$\frac{1}{5}\big((4 + \sqrt{15})^{2n} + (4 - \sqrt{15})^{2n} + 8\big)$ is the sum of 3 consecutive squares

Question

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 $n = 1, 2, 3, 4, \cdots$

Answer 1

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 $$ \big((4+\sqrt{15})^{n}-(4-\sqrt{15})^{n}\big)^2=(2b\sqrt{15})^2=15\cdot4b^2 $$

Answer 2

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.