Is $2013^{2014}+2014^{2015}+2015^{2013}+1$ a prime? (usage of a computer not allowed) 
Prove or disprove: $$2013^{2014}+2014^{2015}+2015^{2013}+1$$ is a prime number, without using a computer.

I tried to transform the expression $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, but couldn't reach useful conclusions.
 A: For the original version, which asked abuot
$$2014^{2015} + 2015^{2016} + 2016^{2014} + 1$$
then no: This is an even number (sum of two evens and two odds), so not prime. It's also divisible by $3$.

For the new question: The number is divisible by $7$, so it is not prime. This can be computed by hand by reducing everything modulo $7$ and using, e.g. successive squaring to compute the residue on the left side. For example,
\begin{align*}
2013 \equiv 4 &\pmod 7 \\
2013^2 \equiv 2 &\pmod 7 \\
2013^4 \equiv 4 &\pmod 7 \\
2013^8 \equiv 2 &\pmod 7
\end{align*}
and so on; it keeps oscillating. Now after writing $2014$ as a sum of powers of $2$, 
\begin{align*}
2013^{2014} &\equiv 4^{2014} \\
&\equiv 4^{1024} \cdot 4^{512} \cdot 4^{256} \cdot 4^{128} \cdot 4^{64} \cdot 4^{16} \cdot 4^8 \cdot 4^4 \cdot 4^2 \\
&\equiv 4\cdot2\cdot4\cdot2\cdot4\cdot4\cdot2\cdot 4 \cdot 2 \\
&\equiv 16384 \equiv 4
\end{align*}
modulo $7$. Hence, the first term is $4$ modulo $7$; similar computations can be done for the other terms, and summing indeed gives $0$.
Alternatively, for a vastly faster computation, Fermat's little theorem implies that $4^6 \equiv 1 \pmod 7$. Writing $2014 = 6 \cdot 335 + 4$, we have
$$2013^{2014} \equiv \Big(4^{6}\Big)^{335} \cdot 4^4 \equiv 4 \pmod 7$$

It's worth mentioning that even without any special form, this is really, really unlikely to be prime. As a consequence of the prime number theorem, if you choose a number on this scale ($\sim 2^{15000}$), then the probability that it's prime is about $1$ in $15000$.
