$\frac{1}{5}\big((4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8\big)$ is the sum of 3 consecutive squares 
Prove that for every positive integer $n$, the number
  $$
\large \frac{(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8}{5}
$$ 
  can be expressed as a sum of squares of three consecutive integers.

Attempt. Since 
$$
(m - 1)^2 + m^2 + (m + 1)^2 = 3m^2 + 2, \forall m \in \mathbb Z^+, m \ge 1,
$$ 
it suffices to show that
$$
(4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} = 15m^2 + 2, \,\,\text{for some}\,\,\, m \in \mathbb Z^+,
$$
$$\implies \sum_{p = 1}^{n - 1}[(4 + \sqrt{15})^{2p} + (4 - \sqrt{15})^{2p}] = 15m^2, \forall m \in \mathbb Z^+, m \ge 1$$
But this couldn't use any mathematical induction, since $m$ hasn't been known to be on a sequence for $n = 1, 2, 3, 4, \cdots$
 A: Note that
$$
\frac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}+8}{5}
=\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2+10}{5}=
\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2}{5}+2
$$
So it suffices to show that
$$
\frac{\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2}{5}=3m^2
$$
for some $m\in\mathbb N$, 
or
$$
\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2=15m^2
$$
But, it is clear (inductive proof) that if
$$
(4+\sqrt{15})^{n}=a+b\sqrt{15}, \quad a,b\in \mathbb N,
$$ 
then
$$
(4-\sqrt{15})^{n}=a-b\sqrt{15}, \quad a,b\in \mathbb N,
$$ 
and hence
$$
\big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2=(2b\sqrt{15})^2=15\cdot4b^2
$$
A: Let $m=\dfrac{(4+\sqrt{15})^n-(4-\sqrt{15})^n}{\sqrt{15}}$.  
$m$ is an integer, because, in the binomial expansions in the numerator, 
terms with even powers of $\sqrt{15}$ cancel, leaving a numerator that is an integer times $\sqrt{15}$.
Furthermore, $m^2=\dfrac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}-2}{15},$ so
$(m-1)^2+m^2+(m+1)^2=3m^2+2=\dfrac{(4+\sqrt{15})^{2n}+(4-\sqrt{15})^{2n}-2}{5}+2,$ as desired.
