How does $ab+ac+bc =0$ imply that $a+b+c = \pm 1$?

Question

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!

Answer 1

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 $a+b+c = \pm 1$.