Crazy Antique

For simplicity, let $T_a=x^a+\dfrac{1}{x^a}$ . We will compute $T_6,T_7$ using the observation that $T_{a}T_1 = T_{a+1}+T_{a-1}$ . Let us tabulate values for $T_1$ to $T_7$ :

$\begin{aligned} T_1 &=n \\ T_2 &=n^2-2 \\ T_3 &= n^3-3n \\ T_4 &= n^4-4n^2+2 \\ T_5 &= n^5-5n^3+5n \\ T_6 &= n^6-6n^4+9n^2-2 \\ T_7 &=n^7-7n^5+14n^3-7n\end{aligned}$

Since $T_7=nT_6$ we have $n^7-7n^5+14n^3-7n=n^7-6n^5+9n^3-2n \\ \Rightarrow n^5-5n^3+5n=0 \Rightarrow n(n^4-5n^2+5)=0$ .

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Realized this later: From both given equations in the question we have $T_7=T_1T_6$ and using $T_{a}T_1 = T_{a+1}T_{a-1}$ we have $T_7=T_5+T_7\Rightarrow T_5=0$ so we directly get $n^5-5n^3+5n=0$ and there was no need to find $T_6 \ , \ T_7$ . Silly me.

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We have one solution as $n=0$ and we need to solve $n^4-5n^2+5$ which can be accomplished using the substitution $a=n^2$ and the quadratic formula we have $a=\dfrac{5\pm\sqrt{5}}{2}\Rightarrow n=\pm\sqrt{\dfrac{5\pm\sqrt{5}}{2}}$ . So there are $5$ solutions for $n$ giving $\boxed{k=5}$ . Sum of squares of solutions simply equal $0+2\left(\dfrac{5+\sqrt{5}}{2}\right)+2\left(\dfrac{5-\sqrt{5}}{2}\right)=\boxed{10}$

Thus sum of squares of solution $+ k = 10+5=\Large{\boxed{15}}$ .

As $x^7+\frac{1}{x^7}=(x+\frac{1}{x})(x^6+\frac{1}{x^6})$ $\implies x^5+\frac{1}{x^5}=0$ As $x+\frac {1}{x} = n$ by Binomial expansion $\implies n^5-5n^3+5n=0 \implies n=0; n^4-5n^2+5=0$ . By Vieta's Theorem the two values of $n^2$ add-up to 5. There are 5 solutions in all including n=0, and +/- n whose squares add up to 5 each. So 5+5=10 is the summation and k=5. The answer is 15.

Let $P_n = x^n + \dfrac{1}{x^n}$ , where $n \in \mathbb N$ . Using Newton sums method, we have:

$\begin{cases} P_2 = P_1^2 - 2 = n^2 - 2 \\ P_n = P_1P_{n-1} - P_{n-2} = nP_{n-1} - P_{n-2} \quad \text{for } n \ge 3 \end{cases}$

Therefore,

$\begin{aligned} P_7 & = nP_6 - P_5 \quad \quad \small \color{#3D99F6}{\text{Since }P_7 = nP_6} \\ \Rightarrow P_5 & = 0 \\ nP_4 - P_3 & = 0 \\ n(nP_3-P_2) - P_3 & = 0 \\ (n^2-1)P_3 - nP_2 & = 0 \\ (n^2-1)(nP_2-P_1) - nP_2 & = 0 \\ nP_2(n^2-2) - n(n^2-1) & = 0 \\ n(n^2-2)^2 - n(n^2-1) & = 0 \\ n(n^4-5n^2 +5) & = 0 \\ \Rightarrow n & = \begin{cases} 0 \\ \pm \sqrt{\frac{5+\sqrt{5}}{2}} \\ \pm \sqrt{\frac{5-\sqrt{5}}{2}} \end{cases} \end{aligned}$

There are $5$ solutions for $n$ and let $n_5 = 0$ , then the other $4$ values of $n$ come from $n^4 + \color{#3D99F6}{0}n^3 \color{#D61F06}{- 5}n^2 + 0n + 5$ , therefore,

$\begin{aligned} n_1^2 + n_2^2 + n_3^2 + n_4^2 & = (\color{#3D99F6}{n_1 + n_2 + n_3 + n_4})^2 - 2(\color{#D61F06}{n_1n_2 + n_1n_3 + n_1n_4 + n_2n_3 + n_2n_4 + n_3n_4}) \\ & = (\color{#3D99F6}{0})^2 - 2(\color{#D61F06}{-5}) \\ & = 10 \end{aligned}$

$\begin{aligned} \Rightarrow \sum_{i=1}^k n_i + k & = 10 + 5 = \boxed{15} \end{aligned}$