Algebra Challenge 2022/07/25

Solution: Let $y=x-\lfloor x \rfloor$ and $z=\lfloor x \rfloor$. Therefore $0\le y<1$, $z$ is an integer, $$(y+z)^2=ay\tag{1}$$ and $$y^2+(2z-a)y+z^2=0\tag{2}$$

Solve $y$ in equation (2), we have: $$y=\dfrac{a-2z\pm\sqrt{a^2-4az}}{2}\tag{3}$$ Because $y$ is a real number, therefore $$a^2-4az\ge 0$$ Because $a\ge 0$, we have $$z\le \dfrac{a}{4} \tag{4}$$

Additionally, because $$4z^2\ge0$$ $$ a^2-4az+4z^2\ge a^2-4az$$ $$(a-2z)^2\ge a^2-4az$$ Because (4), $a-2z\ge 0$, we have $$a-2z\ge\sqrt{a^2-4az}$$ Therefore, $y>=0$ for all $a\ge 0$.

Because $y<1$, according to (3), we have $$a-2z \pm\sqrt{a^2-4az} <2\tag{5}$$ Solve inequality (5), we have $$-\sqrt{a}-1<z<\sqrt{a}-1 \tag{6}$$ Because $$\dfrac{a}{4}-\sqrt{a}+1=(\dfrac{a}{2}-1)^2\ge 0$$ Inequality (6) implies inequality (4). Therefore integer $z$ are in $(-\sqrt{a}-1, \sqrt{a}-1)$. Because each integer $z$ value corresponds to a unique real value of $x$, therefore the number of solutions for $x$ is the number of integers in the range of $(-\sqrt{a}-1, \sqrt{a}-1) \boxed{\phantom{|}}$