# Number Theory Quicke

• Jun 21st 2010, 01:37 PM
Chris11
Number Theory Quicke
This is a chalenge question. It has been approved by Mr.Fantastic

Let $Q(n)=\lfloor\frac{n}{\lfloor{n}^{1/2}\rfloor}\rfloor$, where n is a natural number. Find the natural numbers a for which Q(a)>Q(a+1)

Moderator edit: Approved Challenge question.
• Jun 21st 2010, 02:17 PM
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Spoiler:

Let $k\in\mathbb{Z},k>0$ and consider the sets $S=\{k^2,k^2+1,\dots, (k+1)^2-2\}$ and $S'=\{k^2,k^2+1,\dots, (k+1)^2-1\}$. It is clear that $\forall a\in S':\lfloor\sqrt{a} \rfloor=k$. Therefore $\forall a\in S: Q(a)=\left\lfloor\frac{a}{\lfloor\sqrt{a}\rfloor}\ right\rfloor = \left\lfloor\frac{a}{k}\right\rfloor \le Q(a+1) = \left\lfloor\frac{a+1}{k}\right\rfloor$.

So the only candidates are of the form $a=k^2-1$ for some $k\in\mathbb{Z},k>1$. We know that $\lfloor\sqrt{a} \rfloor=k-1$ and $\lfloor\sqrt{a+1} \rfloor=k$. So we compare

$Q(a)=\left\lfloor\frac{a}{k-1}\right\rfloor=\left\lfloor\frac{k^2-1}{k-1}\right\rfloor=\left\lfloor k+1 \right\rfloor=k+1$

with

$Q(a+1)=\left\lfloor\frac{a+1}{k}\right\rfloor=\lef t\lfloor\frac{k^2}{k}\right\rfloor=k$

Therefore the complete solution set is $\{a|a=k^2-1, k\in\mathbb{Z},k>1\}$.
• Jun 21st 2010, 03:04 PM
Chris11
Good job!