Determine
all possible ordered triplets (x,y,z) of
nonzero integers that satisfy this equation:
1 1 1
-- + ---- + ----- = 1
x xy xyz
Multiply the equation by xyz: yz + z + 1 = xyz
Then move the term yz and factor: (x-1)*y*z = z+1
Since we are working over integers, then we have z divides z+1.
This happens when z=1 (1 divides 2) or z=-1 (-1 divides 0).
Case 1: z=1
Then (x-1)*y=2
There are four possible factorizations of 2: 2*1, 1*2, -1*-2, and -2*-1.
Subcase 1: x-1=2 and y=1. Then (x,y,z) = (3,1,1).
Subcase 2: x-1=1 and y=2. Then (x,y,z) = (2,2,1).
Subcase 3: x-1=-1 and y=-2. Then (x,y,z) = (0,-2,1), but this must be discarded as we seek nonzero values for x, y, and z.
Subcase 4: x-1=-2 and y=-1. Then (x,y,z) = (-1,-1,1).
Case 2: z=-1
Then (x-1)*y = 0
Then either x-1=0 or y=0.
Subcase 1: x-1=0. Then x=1. y then can be any nonzero integer; then (x,y,z) = (1,y,-1).
Subcase 2: y=0. This can be immediately discarded as we seek nonzero values for x, y, and z.
In summary the set of all possible ordered triplets (x,y,z) of nonzero integers which satisfy the equation is (3,1,1), (2,2,1), (-1,-1,1) and (1,y,-1) where y can be any nonzero integer.
Solution
