XLV OM - III - Task 1

Determine all triples $ (x,y,z) $ of positive rational numbers for which the numbers $ x+y+z $, $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} $, $ xyz $ are natural.
Let $ (x,y,z) $ be a triple of rational numbers satisfying the given condition. The numbers

are natural; therefore, the product

is also a natural number. The numbers $ x $, $ y $, $ z $ are roots of the polynomial

whose all coefficients are integers, and the coefficient of the highest power of the variable $ t $ is $ 1 $. It is known (see: *Remark*), that every rational root of such a polynomial is an integer. The numbers $ x $, $ y $, $ z $ are rational by assumption; they must therefore be integers.
Of course, for any natural numbers $ x,y,z \geq 1 $, the numbers $ a $ and $ c $, defined by formulas (1), are also natural. The task thus reduces to finding all triples $ (x,y,z) $ of positive natural numbers for which

is a natural number.
Assume that $ (x,y,z) $ is such a triple. Without loss of generality, we can assume that $ x \geq y \geq z $. Then

which means $ 1 \leq z \leq 3 $. If the numbers $ x $, $ y $, $ z $ are equal, then the relation (2) takes the form $ bx = 3 $, hence

In the further part, we will consider the case when not all three numbers $ x $, $ y $, $ z $ are equal. Then the inequality (3) is strict and we get $ 1 \leq z < 3 $,
which means $ z = 1 $ or $ z = 2 $. Hence $ b =\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq 1 + 1 + \frac{1}{2} < 3 $, and thus also $ b = 1 $ or $ b = 2 $.
Let us rewrite equation (2) in the following equivalent forms:

From the last form, it is clear that the factor $ (bz - 1) $ cannot be zero (which excludes the possibility $ b = z = 1 $). Multiplying both sides by this factor, we get

and by adding the term $ z^2 $ to both sides, we obtain the equation

As we stated above, the numbers $ b $ and $ z $ can only take the values $ 1 $ and $ 2 $, and they cannot be equal to 1 at the same time. We thus have three possibilities:

The equation (5) then takes the form

The first of these equations gives the solution

and the second — the solutions

(the third equation has no solutions in natural numbers $ x \geq y \geq z = 2 $).
Formulas (4) and (6) and (7) represent the general solution of the problem under the assumption that $ x \geq y \geq z $. Dropping this assumption, we obtain the following triples $ (x,y,z) $ as the complete solution of the problem:

*Remark.* Here is the justification of the fact that every rational root of a polynomial of the form

whose coefficients $ a_0, a_1, \ldots, a_{N-1} $ are integers, is also an integer. (In this problem, we are dealing with a polynomial of degree $ N = 3 $.) Let us write the rational root $ x $ of the polynomial $ P(t) $ in the form of an irreducible fraction $ x = r/q $ ($ q, r $ integers, $ q > 0 $). Multiplying the equality $ P(x) = 0 $ by $ q^N $ gives

All terms except $ r^N $ are integers divisible by $ q $. Therefore, $ r^N $ must also be divisible by $ q $. The fraction $ r/q $ is irreducible by assumption. Hence, we conclude that $ q = 1 $.