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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?

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two irrational numbers always give irrational number when added No. (see: $\sqrt2 + (2-\sqrt2)$). – anon Feb 7 '12 at 2:50
I think $(a,b,c)=(1,\frac{1}{2}(3\pm\sqrt{5}),\frac{1}{2}(3\mp\sqrt{5}))$ is a solution—if so, you'd need to re-evaluate your conclusion that the three numbers have to be rational. – Isaac Feb 7 '12 at 2:51
If one of the numbers is irrational, and the other is the negative of that, then, you'll get $2 \times \text{irrational}$, which is an irrational number and not rational as you claim. So, the conclusion is not quite correct. Try harder! – user21436 Feb 7 '12 at 2:56
Hint: Substitute $c = 4 - a - b$ into the second equation and view it is a quadratic equation in $b$. – JavaMan Feb 7 '12 at 4:45
up vote 4 down vote accepted

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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All the points are given by $$ \left( \frac{4}{3}, \; \frac{4}{3}, \; \frac{4}{3} \; \right) + \frac{\sqrt{12}}{3} \; \left(1, \; -1, \; 0 \; \right) \; \cos t + \frac{2}{3} \; \left(1, \; 1, \; -2 \; \right) \; \sin t $$ where the points with a zero and a pair of 2's occur at $t = \frac{\pi}{2},\frac{7 \pi}{6},\frac{11 \pi}{6}, $ and the third component is given by $$ \frac{4}{3} (1 - \sin t) $$ which varies between $0$ and $\frac{8}{3},$ the latter happening at $t = \frac{3\pi}{2}.$

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