8. find all triplets of natural numbers $(a, b, c)$ such that for any natural number $n$ that has no prime divisor less than 2015

$$
n+c \mid a^{n}+b^{n}+n
$$
## Solution:

Lemme - There are infinitely many prime numbers $p$ such that $p \equiv 2(\bmod 3)$.

Proof: By contradiction, let $p_{1}, \ldots, p_{n}$ be the only primes congruent to 2 modulo 3.

If $n$ is even, then $p_{1} p_{2} \ldots p_{n}+1 \equiv 2(\bmod 3)$ and thus this number has at least one prime divisor $q \equiv 2(\bmod 3)$. However, it is not divisible by any of the $p_{i}$.

If $n$ is odd, then $p_{1} p_{2} \ldots p_{n}+3 \equiv 2(\bmod 3)$ and we conclude similarly.

We take a prime $p>\max (|a+b-c|, a, b, 2015)$ and an integer $n$ such that:

$$
\begin{aligned}
& n \equiv -c \quad(\bmod p) \\
& n \equiv 1 \quad(\bmod p-1) \\
& n \equiv 1 \quad\left(\bmod p_{i}\right) \text { for } p_{i}<2015 \text { such that }\left(p-1, p_{i}\right)=1
\end{aligned}
$$

Since $p > \max (|a+b-c|, a, b)$, we have $(a, p) = (b, p) = 1$ and since $n \equiv 1(\bmod p-1)$, we have $a^{n} = a^{k(p-1)+1} \equiv a(\bmod p)$.

Therefore, $|a+b-c|$ is divisible by $p$, but $p > |a+b-c|$, so $a+b=c$.

We repeat the process with a prime $p>\max \left(\left|a^{3}+b^{3}-a-b\right|, a, b, 2015\right), p \equiv 2(\bmod 3)$ and an integer $n$ such that:

$$
\begin{aligned}
& n \equiv -a-b \quad(\bmod p) \\
& n \equiv 3 \quad(\bmod p-1) \\
& n \equiv 1 \quad\left(\bmod p_{i}\right) \text { for } p_{i}<2015 \text { such that }\left(p-1, p_{i}\right)=1
\end{aligned}
$$

Then we have

$$
a^{n}+b^{n}+n \equiv a^{3}+b^{3}+n \equiv a^{3}+b^{3}-a-b \equiv 0 \quad(\bmod p)
$$

Thus, $a+b=a^{3}+b^{3}$ with equality if and only if $a=b=1$. Therefore, the only solution is $(1,1,2)$, which is trivially verified to be a solution.