In the non-zero sequence $x[n-1]+x[n+1]=ax[n]$ and $x[n+4]=-x[n]$
i) Find possible values of $a$.
ii) For what values of $b$ is $b^n$ a solution ($x[n]=b^n$)?
I need to solve this using only A-level knowledge, but have just been unable to do so. Any help/guidance would be appreciated
Start with
$x(n+1) =ax(n)-x(n-1) .$
Then
$\begin{array}\\ x(n+2) &=ax(n+1)-x(n)\\ &=a(ax(n)-x(n-1))-x(n)\\ &=(a^2-1)x(n)-ax(n-1)\\ \end{array} $
$\begin{array}\\ x(n+3) &=ax(n+2)-x(n+1)\\ &=a((a^2-1)x(n)-ax(n-1))-(ax(n)-x(n-1))\\ &=(a^3-2a)x(n)-(a^2-1)x(n-1)\\ \end{array} $
and, finally,
$\begin{array}\\ x(n+4) &=ax(n+3)-x(n+2)\\ &=a((a^3-2a)x(n)-(a^2-1)x(n-1))-((a^2-1)x(n)-ax(n-1))\\ &=(a^4-2a^2-(a^2-1))x(n)-(a^3-a-a)x(n-1)\\ &=(a^4-3a^2+1))x(n)-(a^3-2a)x(n-1)\\ \end{array} $
If $x(n+4) = -x(n)$, then $a^4-3a^2+1 = -1 $ and $a^3-2a = 0 $.
From the second, $0 =a(a^2-2) $, so $a=0$ or $a =\pm \sqrt{2} $.
$a=0$ does not satisfy the first. If $a=\pm \sqrt{2}$, $a^4-3a^2+1 =4-3\cdot 2+1 =-1 $, so both work.
For part (ii), in $x(n+1) =ax(n)-x(n-1) $, substitute $x(n) = b^n$. This becomes $b^{n+1} =ab^n-b^{n-1} $.
If $b \ne 0$, we can divide by $b^{n-1}$ to get $b^2=ab-1$ or $b^2-ab+1 = 0$. This is a standard quadratic, and you should be able to solve for $b$ in terms of $a$.
I am not sure if part (ii) also assumes that $x(n+4)=-x(n)$.