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.

Following Steve Herman's wise lead...

The product of the three integers = xyz(1+1/xyz)^3. Now, writing xyz = a/b in its lowest form, using integers a, b, it follows that gcd(a,b) = 1, and the product becomes (a+b)^3/ba^2. Since  this must be an integer, both a and b must be factors of (a+b) and  therefore a is a factor of b and b is a factor of a, showing that a=b and giving xyz=1.

Steve's final paragraphs can now be appended: 

X+1/yz=2x so 2x must be an integer and therefore, in its lowest form, x must have a denominator of 1 or 2. Same for y and z.

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

Posted by Harry on 2009-03-11 20:29:02