Sequence $a_0,a_1,a_2,...$ satisfies that $a_0=2,a_1=1,a_{n+1}=a_n+a_{n-1}$
Prove that if $p$ is a prime divisor of $a_{2k}-2$,then $p$ is also a prime divisor of $a_{2k+1}-1$
If $x_{1,2}={1\pm\sqrt{5}\over 2}$ then $a_k = x_1^k+x_2^k$ and $a_{2k}-2(-1)^k = (x_1^k-x_2^k)^2$.
The claim follows from
$$(a_{4k+1}-1)^2=5F_{2k+1}^2(a_{4k}-2)\tag1$$ and $$(a_{4k+3}-1)^2=a_{2k+2}^2(a_{4k+2}-2)\tag2$$ where $F_n$ is the $n$-th Fibonacci number defined by $F_0=0,F_1=1,F_{n+1}=F_{n-1}+F_n\ (n\ge 1)$.
Note that our sequence is known as the Lucas sequence.
Proof for $(1)$ :
We use the following facts :
$$a_n^2=a_{2n}+2(-1)^n\tag3$$ $$a_n^2-5F_n^2=4(-1)^n\tag4$$ $$a_{m-n}=\frac 12(-1)^n(a_ma_n-5F_mF_n)\tag5$$ $$a_{m+n}=\frac 12(5F_mF_n+a_ma_n)\tag6$$
Setting $n=2k$ in $(3)$ gives $$a_{2k}^2-4=a_{4k}-2\tag7$$ Setting $n=2k$ in $(4)$ gives $$a_{2k}^2-4=5F_{2k}^2\tag8$$ Setting $m=2k+1,n=2k$ in $(5)$ gives $$1=\frac 12(a_{2k+1}a_{2k}-5F_{2k+1}F_{2k})\tag9$$ Setting $m=2k+1,n=2k$ in $(6)$ gives $$a_{4k+1}=\frac 12(5F_{2k+1}F_{2k}+a_{2k+1}a_{2k})\tag{10}$$ From $(9)(10)$, we have $$a_{4k+1}-1=5F_{2k+1}F_{2k}\tag{11}$$ Finally, from $(11)(8)(7)$, $$(a_{4k+1}-1)^2=5F_{2k+1}^2(a_{4k}-2)\tag1$$ follows.$\quad\blacksquare$
Proof for $(2)$ :
Setting $n=2k+1$ in $(3)$ gives $$a_{2k+1}^2=a_{4k+2}-2\tag{12}$$ Setting $m=2k+2,n=2k+1$ in $(5)$ gives $$1=-\frac 12(a_{2k+2}a_{2k+1}-5F_{2k+2}F_{2k+1})\tag{13}$$ Setting $m=2k+2,n=2k+1$ in $(6)$ gives $$a_{4k+3}=\frac 12(5F_{2k+2}F_{2k+1}+a_{2k+2}a_{2k+1})\tag{14}$$ From $(13)(14)$, we have $$a_{4k+3}-1=a_{2k+2}a_{2k+1}\tag{15}$$ Finally, from $(15)(12)$ $$(a_{4k+3}-1)^2=a_{2k+2}^2(a_{4k+2}-2)\tag2$$ follows.$\quad\blacksquare$
