proof that $\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer

It is easy to prove by induction that $a_n=aF_{n-2}+bF_{n-1}$, where $F_k$ is Fibonacci numbers. I want to prove that $\frac{a_{4n}-a_2}{a_{2n+1}}=F_{2n-2}+F_{2n}$. Therefore, we must prove that $$aF_{4n-2}+bF_{4n-1}-b=(aF_{2n-1}+bF_{2n})(F_{2n-2}+F_{2n})$$ Therefore, we need prove that $F_{4n-2}=F_{2n-1}(F_{2n-2}+F_{2n})$ and $F_{4n-1}-1=F_{2n}(F_{2n-2}+F_{2n})$.These two equalities are easily proved by well-known formulas from Wikipedia.

Hints. Let $\phi$ and $\psi$ be the two roots of $x^2-x-1=0$. Since every Fibonacci sequence is a linear combination of the geometric progressions of $\phi$ and $\psi$, if we define $$ f_n(x)=\frac{x^{4n}-x^2}{x^{2n+1}}=x^{2n-1}-\frac{1}{x^{2n-1}}, $$ it suffices to show that $f_n(\phi)$ is an integer and $f_n(\phi)=f_n(\psi)$ for each $n$. If you want to use mathematical induction, you may need to prove, for each $n$, that $f_n(\phi)=f_n(\psi)\in\mathbb N$ and $g_n(\phi)=g_n(\psi)\in\mathbb N$, where $$ g_n(x)=x^{2n}+\frac{1}{x^{2n}}. $$