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?
 A: There are infinitely many. The complete rational solution to
$$a^2+b^2+c^2=1$$
is given by
$$\left(\frac{p^2-q^2-r^2}s\right)^k+\left(\frac{2pq}s\right)^k+\left(\frac{2pr}s\right)^k=1\tag1$$
where $s=p^2+q^2+r^2$ and $k=2$. But eq $(1)$ is also satisfied for $k=1$ if 
$$p=\frac{q^2+r^2}{q+r}$$
For example, if $q=1,\,r=5$, then,
$$\big({-}\tfrac{5}{31}\big)^k+\big(\tfrac{6}{31}\big)^k+\big(\tfrac{30}{31}\big)^k=1$$
for $k=1,2$, and so on.
A: 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 $c$.
A: $\frac{6}{7}, \frac{3}{7}, -\frac{2}{7}$ is the third solution.
