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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]
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    $\begingroup$ It already did it for you... $\endgroup$
    – rm -rf
    Nov 1, 2012 at 2:51
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    $\begingroup$ Funny that there is a USA Mathematical Talent search problem that is nearly identical: #2 on usamts.org/Tests/Problems_24_2.pdf $\endgroup$
    – 0xFE
    Nov 1, 2012 at 2:56
  • $\begingroup$ 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] $\endgroup$ Nov 1, 2012 at 2:56

1 Answer 1

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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) *)
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