'$$n$$' is a natural number. The number of '$$n$$' for which $$\frac{16(n^2-n-1)^2}{2n-1}$$ is a natural number is ------.

We are given the expression: $$\frac{16(n^2-n-1)^2}{2n-1}$$

Now, let's try to rewrite the numerator as:

$$16(n^2-n-1)^2=(2n-1)Q(n)+r$$ where $$Q(n)$$ is the quotient and $$r$$ is the remainder

So, now to find the remainder, lets put $$n=\dfrac{1}{2}$$ on both sides

$$16(\dfrac{1}{4}-\dfrac{1}{2}-1)^2=(0)Q(\dfrac{1}{2})+r$$

$$25=r$$

So, $$16(n^2-n-1)^2=(2n-1)Q(n)+25$$

Now, $$\dfrac{16(n^2-n-1)^2}{2N-1}=Q(x)+\dfrac{25}{2n-1}$$

Now, $$Q(n)$$ is going to be a cubic function, which will give integral results

Only, term $$\dfrac{25}{2n-1}$$ will be troublesome

So $$2n-1$$ must divide $$25$$. The positive divisors of $$25$$ are $$1, 5, 25$$, giving $$n=1, 3, 13$$, so there are exactly $$3$$ such values of $$n$$

But we still need to verify these values so that we get natural numbers as a solution

Putting $$n=1$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$16$$

Putting $$n=3$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$80$$

Putting $$n=13$$ in $$\frac{16(n^2-n-1)^2}{2n-1}$$, we get :$$15,376$$

All of these are natural numbers, hence there are $$3$$ possible values of $$n$$