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.

Starting off with the initial expression:
$$\frac{16(n^2-n-1)^2}{2n-1}$$
Completing the square:
$$=\frac{4^2(n^2-n-1)^2}{2n-1}$$
$$=\frac{(4n^2-4n-4)^2}{2n-1}$$
$$=\frac{((2n-1)^2-5)^2}{2n-1}$$
Some manipulation:
$$=(2n-1)^3(1-\frac{5}{(2n-1)^2})^2$$
Expanding the expression:
$$=(2n-1)^3-10(2n-1)+\frac{25}{2n-1}$$
Since the first two terms will always be integers,
If this expression is to be an integer, then:
$$(2n-1)= factor\ of\ 25$$
$$n=\frac{1+factor\ of\ 25}{2}$$
The factors of 5 are $\pm 1, \pm 5, \pm 25$
thus,
n=$1\pm1\over 2$
or
n=$1\pm 5 \over 2$
or
n=$1\pm 25 \over 2$
n=1,0,-2,3,-12,13
