## Task 12/84

The solution sought is that of the equation

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

in natural numbers $x ; y ; z$, for which the product $x y z$ is minimal.
Assuming $x ; y ; z$ are solutions of the given equation. Then, by multiplying with $x^{2} y^{2} \neq 0$, we get the equation

$$
x^{2}+y^{2}=x^{2} y^{2} z^{-2}
$$

Since $x^{2} ; y^{2} \in \mathbb{N}$, it follows that $x^{2} y^{2} z^{-2}=n^{2} \in \mathbb{N}$, and the triplet $(x ; y ; n)$ can be represented in the form $\left(x_{0} d ; y_{0} d ; n_{0} d\right)$, where $\left(x_{0} ; y_{0} ; z_{0}\right)$ is a primitive Pythagorean triplet and $d \in \mathbb{N}, d \neq 0$. Therefore, we have

$$
n^{2}=n_{0}^{2} d^{2}=x^{2} y^{2} z^{-2}=x_{0}^{2} d^{2} y_{0}^{2} d^{2} z^{-2} \quad ; \quad z=x_{0} y_{0} n_{0}^{-1} d
$$

Due to the pairwise coprimality of $x_{0} ; y_{0} ; n_{0}$ (primitive Pythagorean triplet), $n_{0}$ must be a divisor of $d$: $d=n_{0} d^{\prime}$ with $d^{\prime} \in \mathbb{N}, d^{\prime} \neq 0$.

Thus, $x y z=\left(x_{0} y_{0} n_{0}\right)^{2} d^{\prime 3}$ is minimal if and only if $d^{\prime}=1$ and $x_{0} ; y_{0} ; z_{0}$ are the smallest numbers that can appear in a primitive Pythagorean triplet: $x_{0}=3, y_{0}=4$ (or vice versa), $n_{0}=d=5$, and thus $(x ; y ; z)=\left(x_{0} d ; y_{0} d ; z_{0} d=x_{0} y_{0}\right)=(3 \cdot 5 ; 4 \cdot 5 ; 3 \cdot 4)=(15 ; 20 ; 12)$ or $(20 ; 15 ; 12)$ (due to the symmetry in $x$ and $y$). Indeed,

$$
15^{-2}+20^{-2}=12^{-2}
$$