# The diophantine equation $m(n-2016)=n^{2016}$

How many natural numbers, $n$, are there such that $$\frac{n^{2016}}{n-2016}$$ is a natural number?

HINT.-There are lots of solutions

HINT.-$\frac{n}{n-2016}=m \iff \frac{2016}{n-2016}=m-1$ and if, for example, $x=a^r\cdot b^s$ then the factors of $x$ are all the $a^i\cdot b^j$ with $0\le i\le r$ and $0\le j\le s$

• You have been quite active here for a significant time, so you should know to add more context to the question. What have you tried? – Rory Daulton Dec 19 '15 at 15:00
• I meant the first part as a compliment. You have been a member of the MSE community for nine months and have asked good questions and given good answers. But despite this, useful context is missing from this question. You give a hint: does that mean you know the answer and this is a challenge, or were you given this as an assignment? And so on. I am sure, with your demonstrated skills, that you could improve the question. And yes, I did vote to close the question, but I will readily remove that vote if you add context. – Rory Daulton Dec 20 '15 at 12:24
• Thank you Sir, I'll give another hint (believe me, however,I swear I am not interested in reputation points, I am just as the nickname of a member of MSE, a Mathlover) – Piquito Dec 20 '15 at 12:42
• @RoryDaulton: Dear friend, my English is so weak to misundertand your courtesy. Believe me, I read good enough and I write badly with the help of Google, but I cannot talk, I can't converse so I don't speak English, – Piquito Dec 20 '15 at 13:19
• @RoryDaulton: I wanted to delete this question but I have been noticed that is bad to delete answered questions. – Piquito Dec 20 '15 at 13:22

Here are some solutions (I think all of them):

$2017$

$2^k + 2016$ for $1\leq k\leq 10080$.

$a^k + 2016$ for $1\leq k\leq 4032$ and $a\in \{3,6,12\}$

$b^k + 2016$ for $1\leq k\leq 2016$ and $b\mid 2016$ and $b\not\in \{1,2,3,4,6,12\}$

You can show that this solutions won't work for $k$ greater than the upper limits I've set. Also if a prime $p$ divides $n - 2016$ but doesn't divide $2016$ then it won't divide $n^{2016}$ (since it doesn't divide $n$). So from $2016 = 2^53^27^1$ we know that any solution $n$ must be of the form $2^r3^s7^t + 2016$.

• See, please, dear friend, the second hint I have added to the question. A part of you reasonning can be change by $E>0$ where $E$ is the given expression. – Piquito Dec 20 '15 at 13:06
• Do you mean $n^{2016}/(n-2016) > 0$. If so, just disregard the last line. – Darth Geek Dec 20 '15 at 13:10
• By definition because the quotient must be a natural number. – Piquito Dec 20 '15 at 13:24

Show, in general, that $\gcd\left(x-k,x^n\right)=\gcd\left(x-k,k^n\right)$ for all integers $x,k,n$ with $n \geq 0$.