So I have been strugling with to prove the following:
$\left|\sum_{k\geq 0}(-1)^k {N-k-1 \choose N-2k}\right| = \begin{cases} 1, & \text{if $N\equiv 1 \bmod 3$} \\ 0, & \text{otherwise} \end{cases} $
I was given the hint of considering the coefficient of $x^N$ on both sides of the equality: $$ \frac{1-x^2}{1+x^3}=\frac{1}{1-\frac{x^2}{1-x}}. $$ My attempt: Break it down to finding the coefficient of $\frac{1-x}{1-x+x^2}.$ This gives me $$ (1-x)\frac{1}{1-(x-x^2)} = (1-x)\sum_{k\geq 0} (x-x^2)^k $$ by geometric series However, I have no idea how to continue from this or if i'm even doing this right. Could someone give me a hint or an explanation please?