Find the set of natural numbers $n \in \mathbb{N}$ for which $\lfloor\sqrt{n+1}\rfloor \neq\lfloor\sqrt{n}\rfloor$

Question

I have to find the set of natural numbers $n \in \mathbb{N}$ for which $\lfloor\sqrt{n+1}\rfloor \neq\lfloor\sqrt{n}\rfloor$.

I have tried writing the formal definition of the floor function and tried squaring the inequalities. But that doesn't seem to get me nowhere

Answer 1

Hint:

With problems like this, it's usually a good idea to check some small numbers $n$. In particular, you could try all numbers from, say, $1$ to $10$.

Out of those numbers, you have $\color{red}{1},\color{red}{2},\color{green}{3},\color{red}{4},\color{red}{5},\color{red}{6},\color{red}{7},\color{green}{8},\color{red}{9},\color{red}{10}$ where the red color means the number does not match your criterion, and the green color means it does.

If you move on, you will see that the numbers up to $50$ that match are $3,8,15,24,35,48$. Can you see a pattern about them?