If $$\alpha$$ is a root of the equation $$x^2 + x + 1 = 0$$ and $$\sum_{k=1}^{n} \left(\alpha^k + \frac{1}{\alpha^k}\right)^2 = 20$$, then n is equal to ______.

The quadratic $$x^{2}+x+1=0$$ has roots that are the non-real cube roots of unity.
Let one root be $$\alpha$$. Then

$$\alpha^{2}+\alpha+1 = 0 \;\; \Longrightarrow \;\; \alpha^{3}=1,\; \alpha\neq 1$$

Because $$\alpha^{3}=1$$, all integral powers of $$\alpha$$ repeat every three steps:
$$\alpha^{k}= \begin{cases} 1 & \text{if } k\equiv 0\pmod{3}\\ \alpha & \text{if } k\equiv 1\pmod{3}\\ \alpha^{2} & \text{if } k\equiv 2\pmod{3} \end{cases}$$

Also $$\alpha^{-1}=\alpha^{2},\;\alpha^{-2}=\alpha,\;\alpha^{-3}=1$$, so $$\alpha^{-k}$$ follows the same cycle.

Define $$T_k=\left(\alpha^{k}+\frac{1}{\alpha^{k}}\right)^{2}$$. Evaluate $$T_k$$ for the three possible residues of $$k$$ modulo $$3$$:

$$\alpha^{k}=1,\;\frac{1}{\alpha^{k}}=1 \quad\Rightarrow\quad T_k=(1+1)^{2}=4$$

$$\alpha^{k}=\alpha,\;\frac{1}{\alpha^{k}}=\alpha^{2} \quad\Rightarrow\quad \alpha+\alpha^{2}=-(1) \quad(\text{from } \alpha^{2}+\alpha+1=0)$$ $$T_k=(-1)^{2}=1$$

$$\alpha^{k}=\alpha^{2},\;\frac{1}{\alpha^{k}}=\alpha \quad\Rightarrow\quad \alpha^{2}+\alpha = -1$$ $$T_k=(-1)^{2}=1$$

Thus the sequence $$T_1,T_2,T_3,\dots$$ is periodic with cycle $$\{1,1,4\}$$, whose sum is

$$1+1+4 = 6$$

Let $$n=3q+r$$ where $$r\in\{0,1,2\}$$. Then

$$\sum_{k=1}^{n} T_k = 6q + \begin{cases} 0 & (r=0)\\ 1 & (r=1)\\ 1+1=2 & (r=2) \end{cases}$$

The given condition is

$$6q + \{0,1,2\}=20$$

Check the three possibilities:
  • $$6q=20$$ (impossible)  
  • $$6q+1=20\;\Rightarrow\;6q=19$$ (not an integer)  
  • $$6q+2=20\;\Rightarrow\;6q=18\;\Rightarrow\;q=3$$

The valid case is $$q=3,\;r=2$$. Hence

$$n = 3q + r = 3\cdot 3 + 2 = 11$$

Therefore, the required value of $$n$$ is $$\mathbf{11}$$.