The discrete heat problem - Part 2: Solving the characteristic equation

In the last post, we came to the conclusion that $E = \sum_{j = 0}^{n-1} \binom{j+n}{2j+1} E_0^{(j)}$. This is an n-1'th degree inhomogenous linear equation. To solve it, we need to solve the characteristic equation $\sum_{j = 0}^{n-1} \binom{j+n}{2j+1} r^j = 0$ for $r$.

It seems that, in general, the solutions to the characteristic equation are $$r_i = -4\cos^2\frac{\pi i}{2 n} (1 \leq i < n)$$ but I don't have a proof for this yet. I found the solutions by chance: by noticing that the solutions for low n involved the same square roots as cos did.

As I don't have a proof yet, I'll give you an example instead. Let $n = 4$. Then the characteristic equation is
$$\begin{eqnarray} 0 &=& \sum_{j = 0}^{3} \binom{j+4}{2j+1} r^j\\ &=& \binom{4}{1} + \binom{5}{3} r + \binom{6}{5} r^2 + \binom{7}{7} r^3\\ &=& 4 + 10r + 6r^2 + r^3\\ &=& (2+r)(2 + 4r + r^2)\\ &=& (2+r)(2+\sqrt{2}+r)(2-\sqrt{2}+r) \end{eqnarray}$$
whence we may read off the solutions $r = -2, -2-\sqrt{2}, -2+\sqrt{2}$. This is just what the formula predicts:
$$\begin{eqnarray} r_1 &=& -4\cos^2\frac{\pi}{8} &=& -4\frac{1}{2}(1+\frac{1}{\sqrt{2}}) &=& -2-\sqrt{2} \\ r_2 &=& -4\cos^2\frac{\pi}{4} &=& -4\frac{1}{2} &=& -2\\ r_3 &=& -4\cos^2\frac{3\pi}{8} &=& -4\frac{1}{2}(1-\frac{1}{\sqrt{2}}) &=& -2+\sqrt{2} \end{eqnarray}$$

In conclusion, the formula is reasonable and will do without a proof for the moment.