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?
 A: Modulo $5$, the residues of $x^2$ take on the values $0, 1, 4, 4, 1$.
So $x^2 + y^2 \equiv 0 \bmod 5$ will be true if both residues are $0$ or if one is $1$ and the other is $4$.
This implies the probability is $\frac{1}{5} \cdot \frac{1}{5} + 2 \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{9}{25}$
Edit: Let's say we wanted to see the remainder after dividing integer $n$ by integer $k$. This means we could write $n = qk + r$ where $0 \leq r \lt k$ where $q$ is an integer. For example $20$ divided by $7$ can be modeled as $20 = 2 \cdot 7 + 6$.
Then:
$$\begin{align}
\frac{x^2+y^2}{5}  &= \frac{(q_1 \cdot 5 + r_1) +(q_2 \cdot 5 + r_2 )}{5} \\
&= \frac{(q_1 + q_2) \cdot 5 + (r_1 + r_2) }{5}  \\
&= (q_1 + q_2) + \dfrac{r_1 + r_2}{5}
\end{align}$$
In other words, after we divide $x^2 + y^2$ by $5$, we'll get some integer $q_1, q_2$ values (that we don't care about), but then we need the sum of the remainders to be divisible by $5$ if $\frac{x^2+y^2}{5}$ is going to be an integer.
A: 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
$$\frac{1}{25}+\frac{4}{25}+\frac{4}{25}=\frac{9}{25}=36\%.$$
