Define two sequences $A_n, B_n$ as follows:
\begin{align*} A_1 &= 1\\ A_2 &= 3\\ A_3 &= 2 \cdot 3+1=7 \\ A_4 &= 2 \cdot 7 + 3 = 17\\ A_5 &= 2 \cdot 17 + 7 = 41\\ A_n &= 2A_{n-1} + A_{n-2}\\ \\ B_1 &= 1\\ B_2 &= 2\\ B_3 &= 2 \cdot 2+1=5\\ B_4 &= 2 \cdot 5 + 2 = 12\\ B_5 &= 2 \cdot 12 + 5 = 29\\ B_n &= 2B_{n-1} + B_{n-2} \end{align*}
Prove by induction that $A_n^2 - 2B_n^2 = (-1)^{n}$
For the base case with $n=1$ we have $A_n^2 - 2B_n^2 = 1-2 = -1 = (-1)^1$. Assume the assertion holds for $n=k$ and $n=k+1$. Then for the case $n = k+2$, $$ A_{k+2}^2 - 2 B_{k+2}^2 = (-1)^{k+2} $$ Working on the left hand side, \begin{align*} \left( 2A_{k+1} + A_{k} \right)^2 - 2 \left( 2 B_{k+1} + B_{k} \right)^2 &= 4A_{k+1}^2 + 4A_{k+1}A_{k} + A_{k}^2 -2 \left( 4B_{k+1}^2 + 4 B_{k+1}B_{k} + B_{k}^2 \right) \\ & = 4A_{k+1}^2 + 4A_{k+1}A_{k} + A_{k}^2 - 8B_{k+1}^2 - 8 B_{k+1}B_{k} - 2B_{k}^2 \\ & = 4A_{k+1}^2 - 8B_{k+1}^2 + A_{k}^2 - 2B_{k}^2 - 8 B_{k+1}B_{k} + 4A_{k+1}A_{k} \\ \end{align*}
I'm not sure how to deal with the last two terms: $- 8 B_{k+1}B_{k} + 4A_{k+1}A_{k}$