Define \(\displaystyle q(n) = \lfloor \frac{n}{\lfloor \sqrt{n} \rfloor}\rfloor\) for \(n=1,2,3,\ldots\). Find all \(n\) satisfying \(q(n) > q(n+1) \).

**Notation:** \( \lfloor \cdot \rfloor \) denotes the floor function.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe solutions are the natural numbers one less than a perfect square, i.e., the solution set is \(S=\{k^2-1\mid k\in\Bbb N\setminus\{1\}\}=\{3,8,15,\ldots\}\)

To prove this, you need to prove that \(q(n)\gt q(n+1)\) when \(n\in S\) and \(q(n)\leq q(n+1)\) otherwise.

For the first part, note that when \(n=k^2-1\), since the square root function is strictly increasing, we have \(\lfloor\sqrt n\rfloor=k-1\) which implies that \(q(n)=k+1\) whereas we have \(\lfloor\sqrt{n+1}\rfloor=k\) which implies that \(q(n+1)=k\) which shows that \(q(n)=k+1\gt k=q(n+1)\) when \(n\in S\).

Now, for the second part, when \(n\notin S\), we note that when \(k^2\leq n\lt (k+1)^2-1\), we have \(\lfloor\sqrt n\rfloor=\lfloor\sqrt{n+1}\rfloor=k\) which implies that \(q(n)=\lfloor\frac nk\rfloor\) and \(q(n+1)=\lfloor\frac{n+1}k\rfloor\). Now, since \(\frac nk\lt\frac {n+1}k\), applying the floor function on both sides of the inequality, we have \(\lfloor\frac nk\rfloor\leq\lfloor\frac{n+1}k\rfloor\) from which we conclude that \(q(n)\leq q(n+1)\).

Log in to reply

+2

Log in to reply

Brilliant Solution!+1

Log in to reply

Comment deleted Oct 15, 2016

Log in to reply

Comment deleted Oct 15, 2016

Log in to reply

Ok @Svatejas Shivakumar.Iam deleting the solution

Log in to reply

I meant \([\dfrac x {\left([a] \right)}]\).

Log in to reply

Log in to reply