<s>[INST] 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.
 [/INST] Let $k = 2n-1$ then $$\dfrac{16(n^2-n-1)^2}{2n-1} = \dfrac{(4n^2-4n-4)^2}{2n-1}$$
$$ = \dfrac{((k+1)^2-2(k+1)-4)^2}{k} = {(k^2-5)^2\over k}$$
So $$k\mid (k^2-5)^2= k^4-10k^2+25$$ $$\implies k\mid 25 \implies k\in\{\pm 1, \pm 5,\pm 25\}$$
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