IMC 2025 · Day 1 · P4

We will show that if $a = 2$ then we must have $x \in [-1, 2) \cup [3, 4)$, otherwise $x \in [-1, a)$.

Suppose that $\lfloor x/a \rfloor = m$. Then $m \le x/a < m + 1$, and $$\tag{2} am \le x < a(m + 1).$$ Let $b = 1$. Then $$\tag{3} \left\lfloor \sqrt[a]{1 + x} \right\rfloor = 1 + \lfloor x/a \rfloor.$$ From (3) it follows that $\bigl\lfloor \sqrt[a]{1 + x} \bigr\rfloor = 1 + m$, or $1 + m \le \sqrt[a]{1 + x} < 2 + m$, or $$\tag{4} (1 + m)^a - 1 \le x < (2 + m)^a - 1,$$ where we have used the obvious fact that $m \ge -1$. Indeed, the number $\sqrt[a]{1 + x}$ and therefore the number $\bigl\lfloor \sqrt[a]{1 + x} \bigr\rfloor = 1 + m$ is not negative. By Bernoulli's inequality, the following two inequalities — which compare the inequalities (2) and (4) — hold: $$\begin{gather} am \le (1 + m)^a - 1, \tag{5} \\ a(m + 1) \le (1 + (m + 1))^a - 1, \tag{6} \end{gather}$$ where in (5), equality holds if and only if $m = 0$, and in (6), equality holds if and only if $m = -1$. From (2), (4)–(6), it follows that $$\tag{7} (m + 1)^a - 1 \le x < a(m + 1).$$ From (7), it follows that $(m + 1)^a - 1 < a(m + 1)$. Therefore $$\tag{8} (m + 1)^a \le a(m + 1).$$ From (8), it follows that $m + 1 \le a^{\frac{1}{a-1}}$. If $a > 2$ then $m + 1 < a^{\frac{1}{a-1}} < 2$, because $2^{a-1} > a$ for $a > 2$ (one can prove this by mathematical induction) and $1 < a^{\frac{1}{a-1}} < 2$ is not an integer. Therefore, if $a > 2$ then $m = 0$ or $m = -1$. If $a = 2$ then $m = -1$, $m = 0$ or $m = 1$. From (7), it follows that the equality (3) holds true only for the values $-1 \le x < 0$ ($m = -1$), $0 \le x < a$ ($m = 0$) if $a > 2$, and $-1 \le x < 0$ ($m = -1$), $0 \le x < 2$ ($m = 0$) and $3 \le x < 4$ ($m = 1$) if $a = 2$.

We will now prove that for these values of $x$, the equality (1) is true for all positive integers $b$. From (7), it follows that $b^a + (m + 1)^a - 1 \le b^a + x < b^a + a(m + 1)$ and $$\tag{9} 1 + \frac{(m + 1)^a - 1}{b^a} \le 1 + \frac{x}{b^a} < 1 + \frac{a(m + 1)}{b^a}.$$ By Bernoulli's inequality $$\tag{10} 1 + \frac{a(m + 1)}{b^a} \le \left( 1 + \frac{m + 1}{b^a} \right)^a,$$ where equality occurs if and only if $m = -1$. It is easy to check that for $m = 1$ (if $a = 2$), $m = 0$ and $m = -1$, the following inequality holds true: $$\tag{11} \left( 1 + \frac{m}{b^a} \right)^a \le 1 + \frac{(m + 1)^a - 1}{b^a}.$$ From (9)–(11), it follows that $$\tag{12} \left( 1 + \frac{m}{b^a} \right)^a \le 1 + \frac{x}{b^a} < \left( 1 + \frac{m + 1}{b^a} \right)^a.$$ From (12), it follows that $$\begin{gather*} b^a + m \le \sqrt[a]{b^a + x} \cdot b^{a-1} < b^a + m + 1, \\ b^a + \lfloor x/a \rfloor \le \sqrt[a]{b^a + x} \cdot b^{a-1} < b^a + \lfloor x/a \rfloor + 1. \end{gather*}$$ Consequently, equality (1) holds.