Inverse of Inverses

Let's try to prove it.

Lemme

The equation

$\frac{1}{x} - \frac{1}{s - x} = \frac{1}{n}$

where $x$ , $s$ are unknown integers and $n$ a given integer, and $x \neq 0$ and $s - x \neq 0$ ,

we have

$(x, s) = (\frac{n(d-n)}{d}, \frac{n(d-n)}{d} + d - n)) \quad for \quad d \quad divisor \quad of \quad n^2 \quad and \quad d \neq n$

So that if we let $s = a + b + c$ we have:

From the first equality : $(a, s) \quad is \quad in \quad \{(3, 15), (2, 6), (5, -15), (12, 6), (-4, -6), (20, 15), (6, -6), (-12, -15)\}$

From the second equality : $(b, s) \quad is \quad in \quad \{(14, -21), (30, 15), (-90, -99), (60, 48), (-10, -15), (-40, -48), (-15, -21), (9, 99), (6, 21), (35, 21), (11, -99), (5, 15), (12, -48), (110, 99), (8, 48), (15, -15)\}$

Then only two value are possible for $s$ 15 or -15

Then

$(a, s) \quad is \quad in \quad \{(3, 15), (5, -15), (20, 15), (-12, -15)\}$

and $(b, s) \quad is \quad in \quad \{ (30, 15), (-10, -15), (5, 15), (15, -15)\}$

From the third equality: $(c, s) \quad is \quad in \quad \{(-8, -15), (105, -15), (120, 15), (7, 15)\}$

if s = -15 then:

$a=5\quad or\quad a=-12\\ b=-10\quad or\quad b=15\\ c=-8\quad or\quad c=105$

No combination adds to -15.

if s = 15 then:

$a=3\quad or\quad a=20\\ b=5\quad or\quad b=30\\ c=7\quad or\quad c=120$

Only a = 3 and b = 5 and c = 7 adds to 15.

CQFD