<s>[INST] System $a+b+c=4$, $a^2+b^2+c^2=8$. find all possible values for $c$. $$a+b+c=4$$$$a^2+b^2+c^2=8$$
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?
 [/INST] We will show that any $c$ in the closed interval $[0,8/3]$ is achievable, and nothing else is. Let $c$ be any real number, and suppose $a$ and $b$ are real numbers such that $(a,b,c)$ satisfies our two equations.  Note that in general
$$2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0. \qquad\qquad(\ast)$$
Put $a^2+b^2=8-c^2$ and $a+b=4-c$. Thus from $(\ast)$,
$$2(8-c^2)-(4-c)^2 =(a-b)^2\ge 0. \qquad\qquad(\ast\ast)$$
So we must have $8c-3c^2\ge 0$. This is only true for $0\le c \le 8/3$.
Now we show that any $c$ in the interval $[0,8/3]$ is achievable.
We want to make $(a-b)^2=8c-3c^2$. If $c$ is in $[0,8/3]$, then $8c-3c^2\ge 0$, so we can take the square root, and obtain
$$a-b=\pm\sqrt{8c-3c^2}.$$
We can now solve the system $a-b=\pm\sqrt{8c-3c^2}$, $a+b=4$ to find the values of $a$ and $b$. We might as well give the solutions $(a,b,c)$ explicitly. They are
$$a=2\pm \frac{1}{2}\sqrt{8t-t^2}, \qquad b=2\mp \frac{1}{2}\sqrt{8t-t^2},\qquad c=t,$$
where  $t$ is a parameter that ranges over the interval $0\le t\le 8/3$.
There is something mildly ugly in the above general solution, since it breaks  symmetry. That can be fixed.
Remark: It is maybe interesting to ask what are the possible values of $c$ under the conditions $c\ge a$, $c\ge b$. Of course we still must have $c\le 8/3$.
For given $a$, $b$, and $c$, look at the monic cubic polynomial $P(x)$ whose roots are $a$, $b$, and $c$. Then 
$$P(x)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc.$$
You found that $ab+bc+ca=4$. So our polynomial has shape
$$P(x)=x^3-4x^2+4x-p,$$
where $p$ is the product of the solutions. From our previous work we know the solutions are $\ge 0$, so $p\ge 0$.
The derivative of $P(x)$ is $3x^2-8x+4$, which vanishes at $x=2/3$ and $x=2$. So $P(x)=0$ has a solution in the interval $[0, 2/3]$, a solution in the interval $[2/3,2]$, and a solution in the interval $[2,8/3)$ (there can be a double root at $2/3$ or at $2$).  Thus at least one of $a$, $b$, $c$ is $\ge 2$.
Let $c$ be a maximal root.  We can have $c=2$ (let $a=0$, $b=2$), and we can have $c=8/3$ ($a=b=2/3$). By continuity, or calculation, one can then see that for any $c$ in the interval $[2,8/3]$, there are $a$ and $b$ such that $(a,b,c)$ is a solution of our system, and $c\ge a$, $c\ge b$.
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