Junior Balkan Mathematical Olympiad 2021 - Problem 1 | Solution - 2 July 2021



Problem 1. Let $n$ ( $n \ge 1$ ) be an integer. Consider the equation $2\cdot \lfloor{\frac{1}{2x}}\rfloor - n + 1 = (n + 1)(1 - nx)$ , where $x$ is the unknown real variable. (a) Solve the equation for $n = 8$ . (b) Prove that there exists an integer $n$ for which the equation has at least $2021$ solutions. (For any real number $y$ by $\lfloor{y} \rfloor$ we denote the largest integer $m$ such that $m \le y$ .)

Solution Let $A = \left\lfloor \frac{1}{2x} \right\rfloor$ -- then the equation gives $x = \frac{2(n-A)}{n(n+1)}$ and now substituting in the definition of $A$ yields $A = \left\lfloor\frac{n(n+1)}{4(n-A)}\right\rfloor.$ The latter equality is a necessary and sufficient condition for the corresponding $x$ to be a solution to the equation. Let us also observe that $A$ is an integer and that $1 \leq A \leq n - 1$ for $n\geq 3$ -- indeed, if $A=0$ , then $0 = \left\lfloor \frac{n+1}{4} \right\rfloor \geq 1$ ; if $A=n$ , the right-hand side is undefined; and if $A<0$ or $A>n$ , then the sides have different signs. a) For $n=8$ we want $A = \left\lfloor \frac{18}{8-A} \right\rfloor$ . By the above, $A$ is an integer between $1$ and $7$ inclusive. A direct verification shows that only $A=3$ and $A=4$ are solutions, with the corresponding $x$ being $x=\frac{5}{36}$ and $x=\frac{1}{9}$ . b) It suffices to have at least $2021$ integer solutions $1\leq A \leq n-1$ to $A \leq \frac{n(n+1)}{4(n-A)} < A + 1$ for some $n$ . The left inequality is equivalent to $(2A - n)^2 + n \geq 0$ and holds for all $A$ . The right inequality is equivalent to $(2A-n+1)^2 < n + 1$ and hence holds precisely for $\frac{n-1 - \sqrt{n+1}}{2} < A < \frac{n-1+\sqrt{n+1}}{2}$ . Observe that this range for $A$ is tighter than $1 \leq A \leq n - 1$ for $n\geq 6$ , as $(n-3)^2 > n + 1$ and $(n-1)^2 > (n+1)$ for these $n$ . Finally, the difference between the endpoints of the interval $\left(\frac{n-1 - \sqrt{n+1}}{2}, \frac{n-1+\sqrt{n+1}}{2}\right)$ is $\sqrt{n+1}$ and hence for sufficiently large $n$ this interval must contain at least $2021$ integers. This completes the proof.;