## Task A-1.1.

Determine the smallest natural number $a$ such that the value of the expression

$$
\frac{n^{8}+n^{6}+n^{4}+n^{2}+a}{n^{2}-1}
$$

for $n=2014$ is an integer divisible by 3.
## Solution.

By simplifying the expression, we get

$$
\begin{aligned}
\frac{n^{8}+n^{6}+n^{4}+n^{2}+a}{n^{2}-1} & =\frac{n^{8}-n^{6}+2\left(n^{6}-n^{4}\right)+3\left(n^{4}-n^{2}\right)+4\left(n^{2}-1\right)+a+4}{n^{2}-1} \\
& =n^{6}+2 \cdot n^{4}+3 \cdot n^{2}+4+\frac{a+4}{n^{2}-1}
\end{aligned}
$$

The remainder when 2014 is divided by 3 is 1, so all powers of 2014 give a remainder of 1 when divided by 3.

Thus, the number $2014^{6}+2 \cdot 2014^{4}+3 \cdot 2014^{2}+4$ gives the same remainder as $1+2+3+4$ when divided by 3, so that remainder is 1. For the entire expression to be divisible by 3, the number $\frac{a+4}{2014^{2}-1}$ must give a remainder of 2 when divided by 3, so the smallest possible $a$ is the one for which $\frac{a+4}{2014^{2}-1}=2$, i.e., $a=2 \cdot 2014^{2}-6$.