Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.

Question

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?

Answer 1

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.

Answer 2

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 $c$.

Answer 3

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