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.

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.

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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 \(q(n)\leq q(n+1)\). – Prasun Biswas · 3 months, 1 week ago

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– Abhishek Alva · 3 months, 1 week ago

+2Log in to reply

– Ayush Rai · 3 months, 1 week ago

Brilliant Solution!+1Log in to reply

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@Svatejas Shivakumar.Iam deleting the solution – Ayush Rai · 3 months, 1 week ago

OkLog in to reply

– Svatejas Shivakumar · 3 months, 1 week ago

I meant \([\dfrac x {\left([a] \right)}]\).Log in to reply

– Abhishek Alva · 3 months, 1 week ago

how to solve the problemLog in to reply