193*. Find all integer values of $x$ and $y$ for which at least one of the numbers

$$
x^{2}-2 x y+2 y^{2}, \quad x^{2}+2 x y+2 y^{2}
$$

is divisible by 5.
$$
\begin{gathered}
\left(x^{2}-2 x y+2 y^{2}\right)\left(x^{2}+2 x y+2 y^{2}\right)=\left(\left(x^{2}+2 y^{2}\right)-2 x y\right)\left(\left(x^{2}+2 y^{2}\right)+2 x y\right)= \\
=\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2}=x^{4}+4 x^{2} y^{2}+4 y^{4}-4 x^{2} y^{2}=x^{4}+4 y^{4}
\end{gathered}
$$

Since the sum $x^{4}+4 y^{4}$ is divisible by 5 by the condition, the numbers $x$ and $y$ are either both divisible or both not divisible by 5.

1) If $x$ and $y$ are divisible by 5, then everything is obvious.
2) Suppose $x$ and $y$ are not divisible by 5. Is this case possible?

Then $x^{4}$ and $y^{4}$, when divided by 5, leave a remainder of 1:

$$
x^{4}=5 k+1, \quad y^{4}=5 l+1
$$

where $k$ and $l$ are non-negative integers. We get:

$$
x^{4}+4 y^{4}=(5 k+1)+4(5 l+1)=5 k+20 l+5
$$

Since the last sum is divisible by 5, this variant is also possible.

Answer: all integers $x$ and $y$, simultaneously divisible or not divisible by 5.