Slovenija 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 $4a^2(5x-x^2)=(a^2-1{)}^2-10x+10a^{2}x+25x^2$ 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 $x=1$.