Proof of ${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4} = -6$ Observe:
\begin{matrix} F(n)|&{F(n)}^{4}& - {4F(n+1)}^{4}& - {19F(n+2)}^{4}&- {4F(n+3)}^{4}&{F(n+4)}^{4}& = -6\\ 1|& 1& -4& -304& -324& 625&=-6\\ 1|& 1& -64& -1539& -2500& 4096&=-6\\ 2|& 16& -324& -11875& -16384& 28561&=-6\\ 3|& 81& -2500& -77824& -114244& 194481&=-6\\ 5|& 625& -16384& -542659& -777924& 1336336&=-6\\ 8|& 4096& -114244& -3695139& -5345344& 9150625&=-6 \end{matrix}
I see that the proof is true but I cant quite grasp the pattern.  I'm more interested in hints rather than a solution. Ive read through this website about Fibonomials a couple times and have a decent understanding but I cant figure how to apply it in this situation. It may not even be necessary. I've tried proving through induction  but ended up with almost exactly the same problem. I think there may be a proof through some sort of recursive sequence but I don't know enough to prove something like that.
I'm more interested in hints rather than a solution.
 A: Here is a fun theorem that's useful for these types of problems: For any integers $d$ and $k$ there exists an integer $N$ such that for any polynomial $p(x_1,x_2,...x_k)$ of degree $d$,  $p(F_n,F_{n+1},...,F_{n+k-1})=0$  for $n = 0,1,2, ... ,N$ implies $p(F_n,F_{n+1},...,F_{n+k-1})=0$ for all $n$.
In your case $d = 4$ and $k = 5$. Depending on how carefully you go through the proof of this theorem you can get $N$ down to say 100 pretty easily, and down to around 18 with a little more work (if my mental calculations are correct).  Either way you can have a computer check that many values pretty easily.
A: Okay, let me give some hints by gesturing at a few things.
On the left-hand side, we have an operator of the form $X^4 - 4X^3 - 19X^2 - 4X  + 1 = (X^2-7 X+1) (X^2+3 X+1)$ applied to the sequence $F(n)^4$.
Let $\phi = \frac{1+\sqrt{5}}{2}$, the positive root of the characteristic equation for the Fibonacci recurrence.
Interestingly enough, $X^2-7 X+1$ is the minimal polynomial for $\phi^4$, and $X^2 + 3X + 1$ is the minimal polynomial for $-\phi^2$.  These things are not coincidences.
If the extremely abstract approach doesn't appeal to you, here's a merely very abstract view: think about the identity $5F_n^2 = L_{2n} + 2\cdot(-1)^n$.  The right-hand side satisfies some recursion related to the Fibonacci sequence.  If you can make sense of this fact, and generalize it to $F_n^4$, you're in the home stretch.
A: This is not the most elegant solution, but its a very straight forward computation using the standard tools we use to solve recurence-relations.
Let $a_n = F_n^4$ (where $F_n$ represents the terms in your recurence relation which we don't yet know is the Fibonacci sequence) then $$a_{n+4} - 4a_{n+3} - 19a_{n+2} - 4a_{n+1} + a_n = -6$$ 
The characteristic polynomial is
$$x^4 - 4x^3 - 19x^2 - 4x + 1 = (x^2+3x+1)(x^2-7x+1)$$
The particular solution is $a_n = C = \frac{6}{25}$ so the full solution reads
$$a_n = \frac{6}{25} + a r_+^n + b r_-^n + c s_+^n + d s_-^n$$
where 
$$r_\pm = \frac{-3 \pm \sqrt{5}}{2} = \left(\frac{1\pm \sqrt{5}}{2}\right)^2$$
$$s_\pm = \frac{7 \pm 3\sqrt{5}}{2} = r_{\pm}^2 =  \left(\frac{1\pm \sqrt{5}}{2}\right)^4$$
are the roots of the characteristic polynomial. Now note that the Fibonacci numbers statisfy
$$F_n = e\left(\frac{1+\sqrt{5}}{2}\right)^n + f\left(\frac{1-\sqrt{5}}{2}\right)^n$$
Fixing the constants $a,b,c,d,e,f$ gives us $a_n = F_n^4$ where $F_n$ here is the Fibonacci sequence.
${\bf Added}:$
Defining $q_\pm = \frac{1\pm\sqrt{5}}{2}$ then $$F_n = \frac{1}{\sqrt{5}}(q_+^n - q_-^n)$$ so
$$F_n^4 = \frac{1}{25}\left((q_+^4)^n + (q^4_-)^n + 4(q_+^3 q_-)^n + 6 (q^2_- q^2_+)^n + 4(q_+q_-^3)^n\right)$$
Now we use $q_+q_- = -1$ to find
$$F_n^4 = \frac{1}{25}\left((q_+^4)^n + (q^4_-)^n - 4(q_+^2)^n - 4(q_-^2)^n + 6\right)$$
or using the notation $s_\pm,r_\pm$ defined above we have
$$F_n^4 = \frac{1}{25}\left(s^n_+ + s^{n}_{-} -4r^{n}_{+} - 4r^{n}_{-} + 6\right)$$
which gives $a=b=-\frac{4}{25}$ and $c=d=\frac{1}{25}$.
A: You already have an accepted answer, but here is an alternative approach.
$${F(n+4)}^{4} - {4F(n+3)}^{4} - {19F(n+2)}^{4} - {4F(n+1)}^{4}+{F(n)}^{4}$$
reduces to
$$-6\left(F(n+1)^2 - F(n+1)F(n) - F(n)^2\right)^2$$ 
by applying Fibonacci recursion.
So what's left to prove is that
$$F(n+1)^2 - F(n+1)F(n) - F(n)^2 \in \{1, -1\}$$
or more specifically:
$$F(n+1)^2 - F(n+1)F(n) - F(n)^2 = \begin{cases} 
+1 \text{ for } 2\mid n \\
-1 \text{ for } 2\not\mid n \\
\end{cases}$$
which follows inductively, given
$$\begin{align}
&  F(n+1)^2 - F(n+1)F(n) - F(n)^2\\
 &= F(n+1)\bigg(F(n+1) - F(n)\bigg) - F(n)^2 \\ 
 &= \bigg(F(n) + F(n-1)\bigg)F(n-1) - F(n)^2 \\
 &= -\left(F(n)^2 - F(n)F(n-1) - F(n-1)^2\right)
\end{align}$$
