How does $ab+ac+bc =0$ imply that $a+b+c = \pm 1$? Show that $$ A=
        \begin{bmatrix}
        a & c & b \\
        b & a & c \\
        c & b & a \\
        \end{bmatrix}
$$ is orthogonal if and only if $a^2+b^2+c^2=1$ and $a+b+c= \pm 1$.
For a matrix to be orthogonal it must satisfy $A^T A^{-1} = I$ where $I$ is the identity matrix.
I calculated $$A^T A^{-1} =\begin{bmatrix}
        a^2+b^2+c^2 & ab+ac+bc & ab+ac+bc \\
        ab+ac+bc & a^2+b^2+c^2 & ab+ac+bc \\
        ab+ac+bc & ab+ac+bc & a^2+b^2+c^2 \\
        \end{bmatrix} $$
So for $A^T A^{-1} = I$ we must have that $$ a^2+b^2+c^2 = 1$$ and $$ab+ac+bc =0 $$
So then I showed that $ a^2+b^2+c^2 = 1$, but through a lot of rough work I can't seem to make $ab+ac+bc =0 $ imply that $a+b+c= \pm 1$. Does anyone have any hints?I feel like i've just about tried every way to sub something into something else!

The comments should be enough,
but
if $a^2+b^2+c^2 = 1$
and
$ab+ac+bc = 0$,
then
$(a+b+c)^2
=a^2+b^2+c^2+2(ab+ac+bc)
=1
$
so
$a+b+c = \pm 1$.
