3A. Determine all pairs of natural numbers $(x, y)$ such that $x$ and $y$ are coprime and

$$
x^{2}+2 y^{2}+334=\operatorname{LCM}\left(x^{2}, y^{2}\right)
$$
Solution. Since $x$ and $y$ are coprime, it follows that

$$
\operatorname{GCD}\left(x^{2}, y^{2}\right)=\operatorname{GCD}(x, y)=1
$$

so $x^{2} y^{2}=\operatorname{LCM}\left(x^{2}, y^{2}\right) \operatorname{GCD}\left(x^{2}, y^{2}\right)=\operatorname{LCM}\left(x^{2}, y^{2}\right)$. According to this, the given equation takes the form $x^{2}+2 y^{2}+334=x^{2} y^{2}$ and we transform it into the equation $y^{2}=1+\frac{336}{x^{2}-2}$. Since $y^{2}$ is a natural number, $x^{2}-2$ must be a divisor of the number $336=2^{4} \cdot 3 \cdot 7$. From this we get

$$
x^{2}-2 \in\{1,2,4,8,16,3,6,12,24,46,7,14,28,56,112,21,42,84,168,336\}
$$

so

$$
x^{2} \in\{3,4,6,10,18,5,8,14,26,48,9,16,30,58,114,23,44,86,170,338\}
$$

In the above set, the only squares of natural numbers are 4, 9, and 16, so $x=2,3,4$. In this case:

- if $x=2$, then $y^{2}=1+\frac{336}{2^{2}-2}=169$, i.e., $y=13$,
- if $x=3$, then $y^{2}=1+\frac{336}{3^{2}-2}=49$, i.e., $y=7$,
- if $x=4$, then $y^{2}=1+\frac{336}{4^{2}-2}=25$, i.e., $y=5$.