Solve the reccurence relation $a_n = a_{n-2}, a_0=a_1 = 1$ This is my first time working through this type of problem and I am looking to see if I have worked it out correctly, thanks!

Solve the reccurence relation $$a_n = a_{n-2}, a_0=a_1 = 1$$

First we get the characteristic equation:
$$x^n = x^{n-2}$$
Dividing by the smallest we get,
$$x^2=1$$
$$x = \pm 1$$
Using the auxiliary equation: general solution
$$a_n = A_1x^{n+1} + A_2x^{n+1}$$
Using conditions in the equation and our solution for $x$ gives,
$$a_0 = 1 = A_1(1) + A_2(-1) = A_1 - A_2$$
$$a_1 = 1 = A_1(1)^2 = A_2(-1)^2 = A_1 + A_2$$
Solving for $A_1, A_2$ I got $A_1 = 1, A_2 = 0$
Therefore plugging into the general solution I got,
$$a_n = 1^{n+1}$$

$$a_{2n}=a_{2n-2}=a_0=1$$
$$a_{2n+1}=a_{2n-1}=a_1=1$$
thus
$$\forall n\in \Bbb N \;a_n=1$$
