Given a positive integer $ n> 2 $. Prove that there exists a natural number $ K $ such that for all integers $ k \ge K $ on the open interval $ \big({{k} ^{n}}, \ {{(k + 1)} ^{n}}\big) $ there are $n$ different integers, the product of which is the $n$-th power of an integer.

Source Ukrainian TST 2011

Progress: Maybe one can choose the smallest prime divisor $q$ of $n$ and then one can choose all $\frac{n}{q}$ powers and among these powers, one can choose $n$ integers whose product is a $q$-th power In this way, we would just have to prove that Between $k^q$ and $(k+1)^q$ we have $n$ integers with their product being a $q$-th power.

  • 2
    $\begingroup$ Have you made any attempts to solve this problem? If so, you should put them in the body of your question so that we can be more helpful in answering. $\endgroup$ May 10, 2020 at 20:43
  • $\begingroup$ Sorry but I already tried a lot of analytical stuff but I couldn't proceed. $\endgroup$
    – Superguy
    May 10, 2020 at 20:46
  • $\begingroup$ Can you please explain the analytical stuff that you tried? $\endgroup$ May 10, 2020 at 20:48
  • $\begingroup$ Can something be done from my progress? $\endgroup$
    – Superguy
    May 10, 2020 at 20:51
  • 1
    $\begingroup$ The first thing I would try is $n=3$ and some small $K$, say $10$. Of course, it could be that the appropriate $K$ is huge, but let's try. Can we find $3$ numbers between $1000$ and $1331$ that multiply to form a cube? This would work if we could find numbers $p,q,r$ such that $p^2q,q^2r,$ and $r^2p$ are in range. Unfortunately, neither $10,11,12$ nor $9,11,12$ work. Can we fix that for larger $K$? $\endgroup$ May 10, 2020 at 21:09

2 Answers 2


From the intuition that the intervals between consecutive $n$th powers contain many $(n-1)th$ powers, we have the following:

If $n+1$ is odd, for the interval in $(k^{n+1},(k+1)^{n+1})$, we can take the following $n+1$ numbers: $x_1=a^n,x_2=a^{n-1}(a+1),\ldots,x_n=a(a+1)^{n-1},x_{n+1}=(a+1)^n$, where $a=\lceil k^{\frac{n+1}{n}} \rceil$. Clearly $a^n$ falls into the interval, and what needs to be checked is that $(a+1)^n<(k+1)^{n+1}$, and we know that $a+1 < k^{\frac{n+1}{n}}+2$. We can see from binomial expansion that for fixed $n$, the first terms cancel and the leading term has degree $n$ on the RHS and degree $\dfrac{n^2-1}{n}<n$ on the LHS, so for sufficiently large $k$, all these numbers fall into the interval.

  • $\begingroup$ I actually think the method is fine. You simply need a different estimate. Well done +1 $\endgroup$ May 11, 2020 at 7:17
  • $\begingroup$ Thanks, that is corrected now. $\endgroup$
    – Aravind
    May 11, 2020 at 7:23
  • $\begingroup$ Perhaps some modification will work for even $n$ as well, which is still pending. $\endgroup$
    – Aravind
    May 11, 2020 at 7:24
  • $\begingroup$ If $n=2\ell$, $\ell$ odd, may be you can similarly squeeze in a sequence of the form $$a^{2(\ell-1)},a^{2(\ell-2)}(a+1)^2,a^{2(\ell-3)}(a+1)^4,\ldots,(a+1)^{2(\ell-1)}?$$ Even if that works (recursively), we would still (only?) be left with the case when $n$ is a power of two. $\endgroup$ May 11, 2020 at 7:51

Okay so finally I think this is the solution $ $

I will just be showing for even $n$

First let $a=\lceil k^{\frac{n}{n-1}} \rceil$ now let $b=a+1$ and $c=a+2$ So setting $x_{i}=a^{n-i}b^{i-1}$ and $y_{i}=c^{n-i}b^{i-1}$ for $i\in [0,n-1]$

We get $k^n<x_{1}<\cdots x_{n-1}<y_{n-1}<\cdots y_{1}<( k^{\frac{n}{n-1}}+3)^{n-1}$ for all $k\geq 3^{n-1}$ Also note that $( k^{\frac{n}{n-1}}+3)^{n-1}<(k+1)^{n}$ for all $k\geq 3^{n-1}$ From here we just have to make cases and choose $n$ numbers out of these $2n-2$ numbers.

For example for the case $n=4m$ the sequence $(x_{1},x_{3},\cdots, x_{4m-1},y_{1},y_{3},\cdots y_{4m-1})$ works.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .