$a+b+c=4$$a^2+b^2+c^2=8$
I'm not sure if my solution is good, since I don't have answers for this problem. Any directions, comments and/or corrections would be appreciated.
It's obvious that $\{a,b,c\}\in[-\sqrt8,\sqrt8]$. Since two irrational numbers always give irrational number when added, if we assume that one of $a,b $ or $c$ is irrational, for example $a$, then one more has to be irrational, for example $b$, such that $a=-b$. That leaves $c=4$ in order to satisfy the first equation, but that makes the second one incorrect. That's why all of $a,b,c$ have to be rational numbers. $(*)$
I squared the first equation and got $ab+bc+ca=4$. Then, since $a=4-(b+c)$, I got quadratic equation $b^2+(c4)b+(c-2)^2=0$. Its' discriminant has to be positive and perfect square to satisfy $(*)$. $ D= -c(3c-8) $ From this, we see that $c$ has to be between $0$ and $8/3$ in order to satisfy definition of square root. Specially, for $c=0$ we get $D=0$ and solution for equation $b=\frac{-c+4}{2}=2$. Since the system is symmetric, we also get $c=2$. Similar, for $c=8/3$ we get $b=2$.
In order for $D$ to be perfect square, one of the following has to be true: $ -c=3c-8 $ $ c=n^2 \land 3c-8=1 $ $ 3c-8=n^2 \land c=1 $
However, only the first one is possible, so $c=2$, and for that we get $b=\frac{(-2+4\pm2)}{2} \Rightarrow b=2 \lor b=0$.
So, all possible values for $c$ are $c\in\{0, 2, \frac{8}3\}$.
EDIT: For $a=b$, there is one more solution: $c=2/3$. Why couldn't I find it with method described above?