TITLE: Is This Sum Obviously Equal to $8$.
QUESTION [9 upvotes]: Question: I write: 
\begin{align} \lambda_1 & = {\sqrt[3]{\frac{1}{2}\bigg(9 + i
 \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt
 {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\  \lambda_2 &
 = -{\bigg(1-i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} -
{2(1+i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i
\sqrt{687}\bigg)}}} \\   \lambda_3 & =
 -{\bigg(1+i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} -
{2(1-i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i
\sqrt{687}\bigg)}}}\\ \end{align}
  Is it obviously true that $$|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2
 =8$$ 
  ?

This question is driven entirely by curiosity and fun. Unless I am missing the obvious it seems if you were to work this out by hand it would be demonstrably difficult ? Here is the origin of the problem and my attempted solution:
Let $A(n)$ be a finite square $n \times n$ matrix with entries $a_{i,j}=1$ if $i+j$ is a perfect power; otherwise equals to $0$. Let $\chi_{A(n)}(X)$ be the characteristic polynomial of $A(n)$. By the fundamental theorem of algebra $\chi_{A(n)}(X)$ has $n$ roots (=eigenvalues). Denote the $n$ eigenvalues by $\lambda_i$ with $1 \leq i \leq n$. Note $A(n)=A(n)^t$ where $^t$ is the matrix transpose and in particular $A(n)$ is a real symmetric matrix consequently it is normal. I bring Schur's Inequality into play:
$$\sum_{i=1}^n |\lambda_i|^2 = \sum_{i=1}^n \sum_{j=1}^n  |a_{ij}|^2$$
Immediately this tells me that the number of $1$'s in $A(n)$ is completely determined by its eigenvalues. Now consider $A(6)$. 
$$A(6)= \text{ }\begin{pmatrix}
0&0&1&0&0&0\\
0&1&0&0&0&1\\
1&0&0&0&1&1\\
0&0&0&1&1&0\\
0&0&1&1&0&0\\
0&1&1&0&0&0\\
\end{pmatrix}$$
I have that $$\chi_{A(6)}(X)=X^6-2X^5-4X^4+8X^3+2X^2-4X-1$$ and that can be factored into $$(X^3-4X-1)(X^2-X-1)(X-1)$$ $A(6)$ has 6 eigenvalues which can be written (courtesy of WOLFRAM ALPHA)
\begin{align}
\lambda_1 & = {\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\
 \lambda_2 & = -{\bigg(1-i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1+i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}} \\ 
 \lambda_3 & = -{\bigg(1+i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1-i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\
\lambda_4 & = -{1\above 1.5 pt \phi}\\
\lambda_5 & = \phi\\
\lambda_6 & = 1
\end{align}
where $\phi= {1+ \sqrt{5}\above 1.5pt 2}$ and is called the golden ratio. Note $\sum_{j=1}^6  a_{ij}=12$ and according to our first equation I can write 
$$1+|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2+ |-{1\above 1.5 pt \phi}|^2 +|\phi|^2 =12$$
I can show that
$$|-{1\above 1.5 pt \phi}|^2 +|\phi|^2=3$$
Cancelling out terms and regrouping yields
$$|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2 =8$$
and we are done.
Outside of hand calculations which I am too lazy to do to I am not aware of any other approach that will obviously and for that matter quickly show the sum equals $8$.

REPLY [5 votes]: It's worth pointing out that showing those are the roots to the polynomial $t^3 - 4t - 1=0$ isn't enough to conclude that $|\lambda_1|^2 + |\lambda_2|^2+|\lambda_3|^2=8$, only that $\lambda_1^2 + \lambda_2^2+\lambda_3^2=8$, and we need to check that the roots you've given are actually real numbers (even if they don't look like it).  The simplest way I can see is to use the intermediate value theorem, checking that there is a sign change between $-\infty$, $-1$, $0$, $\infty$, giving at least three real roots.  Of course this means there are exactly three real roots, and so the modulus signs made no difference.

Let's write $\alpha = \sqrt[3]{\frac{1}{18}\left(9 + i \sqrt{687}\right)} = \frac{\sqrt[3]{\frac{1}{2}\left(9 + i \sqrt{687}\right)}}{3^{2/3}}$ for convenience, and $\omega$ for the cube root of unity, $\omega = e^{2\pi i/3}$.  Then your roots are
$$\begin{align}
\lambda_1 &= \alpha+\frac{4}{3\alpha} \\
\lambda_2 &= \alpha \omega^2 + \frac{4}{3\alpha \omega^2} \\
\lambda_3 &= \alpha \omega +  \frac{4}{3\alpha \omega} \\
\end{align}$$
To recap Cardano's method, to solve a depressed cubic $t^3 = at+b$, we make the substitution $t = u+v$, so $u^3 + v^3 + 3uv(u+v) = a(u+v)+b$, and choose $u^3 + v^3 = b$, $3uv=a$.  We can solve this by writing $v = \frac{a}{3u}$, so $u^3 + \frac{a^3}{27u^3} = b$, producing $u^6 + \frac{a^3}{27} = b u^3$, a quadratic equation in $u^3$.
The value of $\alpha$ given above seems like the cube root of a solution to a quadratic, so let's guess $u=\alpha$ and $v = \frac{4}{3\alpha}$, so we compute $3uv = 4$, and so $a = 4$. Then, $x = u^3$ should satisfy $x^2 -bx +\frac{64}{27}=0$.  We find $x = \frac1{18}(9+i \sqrt{687})$, so $(18x-9)^2 = -687$, and expanding and cancelling gives $x^2 - x + \frac{64}{27}=0$, hence $b=1$.  
Therefore, $\lambda_1$ satisfies $t^3 = 4t+1$.  You can do the same thing for $\lambda_2$ and $\lambda_3$, or just observe that when the solution via Cardano's method is found, we have the choice of three cube roots of $u^3$ to take, corresponding to $\alpha$, $\alpha \omega$ and $\alpha \omega^2$. (The choice of two square roots from the quadratic doesn't give extra solutions, since it would give the $\frac{4}{3\alpha}$ term instead, and you'd get the same answer.)
So, $\lambda_1, \lambda_2, \lambda_3$ are the roots to $t^3 = 4t+1$, and as others have said, Vieta's formulas (or more precisely, Newton's identities) finish off the answer.