Finding the largest $n$ such that an expression is an integer

Question

Find the largest $ n $ such that $ \frac{n^{2016} - 2n^{2015} + 3n^{2014} - \cdots + 2015n^2 - 2016n + 2017}{(2017 + n)} $ is an integer:

My attempt:

\begin{eqnarray*} & & n^{2016} - 2n^{2015} + 3n^{2014} - \cdots + 2015n^2 - 2016n + 2017 \\ &=& n^{2016}\left(1 - \frac{2}{n} + \frac{3}{n^2} - \cdots + \frac{2015}{n^{2014}} - \frac{2016}{n^{2015}} + \frac{2017}{n^{2016}}\right) \\ &= & n^{2016}\left(1 + 2\left(\frac{-1}{n}\right) + 3\left(\frac{-1}{n}\right)^2 + \cdots + 2015\left(\frac{-1}{n}\right)^{2014} + 2016\left(\frac{-1}{n}\right)^{2015} + 2017\left(\frac{-1}{n}\right)^{2016}\right) \\ \end{eqnarray*} Consider: \begin{eqnarray*} 1 + x + x^2 + x^3 + \cdots + x^{2017} &=& \frac{x^{2018} - 1}{x - 1} \\ 1 + 2x + 3x^2 + \cdots + 2017x^{2016} &=& \frac{(x - 1)2018x^{2017} - (x^{2018} - 1)}{(x - 1)^2} \\ &=& \frac{2018x^{2018} - 2018x^{2017} - x^{2018} + 1}{(x - 1)^2} \\ &=& \frac{2017x^{2018} - 2018x^{2017} + 1}{(x - 1)^2} \\ \end{eqnarray*} Put $ x = \frac{-1}{n} $, we get \begin{eqnarray*} 1 + 2x + 3x^2 + \cdots + 2017x^{2016} &=& \frac{2017x^{2018} - 2018x^{2017} + 1}{(x - 1)^2} \\ 1 + 2\left(\frac{-1}{n}\right) + 3\left(\frac{-1}{n}\right)^2 + \cdots + 2017\left(\frac{-1}{n}\right)^{2016} &=& \frac{2017\left(\frac{-1}{n}\right)^{2018} - 2018\left(\frac{-1}{n}\right)^{2017} + 1}{(\left(\frac{-1}{n}\right) - 1)^2} \\ n^{2016}\left(1 + 2\left(\frac{-1}{n}\right) + 3\left(\frac{-1}{n}\right)^2 + \cdots + 2017\left(\frac{-1}{n}\right)^{2016} \right) &=& n^{2016}\left(\frac{2017\left(\frac{-1}{n}\right)^{2018} - 2018\left(\frac{-1}{n}\right)^{2017} + 1}{(\left(\frac{-1}{n}\right) - 1)^2}\right) \\ \end{eqnarray*}Now focus on the right hand side, simplifying \begin{eqnarray*} & & n^{2016}\left(\frac{2017\left(\frac{-1}{n}\right)^{2018} - 2018\left(\frac{-1}{n}\right)^{2017} + 1}{(\left(\frac{-1}{n}\right) - 1)^2}\right) \\ &=& n^{2016}\left(\frac{2017\left(\frac{1}{n}\right)^{2018} + 2018\left(\frac{1}{n}\right)^{2017} + 1}{(\left(\frac{1}{n}\right) + 1)^2}\right) \\ &=& \frac{2017\left(\frac{1}{n^2}\right) + 2018\left(\frac{1}{n}\right) + n^{2016}}{(\left(\frac{1}{n}\right) + 1)^2} \\ &=& \frac{2017 + 2018n + n^{2018}}{(n + 1)^2} \\ &=& \frac{2017 + n + 2017n + n^{2018}}{(n + 1)^2} \\ \end{eqnarray*}

Feel like and Fermat's theorem, but not sure how to move forward from there.

Answer 1

This answer assumes that $n$ is an integer.

From what you've done, we want to find the largest integer $n$ such that $$\frac{n^{2018}+2018n+2017}{(n+2017)(n+1)^2}\tag1$$ is an integer.

By the way, there exists only one set of integers $(a_0,a_1,\cdots,a_{2015},c_0,c_1,c_2)$ such that $$\small n^{2018}+2018n+2017=(n+2017)(n+1)^2(a_{2015}n^{2015}+a_{2014}n^{2014}+\cdots +a_0)+c_2n^2+c_1n+c_0\tag2$$ Now setting $n=-1$ gives $$(-1)^{2018}+2018\times (-1)+2017=c_2\times (-1)^2+c_1\times (-1)+c_0\tag3$$

Also, setting $n=-2017$ gives $$(-2017)^{2018}+2018\times (-2017)+2017=c_2\times (-2017)^2+c_1\times (-2017)+c_0\tag4$$

Differentiating the both sides of $(2)$ and setting $n=-1$ give $$2018\times (-1)^{2017}+2018=2c_2\times (-1)+c_1\tag5$$ From $(3)(4)(5)$, we get $$c_0=m,\quad c_1=2m,\quad c_2=m$$ where $$m=\frac{2017^2(2017^{2016}-1)}{2016^2}$$ Note here that $m$ is integer since it is easy to see that $$(M-1+1)^{M-1}-1=\sum_{k=1}^{M-1}\binom{M-1}{k}(M-1)^k$$ is divisible by $(M-1)^2$.

So, we want to find the largest integer $n$ such that $$\frac{c_2n^2+c_1n+c_0}{(n+2017)(n+1)^2}=\frac{mn^2+2mn+m}{(n+2017)(n+1)^2}=\frac{m}{n+2017}$$ is an integer.

Therefore, the answer is $$m-2017=\color{red}{\frac{2017^2(2017^{2016}-1)}{2016^2}-2017}$$