Originally posted by talzamir
That would just about cover it, yes. 🙂
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