Multiple of $5$ If two natural nos. $x,y$ are selected at random then the probability that $x^2+y^2$ is a multiple of 5?
I think they should give a finite set for selecting natural nos. otherwise there are infinite ordered pairs of $x$ and $y$.
Is the question incomplete or am I missing some trick?

Without loss of generality, you can assume that $x$ and $y$ are restricted to the set $\{0, 1, 2, 3, 4\}$, as pointed out by Lord Shark the Unknown in a comment.  Now,
$$z \equiv 0 \pmod 5 \implies z^2 \equiv 0 \pmod 5$$
$$z \equiv 1 \pmod 5 \implies z^2 \equiv 1 \pmod 5$$
$$z \equiv 2 \pmod 5 \implies z^2 \equiv 4 \pmod 5$$
$$z \equiv 3 \pmod 5 \implies z^2 \equiv 4 \pmod 5$$
$$z \equiv 4 \pmod 5 \implies z^2 \equiv 1 \pmod 5$$
Thus, $x^2 + y^2 \equiv 0 \pmod 5$ can only happen if
$$x \equiv y \equiv 0 \pmod 5$$
or $x \equiv 1 \pmod 5$ and $y \equiv 2 \pmod 5$, or $x \equiv 4 \pmod 5$ and $y \equiv 2 \pmod 5$, or $x \equiv 1 \pmod 5$ and $y \equiv 3 \pmod 5$, or $x \equiv 4 \pmod 5$ and $y \equiv 3 \pmod 5$, or with the roles of $x$ and $y$ interchanged in the last $4$ cases.
Therefore, this gives a probability of
$$\frac{1}{25}+\frac{4}{25}+\frac{4}{25}=\frac{9}{25}=36\%.$$
