Non linear recurrence relation? for $ f: \mathbb N \rightarrow \mathbb N $, 
How do I solve  $ f_n - f_{n+2} = f_{n+3} \times (f_{n+2} - f_{n+4})$
I tried the generating function but it only seems to work for linear relations. any hints on how to make it linear? 
 A: Put
$$f_{2m}=:g_m,\quad f_{2m+1}=:h_m\qquad(m\geq0)\ .$$
In terms of the $g_m$ and the $h_m$ the instances $n=0$ and $n=1$ of the given recurrence read as follows:
$$g_0-g_1=h_1(g_1-g_2),\quad h_0-h_1=g_2(h_1-h_2)\ .$$
In all we obtain the following coupled system of recurrences:
$$g_m-g_{m+1}=h_{m+1}(g_{m+1}-g_{m+2})\qquad(m\geq0)\ ,\tag{1}$$
$$h_m-h_{m+1}=g_{m+2}(h_{m+1}-h_{m+2})\qquad(m\geq0)\ .\tag{2}$$
This system has the trivial constant solutions
$$g_m=g,\quad h_m=h\qquad(m\geq0)$$
with arbitrary $g$, $h\in{\mathbb N}$.
Assume now that $h_r- h_{r+1}\ne0$ for an $r\geq0$. Then from $(2)$ it follows that the quantities $d_m:=|h_m-h_{m+1}|$ are weakly decreasing and $\ne0$ for $m\geq r$, hence ultimately constant $\geq1$. This implies that $g_m=1$ for all large $m$, and from $(1)$, going backwards, we then deduce that $g_m-g_{m+1}=0$ for all $m\geq0$. This shows that necessarily $g_m=1$ for all $m\geq0$. The recurrence $(2)$ then implies that
$$h_m= h_0+m\>d\qquad(m\geq0)$$
for some $d\in{\mathbb N}_{\geq1}$.
Similarly: If $g_r- g_{r+1}\ne0$ for an $r\geq0$ we necessarily have $h_m=1$ for all $m\geq0$, and
$$g_m= g_0+m\>d\qquad(m\geq0)$$
for some $d\in{\mathbb N}_{\geq1}$.
