International Mathematics Competition
for University Students
2025

IMC2025: Day 1, Problem 4

Problem 4. Let \(\displaystyle a\) be an even positive integer. Find all real numbers \(\displaystyle x\) such that

\(\displaystyle \left\lfloor \sqrt[a]{b^{a} + x} \cdot b^{a - 1} \right\rfloor = b^{a} + \left\lfloor x/a\right\rfloor \tag{1} \)

holds for every positive integer \(\displaystyle b\).

(Here \(\displaystyle \lfloor x\rfloor\) denotes the largest integer that is no greater than \(\displaystyle x\).)

Yagub Aliyev, ADA University, Baku, Azerbaijan

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

Suppose that \(\displaystyle \left\lfloor x/a \right\rfloor = m\). Then \(\displaystyle m \leq x/a < m + 1\), and

\(\displaystyle am \leq x < a(m + 1). \tag{2} \)

Let \(\displaystyle b= 1\). Then

\(\displaystyle \left\lfloor \sqrt[a]{1 + x} \right\rfloor = 1 + \left\lfloor x/a \right\rfloor . \tag{3} \)

From (3) it follows that \(\displaystyle \left\lfloor \sqrt[a]{1 + x} \right\rfloor = 1 + m\), or \(\displaystyle 1 + m \leq \sqrt[a]{1 + x} < 2 + m\), or

\(\displaystyle (1 + m)^{a} - 1 \leq x < (2 + m)^{a} - 1, \tag{4} \)

where we have used the obvious fact that \(\displaystyle m \geq - 1\). Indeed, the number \(\displaystyle \sqrt[a]{1 + x}\) and therefore the number \(\displaystyle \left\lfloor \sqrt[a]{1 + x} \right\rfloor = 1 + m\) is not negative. By Bernoulli's inequality, the following two inequalities — which compare the inequalities (2) and (4) — hold:

$$\begin{align*} am & \leq (1 + m)^{a} - 1, \tag{5} \\ a(m + 1) & \leq \left( 1 + (m + 1) \right)^{a} - 1, \tag{6} \end{align*}$$

where in (5), equality holds if and only if \(\displaystyle m = 0\), and in (6), equality holds if and only if \(\displaystyle m = -1\). From (2), (4)–(6), it follows that

\(\displaystyle (m + 1)^{a} - 1 \leq x < a(m + 1). \tag{7} \)

From (7), it follows that \(\displaystyle (m + 1)^{a} - 1 < a(m + 1)\). Therefore

\(\displaystyle (m + 1)^{a}\ \leq \ a(m + 1). \tag{8} \)

From (8), it follows that \(\displaystyle m + 1 \leq a^{\frac{1}{a - 1}}\). If \(\displaystyle a > 2\) then \(\displaystyle m + 1 < a^{\frac{1}{a - 1}} < 2\), because \(\displaystyle 2^{a - 1} > a\) for \(\displaystyle a > 2\) (one can prove this by mathematical induction) and \(\displaystyle 1 < a^{\frac{1}{a - 1}} < 2\) is not an integer. Therefore, if \(\displaystyle a > 2\) then \(\displaystyle m = 0\) or \(\displaystyle m = - 1\). If \(\displaystyle a = 2\) then \(\displaystyle m = - 1\), \(\displaystyle m = 0\) or \(\displaystyle m = 1\). From (7), it follows that the equality (3) holds true only for the values \(\displaystyle - 1\ \leq x < \ 0\ (m = - 1)\), \(\displaystyle 0 \leq x < a\ (m = 0)\) if \(\displaystyle a > 2\), and \(\displaystyle - 1 \leq x < 0\ (m = - 1)\), \(\displaystyle 0 \leq x < 2\ (m = 0)\) and \(\displaystyle 3 \leq x < 4\ (m = 1)\) if \(\displaystyle a = 2\).

We will now prove that for these values of \(\displaystyle x\), the equality (1) is true for all positive integers \(\displaystyle b\). From (7), it follows that \(\displaystyle b^{a} + (m + 1)^{a} - 1 \leq b^{a} + x < b^{a} + a(m + 1)\) and

\(\displaystyle 1 + \frac{(m + 1)^{a} - 1}{b^{a}} \leq 1 + \frac{x}{b^{a}} < 1 + \frac{a(m + 1)}{b^{a}}. \tag{9} \)

By Bernoulli's inequality

\(\displaystyle 1 + \frac{a(m + 1)}{b^{a}} \leq \left( 1 + \frac{m + 1}{b^{a}} \right)^{a}, \tag{10} \)

where equality occurs if and only if \(\displaystyle m = - 1\). It is easy to check that for \(\displaystyle m = 1\) (if \(\displaystyle a = 2\)), \(\displaystyle m = 0\) and \(\displaystyle m = - 1\), the following inequality holds true:

\(\displaystyle \left( 1 + \frac{m}{b^{a}} \right)^{a} \leq 1 + \frac{(m + 1)^{a} - 1}{b^{a}}. \tag{11} \)

From (9)–(11), it follows that

\(\displaystyle \left( 1 + \frac{m}{b^{a}} \right)^{a} \leq 1 + \frac{x}{b^{a}} < \left( 1 + \frac{m + 1}{b^{a}} \right)^{a}. \tag{12} \)

From (12), it follows that

$$\begin{align*} b^{a} + m & \leq \sqrt[a]{b^{a} + x} \cdot b^{a - 1} < b^{a} + m + 1, \\ b^{a} + \left\lfloor x/a \right\rfloor & \leq \sqrt[a]{b^{a} + x} \cdot b^{a - 1} < b^{a} + \left\lfloor x/a \right\rfloor + 1. \end{align*}$$

Consequently, equality (1) holds.