I found this problem in a math magazine:
Given the sequence $(x_n)_{n \in \mathbb{N}}$ defined by: $ x_0 = 0\\ x_1 = 1\\ x_{n+2}+x_{n+1}+2x_{n}=0 $ Prove that $s_n = 2^{n+1}-7x_{n-1}^2, n > 0$ is a square number.
I tried searching a rule between the numbers of the sequence and the square numbers ($s_n$) formed: $ x_2=-1\\ x_3=-1\\ x_4=3\\ x_5=-1\\ x_6=-5\\ x_7=7\\ x_8=3\\ x_9=-17\\ $ $ s_1 = 2^2\\ s_2 = 1^2\\ s_3 = 3^2\\ s_4 = 5^2\\ s_5 = 1^2\\ s_6 = 11^2\\ s_7 = 9^2\\ $ If I rewrite the rule and square it: $ x_{n-1} = -x_{n-2}-2x_{n-3}, n > 3\\ x_{n-1}^2=x_{n-2}^2 + 4x_{n-3}^2 + 4x_{n-2}x_{n-3}\\ $ apply for the next two, $x_{n-2}$, $x_{n-3}$ the same rule: $ x_{n-2}^2=x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-3}x_{n-4}\\ x_{n-3}^2=x_{n-4}^2 + 4x_{n-5}^2 + 4x_{n-4}x_{n-5}\\ $ substitute: $ x_{n-1}^2=x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-3}x_{n-4}+4x_{n-3}^2 + 4x_{n-2}x_{n-3}\\ x_{n-1}^2=5x_{n-3}^2 + 4x_{n-4}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4} \\ x_{n-1}^2=5(x_{n-4}^2 + 4x_{n-5}^2 + 4x_{n-4}x_{n-5}) + 4x_{n-4}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4}\\ x_{n-1}^2=9x_{n-4}^2 + 20x_{n-5}^2 + 4x_{n-2}x_{n-3} + 4x_{n-3}x_{n-4} + 20x_{n-4}x_{n-5}\\ $ It seems to lead to a dead end.
A help would be really appreciated.