China Mathematical Competition (Extra Test)

Proof Define set $A=\{m\sqrt{k+1}\mid m\in {\mathbb{N}}^{\ast },k\in P\}$, where ${\mathbb{N}}^{\ast }$ denotes the set of all positive integers. It is easy to check that for any $k_1,k_2\in P,k_1\ne k_2$, $\frac{\sqrt{k_1+1}}{\sqrt{k_2+1}}$ is an irrational number. Therefore, for any $k_1,k_2\in P$ and positive integers $m_1,m_2$, $m_1\sqrt{k_1+1}=m_2\sqrt{k_2+1}$ implies $m_1=m_2$ and $k_1=k_2$.

Note that $A$ is an infinite set. We arrange the elements in $A$ in ascending order. Then we have an infinite sequence. For any positive integer $n$, suppose the $n$th term of the sequence is $m\sqrt{k+1}$. Any term before the $n$th can be written as $m_i\sqrt{i+1}$, and $$m_i\sqrt{i+1}\le m\sqrt{k+1}.$$ Or equivalently, $m_i\le m\frac{\sqrt{k+1}}{\sqrt{i+1}}$. It is easy to see that there are $\lfloor m\frac{\sqrt{k+1}}{\sqrt{i+1}}\rfloor$ such $m_i$ for $i=1,2,3,4,5$. Therefore, $$n=\sum _{i=1}^5\lfloor m\frac{\sqrt{k+1}}{\sqrt{i+1}}\rfloor =f(m,k).$$