Exercise 6. Determine all triplets of real numbers $(a, b, c)$ satisfying the three equations

$$
a^{2}+b=c^{2}, b^{2}+c=a^{2}, c^{2}+a=b^{2}
$$
Solution to Exercise 6 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 $0=b+c=2b=2c$ so $a=b=c=0$. Conversely, the triplet $(0,0,0)$ satisfies all 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 handle 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 triplets that are solutions are therefore $\{(0,0,0),(0,1,-1),(-1,0,1),(1,-1,0)\}$.

Graders' Comments: This problem was difficult and very few students solved it entirely. However, many managed to make progress and provide very relevant leads to advance, which naturally earned points for their authors.

$\triangleright$ 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 $\sqrt{a}=b-c$.

$\triangleright$ 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 encourage looking at $b=0$ to find other solutions: to earn points for finding solutions, one must often find all of them.

$\triangleright$ Often students factored certain expressions and ended up, for example, with $(c-b)(c+b)=$ $c+b$. Some simplified without hesitation to end up with $c-b=1$, but they should not forget the case where $b+c=0$.

$\triangleright$ 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 having a complete proof and developing relevant ideas to advance.

$\triangleright$ Solutions found must always be verified. In international competitions, this can cost a point.