The Diophantine Equation $m(n-2016)=n^{2016}$

Question

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 $0\le j\le s$

Answer 1

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 $2^r3^s7^t + 2016$.

Answer 2

Show, in general, that $\gcd\left(x-k,x^m\right)=\gcd\left(x-k,k^m\right)$ for all integers $x,k,m$ with $m \geq 0$. Therefore, for positive integers $k$ and $m$, the number of integers $n$ such that $$\dfrac{n^m}{n-k}\in\mathbb{Z}_{> 0}$$ is precisely the number of positive divisors of $k^m$.