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Add to Product Inverse (Posted on 2009-03-06) Difficulty: 3 of 5
Determine all possible triplet(s) (x,y,z) of positive rational numbers, with x ≥ y ≥ z, such that each of x + (yz)-1, y + (zx)-1 and, z + (xy)-1 is an integer.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

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Hints/Tips Outline proof (spoiler) | Comment 3 of 5 |
If all three are integers, then their product must be an integer.  Their product = 3+ xyz +3/xyz + 1/(xyz*xyz).  It looks like their product can only be an integer if xyz = 1, but I haven't proved this yet (I'm on vacation).  Given this start, can anybody prove that xyz must equal 1?

If we can prove this, then 1/yz = x/xyz = x, so 2x must be an integer and it's denominator must be 1 or 2.  Same for y and z.

Then, if none of the denominators are 2, the only solution will be (1,1,1).  (because xyz = 1).  If one of the denominators = 2, the only solution is (2,1,1/2).  If two of the denominators are 2, the only solution is (4,1/2,1/2).  And there are no solutions where all three denominators equal 2. 
 

  Posted by Steve Herman on 2009-03-10 10:42:21
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