Problem Statement: Let $a_{m, n}$ denote the coefficient of $x^n$ in the expansion $(1+x+x^2)^m.$ Prove that for all $k\geq 0$, $$0\leq \sum_{i = 0}^{\left\lfloor 2k/3\right\rfloor} (-1)^ia_{k-i,i} \leq 1$$
I first tried to find a good expression for $a_{m, n}$. We note that if $w = e^{2\pi i/3}$ then $(1+x+x^2)^m = (x-w)^m(x-w^{-1})$ \begin{align} a_{m, n} &= [x^n]\left(\sum_{a = 0}^m\binom{m}{a}(-w)^{m-a}x^a\right)\left(\sum_{b = 1}^m\binom{m}{b}(-w)^{m-b}x^b\right)\\ &=\sum_{a+b = n}\binom{m}{a}\binom{m}{b}(-w)^{n-a-(n-b)}\\ &= \sum_{a = 0}^m\binom{m}{a}\binom{m}{n-a}(-w)^{n-2a}\\ &= \sum_{a = 0}^m(-1)^{n - 2a}\binom{m}{a}^2 w^{n-2a} \end{align}
Thus we have:
\begin{align} \sum_{i = 0}^{\left\lfloor 2k/3\right\rfloor} (-1)^ia_{k-i,i} &= \sum_{i = 0}^{\left\lfloor 2k/3\right\rfloor} (-1)^i\left(\sum_{a = 0}^{k-i}(-1)^{i - 2a}\binom{k-i}{a}^2 w^{i-2a}\right)\\ &=\sum_{i = 0}^{\left\lfloor 2k/3\right\rfloor}\sum_{a = 0}^{k-i}\binom{k-i}{a}^2 w^{i-2a} \end{align}
From here I'm stuck as to how I should proceed. Any hints would be greatly appreciated!
Hint: $a_{k,n} = a_{k, n-1 } + a_{k-1, n-1} + a_{k-2, n-1}$.
$a_{k,n} - a_{k+1,n} =a_{k-2,n-1} - a_{k+1, n}$.
Use this recursively to reduce terms till you're done
E.g. $a_{4,0} - a_{3,1} + a_{2,2} \\ =a_{3,-2} + a_{3,-1} + a_{3,0} - a_{3,1} - a_{2,2} \\ = a_{3,0} - a_{3,1} - a_{2,2} \\ = a_{2,-2} - a_{2,1} + a_{2,2} \\ = - a_{2,1} + a_{2,2} \\ = -a_{1,-1} + a_{1,2} \\ = 0$