# 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]
````
-

1
It already did it for you... – rm -rf♦ Nov 1 '12 at 2:51

3
– 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

## 1 Answer

The Backsubstitution option will help here.

````Reduce[
 1 + x + y + x y + z + x z + y z - x y z == 0 && x >= y >= z >= 1, {x,
   y, z}, Integers, Backsubstitution -> True]

(* (x == 5 && y == 4 && z == 3) || (x == 7 && y == 6 && 
   z == 2) || (x == 8 && y == 3 && z == 3) || (x == 9 && y == 5 && 
   z == 2) || (x == 15 && y == 4 && z == 2) *)
````
-
lang-mma
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