When does $2n-1$ divide $16(n^2-n-1)^2$? Find all integers $n$ such that 
$\dfrac{16(n^2-n-1)^2}{2n-1}$ 
is an integer.

Hint $\!\overbrace{{\rm numerator} = f(\color{#c00}{2n})}^{\textstyle\begin{align}{\ \ \ \ \ (4(n^2\,-\ n\ -\ 1))^2\\ = ((\color{#c00}{2n})^2\!-\!2(\color{#c00}{2n})\!-\!4)^2}\end{align}_{\phantom{|}}}\! $  & $\bmod  \overbrace{2n-1}^{\textstyle\color{#c00}{2n\equiv 1}}\!:\ f(\color{#c00}{2n})\equiv f(\color{#c00}1)\,$ by Polynomial Congruence Rule.
Remark $ $ More generally we can use modular fractions and simply evaluate $\,f(x)\,$ at $\,\color{#c00}{n\equiv 1/2}\,$ or, equivalently we can use the nonmonic division algorithm to work fraction-free. See also here.
