Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.
I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?

Here's a start that shows that
any other solutions
would have to have
distinct $a, b, $ and $c$.
In
$a^2+b^2+c^2 =1$
and
$a+b+c= \pm 1$,
if $a=b$,
these become
$2a^2+c^2 = 1,
2a+c = \pm 1$.
Then
$c = -2a\pm 1$,
so
$1
= 2a^2+(-2a\pm 1)^2
=2a^2+4a^2\pm 4a+1
=6a^2\pm 4a+1
$
so
$0 = 6a^2\pm 4a
=2a(3a\pm 2)
$.
Therefore
$a=0$ or
$a = \pm \frac23$.
If $a=b=0$,
then
$c = \pm 1$;
if $a=b=\pm \frac23$,
then
$c = -2a\pm 1
=\mp \frac43 \pm 1
=\pm \frac13
$
and these are the solutions
that you already have.
Therefore
any other solutions
would have to have
distinct $a, b, $ and $c$.
