*Originally posted by talzamir*

**That would just about cover it, yes. ðŸ™‚
**

1/x + 1/y = 1/11

0 = xy - 11x - 11y

121 = xy - 11x - 11y + 121

121 = (x - 11) (y - 11)

121 can be a factor of two integers as

11 x 11 ; which gives 1 / 22 + 1 / 22 = 1 / 11

121 x 1 ; which gives 1 / 132 + 1 / 12 = 1 / 11

(-121) x (-1) ; which gives 1 / (-110) + 1 / 10 = 1 / 11

(-11) x (-11) ; which gives e a bit trickier as there are so many ways to derive them by multiplying two integers together.

rearrange to y = 11x/(x-11)

if x is n*11 we have:

Y = 11*11*n/(n-1)*11

= 11n/(n-1)

n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:

x = 0

x = 22

x = -110

x = 132

let us try for solutions where x = n*11 + b, where b is between 1 and 10

then y = 11*(n*11 + b)/(n*11+b - 11)

Here the top is clearly a multiple of11, and the bottom is clearly NOT a multiple of 11, so this only has integer solutions when the bottom is plus or minus 1

so this gives us the solutions

n=0, b = 10, x = 10

n=1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132