Bulgarian Mathematical Olympiad

Solution:
Set
$$\frac{a(a-b)+b(b-c)+c(c-a)}{2} = d^{2}$$
where $d$ is an integer, $x = a-b$, $y = b-c$ and $z = c-a$. Then we have
$$x + y + z = 0, \quad x^{2} + y^{2} + z^{2} = 4 d^{2}$$
Since any square is congruent to $0$ or $1$ modulo $4$, it follows from (1) that the integers $x$, $y$ and $z$ are even. Set $x_{1} = \frac{x}{2}$, $y_{1} = \frac{y}{2}$ and $z_{1} = \frac{z}{2}$. Then (1) gives
$$x_{1} + y_{1} + z_{1} = 0, \quad x_{1}^{2} + y_{1}^{2} + z_{1}^{2} = d^{2}$$
and we conclude as above that $x_{1}$, $y_{1}$, $z_{1}$ and $d$ are even integers. Repeating the same argument we see that $2^{n}$ divides $x$, $y$ and $z$ for every positive integer $n$. Therefore $x = y = z = 0$, i.e. $a = b = c$.