<s>[INST] How to prove that the number $1+4a_{n}a_{n+1}$ is a perfect square. 
A sequence of integer $\{a_{n}\}$ is given by the conditions $a_{1}=1,
a_{2}=12,a_{3}=20$,and $$a_{n+3}=2a_{n+2}+2a_{n+1}-a_{n}$$
show that
for every postive integer $n$, the number $1+4a_{n}a_{n+1}$ is a perfect square.

since the
$$r^3=2r^2+2r-1$$
then
$$r^3-2r^2-2r+1=$$
$$(r+1)(r^2-3r+1)=0$$
so
$$r_{1}=-1,r_{2,3}=\dfrac{3\pm \sqrt{5}}{2}$$
so $$a_{n}=A(r_{1})^n+B(r_{2})^n+C(r_{3})^n$$
where $A,B,C$ is constant.
Then I fell follow is very ugly,maybe someone have simple methods.Thank you
 [/INST] Given the sequence $a_{n+3} = 2 a_{n+2} + 2 a_{n+1} - a_{n}$ it is evident that a solution of the form $a_{n} \approx r^{n}$ leads to $r^{3} - 2 r^{2} - 2r + 1 = 0$, or $(r+1)(r^{2} - 3r + 1) = 0$ has roots $r_{1} = -1$, $r_{2,3} = (3 \pm \sqrt{5})/2$. Now it is seen that $r_{2,3} = (1 \pm \sqrt{5})^{2}/4$ and
\begin{align}
a_{n} = A (-1)^{n} + B \left( \frac{1+\sqrt{5}}{2} \right)^{2n} + C \left( \frac{1 - \sqrt{5}}{2} \right)^{2n}.
\end{align}
Since the coefficients associated with B and C are elements of the Fibonacci and Lucas numbers the expression for $a_{n}$ can be seen in the form
\begin{align}
a_{n} = A(-1)^{n} + B \ F_{2n} + C \ L_{2n}.
\end{align}
It is easier now to find the coefficients by using $a_{1} = 1$, $a_{2} = 12$, and $a_{3} = 20$ it is seen that 
\begin{align}
a_{n} = 3 (-1)^{n} - \frac{F_{2n}}{2} + \frac{3 L_{2n}}{2} = \frac{1}{2} \left( 3 L_{n}^{2} - F_{2n} \right) = \frac{1}{2} \ L_{n}(3 L_{n}-F_{n}) = L_{n} (L_{n} +F_{n-1}).
\end{align}
Now let $S_{n} = 1 + 4 a_{n} a_{n+1}$ for which:
\begin{align}
S_{n} = \left[ 2 L_{n} (L_{n+1} + F_{n}) + (-1)^{n} \right]^{2}
\end{align}
is an integer squared. 
</s>