Determine all triplets of real numbers $(a, b, c)$ satisfying the three equations

$$
a^{2}+b=c^{2}, b^{2}+c=a^{2}, c^{2}+a=b^{2}
$$
We start by summing the three equations. We obtain

$$
a^{2}+b+b^{2}+c+c^{2}+a=c^{2}+a^{2}+b^{2}
$$

which, after simplification, gives $a+b+c=0$.

On the other hand, the third equation can be rewritten as:

$$
a=b^{2}-c^{2}=(b-c)(b+c)
$$

Using that $b+c=-a$, we find that $a=-a(b-c)$, or $a(b-c+1)=0$. Therefore, we have $a=0$ or $b=c-1$. Similarly, we have $b=0$ or $c=a-1$ and $c=0$ or $a=b-1$.

Suppose that $a=0$. The third equation becomes $c^{2}=b^{2}$, or $b= \pm c$. If $b=c$, then $0=b+c=2 b=2 c$ so $a=b=c=0$. Conversely, the triplet $(0,0,0)$ satisfies the three equations. If $b=-c$, the first equation gives $c^{2}=-c$, or $0=c^{2}+c=c(c+1)$. If $c=0$, then $b=-a-c=0$ and we have found the triplet $(0,0,0)$ again. If $c+1=0$, then $c=-1=-b$. Conversely, the triplet $(0,1,-1)$ satisfies the equations.

We treat the case where $b=0$ in the same way, which gives the two triplets $(0,0,0)$ and $(-1,0,1)$, which we verify are indeed solutions to the problem, as well as the case where $c=0$, which gives the two triplets $(0,0,0)$ and $(1,-1,0)$, which we verify are indeed solutions to the problem.

Now suppose that none of the real numbers $a, b$, or $c$ is zero. Then we have $c=b+1, b=a+1$ and $a=c+1$. But then $a=c+1=b+2=a+3$, which is absurd.

The only triplet solutions are therefore $\{(0,0,0),(0,1,-1),(-1,0,1),(1,-1,0)\}$.

Comment from the graders: This problem was difficult and very few students solved it entirely. However, many managed to make progress and provide very relevant leads, which naturally earned points for their authors.

- The concept of square root is often misused. Some students use that $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$, which is false, especially when some quantities could be negative. For example, $a=b^{2}-c^{2}$ does not imply that $\sqrt{a}=b-c$.
- Students often found $(0,0,0)$ as a solution. But rarely did they see that $(1,0,-1)$, for example, was a solution. Finding $(0,0,0)$ as a solution should have encouraged them to look at $b=0$ to find other solutions: to get points for finding solutions, one must often find them all.
- Often students factored certain expressions and ended up, for example, with $(c-b)(c+b)=c+b$. Some simplified without hesitation to get $c-b=1$, but they should not forget the case where $b+c=0$.
- A good number of students managed to prove that $a+b+c=0$. But often, the reasoning abruptly stops by saying that necessarily $(a, b, c)=(0,0,0)$. One must be rigorous and honest: hastily concluding that $(a, b, c)=(0,0,0)$ prevents a complete proof and the development of relevant ideas to move forward.
- Solutions found must always be verified. In international competitions, this can cost a point.