if $n$ is not divisible by any prime less than $2014$, then $n+c$ divides $a^n+b^n+n$ 
Find all triples $(a,b,c)$ of positive integers such that if $n$ is not divisible by any prime less than $2014$, then $\color{red}n+\color{red}c$ divides $\color{red}a^n+\color{red}b^n+\color{red}n$

This problem is from the New Zealand TST 2014 and there is only one solution:
$\textbf{answer}$:$\color{red}(\color{red}a,\color{red}b,\color{red}c)\color{red}=(1,1,2)$
Can you someone help explain why there is only one solution?
 A: Let q be the product of all primes less than 2014, and :
$$\begin{align}R&=&\{r\in \mathbb{N}/ r=1\mod q\} \\
P&= &\{p\in \mathbb{P}/ p=1\mod q\}\end{align}$$
where $\mathbb{N}$ denote  the set of positive integers and $\mathbb{P}$ the set of prime numbers.

Let $(a,b,c)$ be a triple of positive integers such that if $n$ is not divisible by any prime less than $2014$, then $n+c$ divides $a^n+b^n+n$

Given $r\in R,p\in P$, Let $k=(p-1)(r+c)+r$ so that  : 
$$n=1\mod q $$
( because $q$ divides $p-1$ and $r=1\mod q$)
hence $k$ is not divisible by any prime less than $2014$, then $k+c$ divides $a^k+b^k+k$ which is equivalent to $k+c$ divides $a^k+b^k-c=(a^k+b^k+k)-(k+c)$.
Now observe that $p$ divides $k+c$ hence $p$ divides $a^k+b^k-c$. The definition of $k$ gives us also $\begin{align} n&=r\mod (p-1)\end{align}$, using Fermat's little theorem we obtain:
 $$\begin{align} a^r+b^r-c&=0\mod p \end{align}$$ 
This result is true for all primes $p\in P$ and because $P$ is infinite as a consequence of Dirichlet's theorem:
$$a^r+b^r-c=0 $$
(notice that a positive constant cannot be divisible by an infinity of primes)  
and also this result is true for all integers $r\in R$ in particular for $1$ and $q+1$:
$$\begin{align} a+b&=c &(r=1) \\ a^{q+1}+b^{q+1}&=c &(r=q+1) \end{align}$$
Using the inequality:$$2^{q}(a^{q+1}+b^{q+1})\geq (a+b)^{q+1} $$ 
and hence $ 2^kc\geq c^{k+1}$ which implies $2\geq c$ and because $c=a+b$ and $c,a,b$ are positive we conclude that $c=2$ and $a=b=1$.
It remains to check that $(1,1,2)$ verifies the given propriety which is obvious.
