Slovenian Mathematical Olympiad, 2008

Let us find an estimate for the value of this expression. Evidently, $1 - x^2 \le 1$ and $5x - x^2$ is bounded by $5x - x^2 = \frac{25}{4} - (x - \frac{5}{2})^2 \le \frac{25}{4}$, so
$$\sqrt{1-x^2} + \sqrt{5x-x^2} \le \sqrt{1} + \sqrt{\frac{25}{4}} = 1 + \frac{5}{2} = 3 + \frac{1}{2}.$$
On the other hand, the expression is non-negative and cannot be equal to 0 because that would imply $1 - x^2 = 0$ and $5x - x^2 = 0$, which is impossible. So, the only integer values the expression can take are 1, 2 or 3. Let $\sqrt{1-x^2} + \sqrt{5x-x^2} = a$ and eliminate the square roots. By squaring both sides of $\sqrt{1-x^2} = a - \sqrt{5x-x^2}$, we get $1 - x^2 = a^2 - 2a\sqrt{5x-x^2} + 5x - x^2$ or $2a\sqrt{5x-x^2} = a^2 - 1 + 5x$. After squaring both sides once again, we get $4a^2(5x - x^2) = (a^2 - 1)^2 - 10x + 10a^2x + 25x^2$ or
$$x^2(25 + 4a^2) + x(-10a^2 - 10) + (a^2 - 1)^2 = 0.$$
If $a = 1$ we have $x(29x - 20) = 0$, so $x = 0$ or $x = \frac{20}{29}$. Inserting these values of $x$ into the initial expression we see that only $x = 0$ works.

When $a = 2$ the equation becomes $(x - 1)(41x - 9) = 0$, so $x = 1$ or $x = \frac{9}{41}$. In both cases a short calculation confirms that the value of the expression is indeed 2.

Finally, let us consider the case $a = 3$. The quadratic equation $61x^2 - 100x + 64 = 0$ has a negative discriminant $D = 100^2 - 4 \cdot 61 \cdot 64 = 100^2 - (2 \cdot 64)(2 \cdot 61) = 100^2 - 128 \cdot 122$, so there are no real solutions.

We conclude that the value of the expression is an integer only when $x = 0$, $x = \frac{9}{41}$ or $x = 1$.