Find all natural numbers $n$ such that $\frac{\sqrt{n}}{2}+\frac{10}{\sqrt{n}}$ is a natural number.
Problem
Source: Kosovo Math Olympiad 2025, Grade 9, Problem 2
Tags: Kosovo, national olympiad, number theory
17.11.2024 20:17
For $n \in \mathbb{N}$, $n \neq 0$, if $\frac{\sqrt{n}}{2} + \frac{10}{\sqrt{n}} \in \mathbb{N}$, find all possible values of $n$. We know that $n$ is a natural number and the result is also a natural number. Now, we must search for the divisors of 10: $b | 10$, so $b \in \{1, 2, 5, 10\}$. Therefore, from this part, $n \in \{1, 4, 25, 100\}$. Trying each one: 1. For $n = 1$, $\frac{1}{2}$ is not a natural number, so the final result will not be natural either. Therefore, $n = 1$ is a contradiction. 2. $n = 4$ is true. 3. $n = 25$ is not true because dividing by 2 will not give a natural number. 4. $n = 100$ is true. Therefore, the true answers are $n \in \{4, 100\}$.
17.11.2024 20:19
is that really olympiad problem?
17.11.2024 20:49
Mhremath wrote: is that really olympiad problem? I mean, it's a Grade 9 problem. Higher grades (e.g., 12th) have actually quite nice problems, take a look
17.11.2024 21:28
18.11.2024 09:58
Let $n = k^2$ (this can be checked to be necessary), then $2k \mid k^2 + 20 \implies 2k \mid 80 \implies k \mid 40$. Now manually check to see which work.
19.11.2024 17:20
DensSv wrote: 2)$k=2 \Leftrightarrow 6\in \mathbb{N}$ which is true. 5)$k=10\Leftrightarrow 6\in\mathbb{N}$, which is true. Nice that when it is an integer, it is equal to six