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

$$
a^{2} b+c=b^{2} c+a=c^{2} a+b \text {. }
$$

Solution:
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 $\left(-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3},-\frac{1}{3} \sqrt{3}\right)$.

From the problem statement $a b=1-b c-c a$ and thus $b^{2} c+a=a^{2} b+c=$ $a-a b c-a^{2} c+c, c\left(b^{2}+a^{2}+a b-1\right)=0$. If $c=0$ then $a b=1$ and $a^{2} b=b$, which implies $a=b= \pm 1$. Otherwise $b^{2}+a^{2}+a b=1$. Cases $a=0$ and $b=0$ are completely analogous to $c=0$, so we may suppose that $a, b, c \neq 0$. In this case we end up with

$$
\left\{\begin{array}{l}
a^{2}+b^{2}+a b=1, \\
b^{2}+c^{2}+b c=1, \\
c^{2}+a^{2}+c a=1, \\
a b+b c+c a=1
\end{array}\right.
$$

Adding first three equations and subtracting the fourth yields $2\left(a^{2}+b^{2}+c^{2}\right)=2=$ $2(a b+b c+c a)$. Consequently, $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$. Now we can easily conclude that $a=b=c= \pm \frac{1}{\sqrt{3}}$.

We have

$$
c\left(1-b^{2}\right)=a(1-a b)=a(b c+c a)=c\left(a b+a^{2}\right),
$$

and so

$$
c\left(a^{2}+a b+b^{2}-1\right)=0 .
$$

Similarly, we have

$$
b\left(a^{2}+a c+c^{2}-1\right)=0 \quad \text { and } \quad a\left(b^{2}+b c+c^{2}-1\right)=0
$$



If $c=0$, then we get $a b=1$ and $a^{2} b=a=b$, which give us $a=b=1$, or $a=b=-1$. Similarly, if $a=0$, then $b=c=1$, or $b=c=-1$, while if $b=0$, then $a=c=1$, or $a=c=-1$.

So assume that $a b c \neq 0$. Then

$$
a^{2}+a b+b^{2}=b^{2}+b c+c^{2}=c^{2}+c a+a^{2}=1
$$

Adding these gives us

$$
2\left(a^{2}+b^{2}+c^{2}\right)+a b+b c+c a=3
$$

and using the fact that $a b+b c+c a=1$, we get

$$
a^{2}+b^{2}+c^{2}=1=a b+b c+c a .
$$

Hence

$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=0
$$

and so $a=b=c= \pm \frac{1}{\sqrt{3}}$.

Therefore, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$

It is $a b+b c+c a=1$ and

$$
a^{2} b+c=b^{2} c+a=c^{2} a+b . \tag{1}
$$

We have

$$
\begin{aligned}
a^{2} b+c=b^{2} c+a & \Longleftrightarrow a^{2} b-a=b^{2} c-c \\
& \Longleftrightarrow a(a b-1)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow a(-b c-a c)=c\left(b^{2}-1\right) \\
& \Longleftrightarrow-a c(a+b)=c\left(b^{2}-1\right)
\end{aligned}
\tag{2}
$$

First, consider the case where one of $a, b, c$ is equal to 0 . Without loss of generality, assume that $a=0$. Then $b c=1$ and $b=c$ from (1), and so $b^{2}=1$ giving us $b=1$ or -1 . Hence $b=c=1$ or $b=c=-1$.

Therefore, $(a, b, c)$ equals one of the triples $(0,1,1),(0,-1,-1)$, as well as their rearrangements $(1,0,1)$ and $(-1,0,-1)$ when $b=0$, or $(1,1,0)$ and $(-1,-1,0)$ when $c=0$.

Now consider the case where $a \neq 0, b \neq 0$ and $c \neq 0$. Then (2) gives us

$$
-a(a+b)=b^{2}-1 \Longleftrightarrow-a^{2}-a b=b^{2}-1 \Longleftrightarrow a^{2}+a b+b^{2}-1=0 .
$$

The quadratic $P(x)=x^{2}+b x+b^{2}-1$ has $x=a$ as a root. Let $x_{1}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0). From Vieta's formulas we get

$$
\left\{\begin{aligned}
x_{1}+a=-b & \Longleftrightarrow x_{1}=-b-a, \text { and } \\
x_{1} a=b^{2}-1 & \Longleftrightarrow x_{1}=\frac{b^{2}-1}{a} .
\end{aligned}\right.
$$



