Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1 & =\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}+\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}-2\\
&=\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{2n+1}\right)^2
\end{align}
$$
I don't know how to get the first equal sign, is it mean
$$
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor = \left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}-\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\} $$
But how to prove the decimal part of $\phi^{4n+2}$
$$ \left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2} \right\} =1-\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}$$
 A: Let $\varphi:=\frac{1+\sqrt5}2$ and $\bar\varphi:=\frac{1-\sqrt5}2.$ The hint you wonder about,
$$\left\lfloor\varphi^{2m}\right\rfloor=\varphi^{2m}+\bar\varphi^{2m}-1,$$
is due to the fact that $\varphi^k+\bar\varphi^k$ is an integer (the $k$-th Lucas number) and
$$\varphi^{2m}+\bar\varphi^{2m}-1\le\varphi^{2m}<\varphi^{2m}+\bar\varphi^{2m},$$
i.e.
$$0<\bar\varphi^{2m}\le1.$$
A: By considering the binomial expansions of $\left(\frac{1+\sqrt5}2\right)^{4n+2}$ and $\left(\frac{1-\sqrt5}2\right)^{4n+2}$ it can be all non-integer terms cancel out when the two are added together, giving an integer result. Therefore we can say that $$\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}\in\mathbb{Z}$$
However, since $\left|\frac{1-\sqrt5}2\right|<1$, $\left|\left(\frac{1-\sqrt5}2\right)^{4n+2}\right|<1$. Since the exponent of $\frac{1-\sqrt5}2$ is even, $\left(\frac{1-\sqrt5}2\right)^{4n+2}>0$. Therefore, $$\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}=\left\lceil\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rceil$$
If $m\notin\mathbb{Z}, \lceil m\rceil=\lfloor m\rfloor+1$.
$$\therefore\left(\frac{1+\sqrt5}2\right)^{4n+2}+\left(\frac{1-\sqrt5}2\right)^{4n+2}-1=\left\lceil\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rceil-1=\left\lfloor\left(\frac{1+\sqrt5}2\right)^{4n+2}\right\rfloor,$$
as required.
A: Let $$X_n=(\frac{1+\sqrt{5}}{2})^{4n-2}=(\frac{3+\sqrt{5}}{2})^{2n-1}=I+f, I \in I^+, 0 < f <1,0 <f'=(\frac{3-\sqrt{5}}{2})^{2n-1}<1$$
$$X_n+f'=I+f+f'
=(\frac{3+\sqrt{5}}{2})^{2n-1}+(\frac{3-\sqrt{5}}{2})^{2n-1}$$
$$= (7+3\sqrt{5})^{n}\frac{2}{3+\sqrt{5}}+(7-3\sqrt{5})^{n}\frac{2}{3-\sqrt{5}}=3A_n-5B_n=K$$
Where $(7+3\sqrt{5})^n=A_n+B_n\sqrt{5}, A_n. B_n \in I^+$
$K$ is integer,so is $I+f+f'$. This implies that $f+f'$ is also an integer as $0<f+f'<2$ and it equals 1. Hence $I=K-1.$
Check that $$K=\left((\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}\right)^2+2. $$
$$\implies I-1=\left((\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}\right)^{2}.$$
Check that $$(\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}=(3+\sqrt{5})^n\frac{2}{1+\sqrt{5}}+(3-\sqrt{5})^n\frac{2}{1-\sqrt{5}}$$
$$=5D_n-C_n,~ \text{where} ~(3+\sqrt{5})^n=C_n+D_n\sqrt{5}, C_n, D_n \in I^{+}.$$
Finally we have $I-1=(5D_n-C_n)^2, C_n, D_n \in I^+$
