Exercise 5. Find all triplets of non-negative real numbers $(a, b, c)$ such that

$$
\left\{\begin{array}{l}
a^{2}+a b=c \\
b^{2}+b c=a \\
c^{2}+c a=b
\end{array}\right.
$$
## Solution to Exercise 5

First, assume one of the real numbers is zero, without loss of generality, let $a=0$. Then the first equation gives $\mathrm{c}=0$ and the last equation gives $\mathrm{b}=0$. Conversely, the triplet $\mathrm{a}=\mathrm{b}=\mathrm{c}=0$ is indeed a solution. Now, assume all three real numbers are non-zero.

Next, assume two of the real numbers are equal, without loss of generality, let $\mathrm{a}=\mathrm{b}$. We can then subtract the second and third equations to get $\mathrm{b}^{2}-\mathrm{c}^{2}=0$. Since $\mathrm{b}$ and $\mathrm{c}$ are both positive, we have $\mathrm{a}=\mathrm{b}=\mathrm{c}$. In the first equation, we then have $2 \mathrm{a}^{2}=\mathrm{a}$. Since $\mathrm{a}$ is non-zero, we can divide by $2a$ to get $a=\frac{1}{2}$. Conversely, the triplet $a=b=c=\frac{1}{2}$ is indeed a solution. Now, assume all three real numbers are distinct.

Multiply the equations to obtain

$$
a(a+b) b(b+c) c(c+a)=a b c
$$

We assumed previously that the real numbers are all non-zero, so we can divide by $abc$ to get

$$
(a+b)(b+c)(c+a)=1
$$

By subtracting the first two equations, we get $a^{2}-b^{2}+b a-b c=c-a$, which factors to

$$
(a-b)(a+b)=(c-a)(b+1)
$$

Similarly, by subtracting the second and third equations, and the third and first equations, we get

$$
\left\{\begin{array}{l}
(b-c)(b+c)=(a-b)(c+1) \\
(c-a)(c+a)=(b-c)(a+1)
\end{array}\right.
$$

By multiplying these three equations and dividing both sides by $(a-b)(b-c)(c-a)$ (which is non-zero because we assumed $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are distinct), we get

$$
(1+a)(1+b)(1+c)=(a+b)(b+c)(c+a)
$$

But we previously showed that the second term equals 1, which is absurd because

$$
(1+a)(1+b)(1+c)>1 \cdot 1 \cdot 1=1
$$

Thus, the only solutions are those found at the beginning of the solution, namely $\mathrm{a}=\mathrm{b}=\mathrm{c}=0$ and $\mathrm{a}=\mathrm{b}=\mathrm{c}=\frac{1}{2}$.

Grader's Comment: The problem was approached and fairly well executed. However, several recurring errors are noted:

- Some students find the two triplets as solutions but do not verify that they are indeed solutions. When reasoning with implications, it is not at all certain that the triplets found are solutions.
- Many assume, for example, $a>b>c$. Two issues arise: they forget that multiple variables can be equal. More importantly, to make this assumption, the system must be symmetric, meaning that if two variables, such as $a$ and $b$, are swapped, the system remains unchanged. This is not the case here: the system is completely altered.