Slovenia — National Math Olympiad

Assume that such $n$ exists. From $x^4 - 2011x^2 + n = 0$ we deduce that
$$x^2 = \frac{2011 \pm \sqrt{2011^2 - 4n}}{2}.$$
This has to be an integer, so $2011^2 - 4n$ has to be a perfect square. We can write $2011^2 - 4n = m^2$ for some odd positive integer $m$ or $n = \frac{2011^2 - m^2}{4}$. So, $x^2 = \frac{2011 \pm m}{2}$. The numbers $\frac{2011+m}{2}$ and $\frac{2011-m}{2}$ are perfect squares and their sum is 2011. Let us show that 2011 cannot be written as a sum of two perfect squares. When divided by 4 a perfect square can only give the remainder 0 or 1. A sum of two perfect squares can therefore only give the remainder 0, 1 or 2. Since 2011 gives the remainder 3, it cannot be the sum of two perfect squares. Hence, an integer $n$ with the required properties does not exist.