# How to solve this equation with positive integers as a solutions?

This is a problem of United Kingdom Mathematical Olympiad. Find all triples $(x,y,z)$ of positive integers such that $$\biggl(1+\dfrac{1}{x}\biggr)\cdot \biggl(1+\dfrac{1}{y}\biggr)\cdot \biggl(1+\dfrac{1}{z}\biggr)=2.$$ I tried

Reduce[ (1 + 1/x)(1 + 1/y)(1 + 1/z) == 2 && x > 0 && y > 0 && z > 0,
{x, y, z}, Integers]


And I get

(x | y | z) ∈ Integers && x >= 2 && y > (1 + x)/(-1 + x) &&
z == (1 + x + y + x y)/(-1 - x - y + x y)


How do I tell Mathematica to do that?

O.K,

Reduce[ (1 + 1/x)(1 + 1/y)(1 + 1/z) == 2 && x > 0 && y > 0 && z > 0 &&
x >= y && y >= z, {x, y, z}, Integers]

• It already did it for you... – rm -rf Nov 1 '12 at 2:51
• Funny that there is a USA Mathematical Talent search problem that is nearly identical: #2 on usamts.org/Tests/Problems_24_2.pdf – 0xFE Nov 1 '12 at 2:56
• Thank you very much. Without loss of generality, we may assume $x \geqslant y \geqslant z$. And I tried Reduce[(1 + 1/x)*(1 + 1/y)*(1 + 1/z) == 2 && x > 0 && y > 0 && z > 0 && x >= y && y >= z, {x, y, z}, Integers] – minthao_2011 Nov 1 '12 at 2:56

Reduce[