For what values of $k$ is there a perfect $x^p$ in {$n, n+1, n+2, ... ,kn$}? I was thinking about this the other day, and I can't seem to find the answer.
It is a fairly simple proof by induction to show that for all $n ∈ ℕ$, there is a perfect square in ${n, n+1, n+2, ... , 2n}$.
I am trying to generalize this question. For what values of $k$ is there always perfect cube in ${{n, n+1, n+2,..., kn}}$?
More generally, for what values of $k$ is there always a perfect $x^p$ in ${n, n+1, n+2, ... , kn}$? I'm not sure how to approach a proof of this. Maybe it's way above me?
Thanks.
 A: For all $k\ge2$ and for all $n\ge1$, if there is a perfect $p$-th power in $\{n+1,\cdots, kn\}$, then the same $p$-th power is in $\{n+1,\cdots, kn+k\}$. Therefore, the hard inductive steps are when you have the hypothesis that there is a perfect $p$-th power in $\{n,n+1,\cdots,kn\}$, but not in $\{n+1,\cdots,kn\}$: in other words, you only need to find an integer $k_p$ such that for all $n$ there is a perfect $p$-th power in $\{n^p+1,\cdots, k_p(n^p+1)\}$
In other words, you want $k_p$ such that, for all $n\ge1$, $(n+1)^p\le k_p(n^p+1)$. Or, equivalently, $$k_p=\left\lceil\sup_{n\ge1}\frac{(n+1)^p}{n^p+1}\right\rceil$$
Now, call $g(x)=\frac{(x+1)^p}{x^p+1}$. We have $g'(x)=\frac{p(x+1)^{p-1}(1-x^{p-1})}{(x^p+1)^2}$ and therefore $\sup_{n\ge 1}\frac{(n+1)^p}{n^p+1}$ is actually for $n=1$. Therefore $k_p=2^{p-1}$ (but of course any value of $k$ larger than $k_p$ will do).

Added: Let's straighten the exposition. Suppose that $k$ is a natural number such that $k\ge\frac{(n+1)^p}{n^p+1}$ for all $n\in\Bbb N\setminus \{0\}$. I claim that for all $n\in\Bbb N\setminus\{0\}$ there is an integer $x$ such that $n\le x^p\le kn$. In fact, by induction:

*

*$1\le 1^p\le k$;

*if there is some $x$ such that $n\le x^p\le kn$, then two cases: either $n+1\le x^p\le kn$, in which case $n+1\le x^p\le kn+k$, or $n=x^p$. But if $n=x^p$, then, by hypothesis on $k$, $\frac{(x+1)^p}{x^p+1}\le k$ and therefore $(x+1)^p\le k(x^p+1)=k(n+1)$. The inequality $n+1\le (x+1)^p$ is just $n=x^p<(x+1)^p$, completing the proof that $n+1\le (x+1)^p\le k(n+1)$ for this case. Either way, there is a perfect power in $\{n+1,\cdots, k(n+1)\}$.

Now, the question is: is there such a $k$? As a remark, the condition is necessary, because if there is some $n$ such that $\frac{(n+1)^p}{n^p+1}>k$, then $n^p<n^p+1$ and $(n+1)^p>k(n^p+1)$, and therefore there is no perfect power in $\{n^p+1, n^p+2,\cdots, k(n^p+1)\}$.
Fortunately, there are such $k$-s: the function $\frac{(n+1)^p}{n^p+1}$ is decreasing for $n\ge1$, and therefore any $k\ge2^{p-1}$ will do.
A: For $n = 1$, since $1$ is a $p$'th power, any $k$ will do. For $n = 2$, the next largest $p$'th power if $2^p$, so
$$kn \ge 2^p \implies k \ge 2^{p-1} \tag{1}\label{eq1A}$$
I will show the minimum allowed value of $k = 2^{p-1}$ always works. For any $2 \le n \le 2^p$, this value of $k$ works. Also, if $n = m^p$ for any integer $m$, then this $k$ also works. Next, consider
$$m^p \lt n \lt (m+1)^p \tag{2}\label{eq2A}$$
for an integer $m \ge 2$.
For $p = 1$, we have $k = 1$, with this working since each value is it's own first power. For $p = 2$, we have $k = 2$, which you've stated you can prove using induction. Thus, consider $p \ge 3$. Using \eqref{eq2A}, the Binomial theorem expansion with $x \gt 0$ giving $(1 + x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \ldots + x^p \gt 1 + px$, and $m \ge 2 \implies m + m \ge m + 2 \implies 2m - 1 \ge m + 1$, we then get
$$\begin{equation}\begin{aligned}
\left(2^{p-1}\right)n & \gt \left(2^{p-1}\right)m^p \\
& = \left(\frac{1}{2}\right)\left(2^{p}\right)m^p\left(\frac{(m+1)^p}{(m+1)^p}\right) \\
& = \left(\frac{1}{2}\right)\left(\frac{2m}{m+1}\right)^p(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(1 + \frac{m - 1}{m+1}\right)^p(m+1)^p \\
& \gt \left(\frac{1}{2}\right)\left(1 + \frac{p(m - 1)}{m+1}\right)(m+1)^p \\
& \ge \left(\frac{1}{2}\right)\left(1 + \frac{3(m - 1)}{m+1}\right)(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(\frac{4m - 2}{m+1}\right)(m+1)^p \\
& = \left(\frac{2m - 1}{m+1}\right)(m+1)^p \\
& \ge (m+1)^p
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, $(m+1)^p \in [n,kn]$, confirming $k = 2^{p-1}$ (as well as any larger $k$) will always work.
