Post by ***@gmail.comFor what integers (x,y)
is 1/22
expressible as ( 1/x + 1/y ) ?
[spoiler space]
Both x and y must be non-zero. (If x is zero, then 1/x is infinitely
large, so 1/x + 1/y is also infinitely large. Similar for y.)
At least one of x, y must be positive. (If both are negative, then 1/x
and 1/y are also both negative, so 1/x + 1/y is also negative.)
Without loss of generality, assume that x > 0 and x >= y. (Each
resulting pair can be swapped.) Then consider:
1/22 = 1/x + 1/y
xy = 22y + 22x
(x - 22)y = 22x
y = 22x/(x - 22)
Now we can just look at x = 1, 2, 3, etc. and see which ones result in
y being an integer:
11, -22 -> 22/(22*11) - 11/(22*11)
18, -99 -> 22/(22*18) - 4/(22*18)
20, -220 -> 22/(22*20) - 2/(22*20)
21, -462 -> 22/(22*21) - 1/(22*21)
23, 506 -> 22/(22*23) + 1/(22*23)
24, 264 -> 22/(22*24) + 2/(22*24)
26, 143 -> 22/(22*26) + 4/(22*26)
33, 66 -> 22/(22*33) + 11/(22*33)
44, 44 -> 22/(22*44) + 22/(22*44)
Based on the right side, we can reverse-engineer a general pattern of
1/22 = 22/(22*x) + (x-22)/(22*x)
Now 22/(22*x) reduces to 1/x, leaving
1/y = (x-22)/(22*x)
y = (22*x)/(x-22)
i.e. we need 22*x to be divisible by x-22 (or, equivalently, by the
absolute value of x-22). So any prime factors of x-22 that aren't
covered by the prime factors of 22 (those being 2 and 11) must
instead be covered by the prime factors of x. But if something is a
prime factor of both x-22 and x, then it must also be a prime factor
of their difference (22), so again 2 or 11. And indeed this is what
we see play out above; the valid values of y (not including swapping
pairs around) are a subset of
(+1 or -1) * 2^a (0 <= a <= 2) * 11^b (0 <= b <= 2)