
Problem 1 (Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and

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


Solution 1. First suppose that $a=0$. Then we have $b c=1$ and $c=b^{2} c=b$. So $b=c$, which implies $b^{2}=1$ and hence $b= \pm 1$. This leads to the solutions $(a, b, c)=(0,1,1)$ and $(a, b, c)=(0,-1,-1)$. Similarly, $b=0$ gives the solutions $(a, b, c)=(1,0,1)$ and $(a, b, c)=(-1,0,-1)$, while $c=0$ gives $(a, b, c)=(1,1,0)$ and $(a, b, c)=(-1,-1,0)$.

Now we may assume that $a, b, c \neq=0$. We multiply $a b+b c+c a=1$ by $a$ to find $a^{2} b+a b c+c a^{2}=a$, hence $a^{2} b=a-a b c-a^{2} c$. Substituting this in $a^{2} b+c=b^{2} c+a$ yields $a-a b c-a^{2} c+c=b^{2} c+a$, so $b^{2} c+a b c+a^{2} c=c$. As $c \neq=0$, we find $b^{2}+a b+a^{2}=1$.

Analogously we have $b^{2}+b c+c^{2}=1$ and $a^{2}+a c+c^{2}=1$. Adding these three equations yields $2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3$, which implies $a^{2}+b^{2}+c^{2}=1$. Combining this result with $b^{2}+a b+a^{2}=1$, we get $1-a b=1-c^{2}$, so $c^{2}=a b$.

Analogously we also have $b^{2}=a c$ and $a^{2}=b c$. In particular we now have that $a b, b c$ and $c a$ are all positive. This means that $a, b$ and $c$ must all be positive or all be negative. Now assume that $|c|$ is the largest among $|a|,|b|$ and $|c|$, then $c^{2} \geq|a b|=a b=c^{2}$, so we must have equality. This means that $|c|=|a|$ and $|c|=|b|$. Since $(a, b, c)$ must all have the same sign, we find $a=b=c$. Now we have $3 a^{2}=1$, hence $a= \pm \frac{1}{3} \sqrt{3}$. We find the solutions $(a, b, c)=\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $(a, b, c)=\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.

We conclude that all possible triples $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1)$, $(1,1,0),(-1,-1,0),\left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)$ and $\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.
