jasperE3 wrote: Let $k\ge2$ be an integer. The function $f:\mathbb N\to\mathbb N$ is defined by $$f(n)=n+\left\lfloor\sqrt[k]{n+\sqrt[k]n}\right\rfloor.$$Determine the set of values taken by the function $f$. Let $a_m=m^k-m+1$, where $m$ is a positive integer. $a_m$ is an increasing sequence ($k\ge 2$) 1) $a_m> (m-1)^k$ and so $a_m+\sqrt[k]{a_m}>a_m+m-1=m^k$ and so $\sqrt[k]{a_m+\sqrt[k]{a_m}}>m$ 2) $a_m-1=m^k-m<m^k$ and so $a_m-1+\sqrt[k]{a_m-1}<a_m-1+m=m^k$ and so $\sqrt[k]{a_m-1+\sqrt[k]{a_m-1}}<m$ And so $\forall n\in[a_m,a_{m+1}-1]$ : $\sqrt[k]{n+\sqrt[k]{n}}\in(m,m+1)$ and so $\left\lfloor\sqrt[k]{n+\sqrt[k]n}\right\rfloor=m$ And so $f(n)=n+m$ $\forall n\in[a_m,a_{m+1}-1]$ And so $f([a_m,a_{m+1}-1])=[a_m+m,a_{m+1}-1+m]$ $=[m^k+1,(m+1)^k-1]$ Hence the answer $\boxed{f(\mathbb Z_{>0})=\mathbb Z_{>0}\setminus\{n^k\quad\forall n\in\mathbb Z_{>0}\}}$