3. Find all natural numbers $n$ for which $2^{n}+n^{2016}$ is a prime number.
Solution. Let's consider three cases.

- If $n$ is even, then the given number is also even (and greater than two for $n>0$).
- If $n$ is odd and not divisible by 3, then $2^{n}$ gives a remainder of 2 when divided by 3, and $n^{2016}=\left(n^{504}\right)^{4}$ gives a remainder of 1 when divided by 3, so the sum is divisible by 3 (and greater than three for $n>1$). For $n=1$, the result is 3, which is a prime number.
- Finally, let $n$ be divisible by 3 (and odd, which is not used). Then the number in question is the sum of cubes: if $n=3k$, then

$2^{n}+n^{2016}=\left(2^{k}\right)^{3}+\left(n^{672}\right)^{3}=\left(2^{k}+n^{672}\right) \cdot\left(2^{2k}-2^{k} \cdot n^{672}+n^{2 \cdot 672}\right)-$ a composite number (it is obvious that $1<2^{k}+n^{672}<2^{n}+n^{2016}$ for $n \geqslant 3$).

Answer: $n=1$.