Using $a^{2} b+c=c^{2} a+b$ we obtain $b\left(a^{2}-1\right)=c(a c-1)$ yielding $a^{2}+a c+c^{2}-1=0$ in a similar way. The quadratic $Q(x)=x^{2}+c x+c^{2}-1$ has $x=a$ as a root. Let $x_{2}$ be its second root (which could be equal to $a$ in the case where the discriminant is 0). From Vieta's formulas we get

$$
\left\{\begin{aligned}
x_{2}+a=-c & \Longleftrightarrow x_{2}=-c-a, \text { and } \\
x_{2} a=c^{2}-1 & \Longleftrightarrow x_{2}=\frac{c^{2}-1}{a} .
\end{aligned}\right.
$$

Then

which give us

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-b-a-c-a, \text { and } \\
x_{1}+x_{2}=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a}
\end{array}\right.
$$

$$
\begin{aligned}
-(2 a+b+c)=\frac{b^{2}-1}{a}+\frac{c^{2}-1}{a} & \Longleftrightarrow-2 a^{2}-b a-c a=b^{2}+c^{2}-2 \\
& \Longleftrightarrow b c-1-2 a^{2}=b^{2}+c^{2}-2 \\
& \Longleftrightarrow 2 a^{2}+b^{2}+c^{2}=1+b c .
\end{aligned} 
\tag{3}
$$

By symmetry, we get

$$
\begin{aligned}
& 2 b^{2}+a^{2}+c^{2}=1+a c, \text { and } \\ \tag{4}
\end{aligned}
$$
$$
\begin{aligned}
& 2 c^{2}+a^{2}+b^{2}=1+b c
\end{aligned} \tag{5}
$$

Adding equations (3), (4), and (5), we get

$$
4\left(a^{2}+b^{2}+c^{2}\right)=3+a b+b c+c a \Longleftrightarrow 4\left(a^{2}+b^{2}+c^{2}\right)=4 \Longleftrightarrow a^{2}+b^{2}+c^{2}=1
$$

From this and (3), since $a b+b c+c a=1$, we get

$$
a^{2}=b c=1-a b-a c \Longleftrightarrow a(a+b+c)=1
$$

Similarly, from (4) we get

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

and from (4),

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

Clearly, it is $a+b+c \neq 0$ (for otherwise it would be $0=1$, a contradiction). Therefore,

$$
a=b=c=\frac{1}{a+b+c}
$$

and so $3 a^{2}=1$ giving us $a=b=c= \pm \frac{1}{\sqrt{3}}$.

In conclusion, the solutions $(a, b, c)$ are $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$, $(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$, and $\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$.

First, homogenize the condition $a^{2} b+c=b^{2} c+a=c^{2} a+b$ by replacing $c$ by $c(a b+b c+c a)$ (etc.), yielding

$$
a^{2} b+c=a^{2} b+a b c+b c^{2}+c^{2} a=a b c+\sum_{c y c} a^{2} b+\left(c^{2} b-b^{2} c\right)=a b c+\sum_{c y c} a^{2} b+b c(c-b) .
$$



Thus, after substracting the cyclicly symmetric part $a b c+\sum_{c y c} a^{2} b$ we find the condition is eqivalent to

$$
D:=b c(c-b)=c a(a-c)=a b(b-a) .
$$

Ending 1. It is easy to see that if e.g. $a=0$ then $b=c= \pm 1$, and if e.g. $a=b$ then either $a=b=c= \pm \frac{1}{\sqrt{3}}$ or $a=b= \pm 1, c=0$, and these are indeed solutions. So, to show that these are all solutions (up to symmetries), we may assume by contradiction that $a, b, c$ are pairwise different and non-zero. All conditions are preserved under cyclic shifts and under simultaenously switching signs on all $a, b, c$, and by applying these operations as necessary we may assume $a<b<c$. It follows that $D^{3}=a^{2} b^{2} c^{2}(c-b)(a-c)(b-a)$ must be negative (the only negative term is $a-c$, hence $D$ is negative, i.e. $b c, a b<0<a c$. But this means that $a, c$ have the same sign and $b$ has a different one, which clearly contradicts $a<b<c$ ! So, such configurations are impossible.

Ending 2. Note that $3 D=\sum c^{2} b-\sum b^{2} c=(c-b)(c-a)(b-a)$ and $D^{3}=a^{2} b^{2} c^{2}(c-$ $b)(a-c)(b-a)=-3 a^{2} b^{2} c^{2} D$. Since $3 D$ and $D^{3}$ must have the same sign, and $-3 a^{2} b^{2} c^{2}$ is non-positive, necessarily $D=0$. Thus (up to cyclic permutation) $a=b$ and from there we immediately find either $a=b= \pm 1, c=0$ or $a=b=c= \pm \frac{1}{\sqrt{3}}$.

Final answer: $(0,1,1),(0,-1,-1),(1,0,1),(-1,0,-1),(1,1,0)$,$(-1,-1,0),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$,$\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$