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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?

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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.

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 $c$.

$\frac{6}{7}, \frac{3}{7}, -\frac{2}{7}$ is the third solution.

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