## Task 35/70

Determine all triples $(x ; y ; z)$ of natural numbers with the smallest $z$ that satisfy the equation

$$
\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{z^{2}}
$$
Multiplying the given equation

$$
\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{1}{z^{2}}
$$

by \(x^{2} y^{2} z^{2}\), we obtain the equation \((y z)^{2}+(x z)^{2}=(x y)^{2}\). By substituting \(u=y z, v=x z, w=x y\), we get \(u^{2}+v^{2}=w^{2} \quad(3)\).

Since \(x, y\), and \(z\) are supposed to be natural numbers according to the problem statement, \(u\), \(v\), and \(w\) are also natural numbers after the substitution, and \((u ; v ; w)\) is a Pythagorean triple. Therefore, each solution of equation (1) can be uniquely associated with a Pythagorean triple through the substitution.

To find solutions of (1) in natural numbers, one can start from Pythagorean triples. Once such a triple is found, dividing (3) by \(u^{2} v^{2} w^{2}\) yields the equation

$$
\frac{1}{v^{2} w^{2}}+\frac{1}{u^{2} w^{2}}=\frac{1}{u^{2} v^{2}}
$$

from which we get \(x=v w, y=u w, z=u v\). The number \(z\) takes the smallest possible value precisely when the Pythagorean triple in (3) with the smallest numbers is chosen, which are the triples \((3 ; 4 ; 5)\) and \((4 ; 3 ; 5)\). They provide the solution triples \((20 ; 15 ; 12)\) and \((15 ; 20 ; 12)\) of equation (1) with the smallest \(z\).