Roots of a cubic equation with coefficients based on unknown values $a$, $b$ and $c$. I want to find the eigenvalues of the following matrix:
$$
 \left( \begin{array}{ccc}
0 & a & b \\
a & 0 & c \\
b & c & 0 \end{array} \right)
$$
So, I found the characteristic equation, $\lambda^3-(a^2+b^2+c^2) \lambda - 2abc = 0$.
But I am unable to find the roots by observation. Is there a way I can find the roots of the equation?
If you are interested, my goal is to find distinct values of $a$, $b$, and $c$ such that the matrix $M$ has integer eigenvalues. Also, simple "$a = -b$" type trivial solutions are not allowed.
Thanks,
Naren
 A: Set $a = 0$, then we have
$$ \pmatrix{0 & 0 & b \\ 0 & 0 & c \\ b & c & 0}. $$
The eigenvalues are defined by $ -\lambda(\lambda^2 - c^2) + b^2\lambda = 0 $, so we have $\lambda = 0$ or $\lambda = \pm\sqrt{b^2 + c^2}$. 
Choose Pythagorean triple values such as $b = 3$ and $c = 4$.
Then we have $\lambda = 0$ and $\lambda = \pm5$.
Does this answer your question?
Note that this is the kind of solution you get if you do not reduce the domain of the problem. 
EDIT:
Also, you might also be interested in Vieta's Formula for cubic polynomials. 
A: Note that the sum of the eigenvalues is 0. If all eigenvalues are integers, then all coefficients of the characteristic polynomial are integers. Set $d=a^2+b^2+c^2$, an integer.
Suppose $n$ is an integer eigenvalue. Dividing $\lambda-n$ into the char poly gives $$ \lambda^2+n\lambda+(n^2-d).$$
This has roots $$\frac{-n\pm \sqrt{4d-3n^2}}{2}.$$
These are integers precisely when $\sqrt{4d-3n^2}=2k+n$ for some integer $k$. In which case the eigenvalues are $k,k-n,n$. These sum to zero precisely when $k=0$. Equivalently, $d=n^2$.
Whence we have all integer eigenvalues if and only if $a^2+b^2+c^2=n^2$ is a perfect square, in which case the eigenvalues are $0,\pm n$.
You then find that $2abc=0$ holds, whence one of $a,b, c$ is zero. The  remaining two solve a (signed) right triangle with hypotenuse $n$.
A: 
Result from MATLAB, using symbolic manipulation on MuPad (Source: Guodong Xie).  
