & y^2+2y+4 = 0 & \mbox{so} & y^2+2y = 3 & y=1 & \mbox{or} & y=4 \\ \mbox{when} & x=3 & y^2+3y+2 = 0 & \mbox{so} & y^2+3y = 5 & y=6 & \mbox{or} & y=5 \\ \mbox{when} & x=4 & y^2+4y+2 = 0 & \mbox{so} & y^2+4y = 5 & y=1 & \mbox{or} & y=2 \\ \mbox{when} & x=5 & y^2+5y+1 = 0 & \mbox{so} & y^2+5y = 3 & y=3 & \mbox{or} & y=6 \\ \mbox{when} & x=6 & y^2+6y+1 = 0 & \mbox{so} & y^2+6y = 6 & y=3 & \mbox{or} & y=5 \\ \end{tabular} $$

We have found the six solution pairs $(1,2)$, $(1,4)$, $(2,4)$, $(3,6)$, $(3,5)$, $(5,6)$. The equation is symmetric in $x$ and $y$ so there are six corresponding solutions when we exchange $x$ and $y$.

We don't allow negative solutions (e.g. $x=1,\ y=-3$) because we are working entirely in the set $\{0,1,2,3,4,5,6\}$.

If we work with real numbers, we think of solving the quadratic equation $x^2+yx+y^2=0$ as an equation in $x$. The formula for the solution of this quadratic equation involves $y^2-4y^2=-3y^2< 0$ which has no real values, therefore there is no real solution for $x$.