I am supposed to solve for the general solution of $f(n+2)=2(f(n+2))^2 -f(n+2)f(n)-2012$. I tried the method of generating functions but I am stuck with the power $2$ on the RHS. any other methods or idea on how to proceed?
Edit: $f:\mathbb N \to \mathbb N$
On
As it is formulated now, your problem has no solution.
Note that the relation that you give (that does not specify the initial terms $f(0)$ and $f(1)$ - but this has no consequences over our reasoning) can be rewritten as
$$2012 = f(n+2) \Big( 2 f(n+2) - f(n) - 1 \Big) ,$$
which shows that $f(n+2) \mid 2012, \ \forall n \ge 0$. But $2012 = 4 \cdot 503$, which means that $f(n+2) \in \{ 1, 2, 4, 503, 1006, 2012 \} \ \forall n \ge 0$.
On the other hand, the given relation can be rearranged as a 2nd degree equation in $f(n+2)$ as
$$2 f(n+2) ^2 - \Big( f(n) + 1 \Big) f(n+2) - 2012 = 0 ,$$
which has for discriminant the quantity $\Big( f(n) + 1 \Big)^2 + 16096$. In order for the "solution" $f(n+2)$ to be a natural number, the discriminant must be a perfect square. If $n \ge 2$ then $f(n)$ must be one of the six divisors listed above, but none of them gives a perfect square when plugged in the expression of the discriminant, which shows that your problem has no solution. Are you sure that you have copied it correctly?
Let $f:\mathbb N\to\mathbb N$ satisfy $$2f(n+2)^2-\big(f(n)+1\big)f(n+2)-2012=0$$ for every natural number $n$. Since $f(n+2)$ can't be negative, we have: $$f(n+2)=\frac{f(n)+1+\sqrt{\big(f(n)+1\big)^2+16096}}4$$ Now, $f(n+2)$ must be a natural number, so $\big(f(n)+1\big)^2+16096$ must be a perfect square. Hence there is a natural number $k$ such that: $$\big(f(n)+1\big)^2+16096=k^2$$ $$\therefore\quad k^2-\big(f(n)+1\big)^2=16096$$ $$\therefore\quad \big(k-f(n)-1\big)\big(k+f(n)+1\big)=2^5\cdot503$$ Because $k-f(n)-1$ and $k+f(n)+1$ have the same parity and their product is even, therefore both of them have to be even. Also because $f(n+2)=\frac{f(n)+1+k}4$, therefore $k+f(n)+1$ must be divisible by $4$. Obviously $k+f(n)+1>k-f(n)-1$, so by the above factorization, $\big(k-f(n)-1,k+f(n)+1\big)\in\{(8,2012),(4,4024),(2,8048)\}$, which yields $\big(f(n),f(n+2)\big)\in\{(1001,503),(2009,1006),(4022,2012)\}$. But this is impossible since $\big(f(n+2),f(n+4)\big)$ must be an element of the same set and hence $f(n+2)\in\{1001,2009,4022\}\cap\{503,1006,2012\}=\varnothing$, which leads to a contradiction. So no such function exists.