2. The natural number $a$ is not divisible by 5. Determine the remainder obtained when the number $a^{4}$ is divided by the number 5.
Solution. A number a that is not divisible by 5 can be written in the form $5 k \pm 1$ or $5 k \pm 2$. We have

$$
\begin{aligned}
& a^{2}=(5 k \pm 1)^{2}=25 k^{2} \pm 10 k+1=5\left(5 k^{2} \pm 2 k\right)+1=5 m+1 \text { or } \\
& a^{2}=(5 k \pm 2)^{2}=25 k^{2} \pm 20 k+4=5\left(5 k^{2} \pm 4 k\right)+4=5 n+4
\end{aligned}
$$

If we square again, we get

$$
\begin{aligned}
& a^{4}=(5 m+1)^{2}=5\left(5 m^{2}+2 m\right)+1=5 M+1 \text { or } \\
& a^{4}=(5 n+4)^{2}=5\left(5 n^{2}+8 n+3\right)+1=5 N+1
\end{aligned}
$$

so the desired remainder is 1.