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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.

VividD
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    If it were a prime, it would be unreasonable to ask for a proof (even a computer would struggle to write one, and it wouldn't be easy for a human to read it). Therefore, it's not a prime. – Rebecca J. Stones Jul 14 '14 at 06:41
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    @Rebecca What out-of-the-box thinking! If you put this as an answer, I would be seriously tempted to mark it as the best. – VividD Jul 14 '14 at 06:42
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    In case you're interested, $1765^{1766} + 1766^{1767} + 1767^{1765} + 1$ is prime, according to Mathematica. – Dejan Govc Jul 14 '14 at 06:56
  • @DejanGovc, yes, thank you, I am interested. I did some experiments too. – VividD Jul 14 '14 at 06:58
  • Did you changed the question? I notice a difference in what is posted against what @T.Bongers used – MonK Jul 14 '14 at 08:11
  • @Sid Yes, for a brief time it was a different formulation involving 2014, 2015, 2016, because of my mistake, but this is a final formulation, what is now in the question. – VividD Jul 14 '14 at 08:14

1 Answers1

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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 $15000$.

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    I apologize, original formulation involved 2013, 2014, 2015, and I naivelly wanted to avoid 13 as unlucky number, changed to 2014, 2015, 2016, and that lead to your answer. +1 anyway for quick reaction! Sorry again, my bad. – VividD Jul 14 '14 at 06:35
  • @VividD I've edited my answer to address the new question too. –  Jul 14 '14 at 06:36
  • OK! :) (I will not change the question this time haha ;)) – VividD Jul 14 '14 at 06:39
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    If the downvoter would like to offer a suggestion or improvement, I'd appreciate it greatly. –  Jul 14 '14 at 06:43