Balti Tee 2010 · Ülesanne 1

There are two solutions: $(0, 0, 0, 0)$ and $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3})$. If $(a, b, c, d)$ satisfies the equations, then we may as well assume $a \le b \le c \le d$. These are non-negative because an even power of a real number is always non-negative. It follows that

$$b+c+d \ge a+c+d \ge a+b+d \ge a+b+c$$

and since $x \mapsto x^{2010}$ is increasing for $x \ge 0$ we have that

$$3a = (b+c+d)^{2010} \ge (a+c+d)^{2010} \ge (a+b+d)^{2010} \ge (a+b+c)^{2010} = 3d.$$

We conclude that $a = b = c = d$ and all the equations take the form $(3a)^{2010} = 3a$, so $a = 0$ or $3a = 1$. Finally, it is clear that $a = b = c = d = 0$ and $a = b = c = d = \frac{1}{3}$ solve the system.