Find all positive integer solutions of the equation (1+1/x)(1+1/y)(1+1/z) = 3.

Level: Senior

Assume that 0 < x <= y <= z. Using these inequalities, the given equation becomes
3 = (1+1/x)(1+1/y)(1+1/z) <= (1+1/x)^3,
3^(1/3)-1 <= 1/x,
x <= 1/[3^(1/3)-1] ~ 2.26,
x = 1 or 2.
(i) Assume x = 1. Then the given equation becomes
(1+1/1)(1+1/y)(1+1/z) = 3,
(1+1/y)(1+1/z) = 3/2,
2(y+1)(z+1) = 3yz,
yz-2y-2z-2 = 0,
(y-2)(z-2) = 6.
Since the only factorizations of 6 are 1*6 and 2*3, and y-2 <= z-2, so
y-2 = 1, 2 and z-2 = 6, 3,
y = 3, 4 and z = 8, 5.
(ii) Assume x = 2. Then the given equation becomes
(1+1/2)(1+1/y)(1+1/z) = 3,
(1+1/y)(1+1/z) = 2,
(y+1)(z+1) = 2yz,
yz-y-z-1 = 0,
(y-1)(z-1) = 2.
Since the only factorizations of 2 are 1*2, and y-1 <= z-1, so
y-1 = 1 and z-1 = 2,
y = 2 and z = 3.
In the end, the positive solutions are (x, y, z) = (1, 3, 8), (1, 4, 5), (2, 2, 3), and their permutations of x, y, z.