# There exist $n$ different integers in the interval $\big(k^n,(k+1)^n\big)$ whose product is a perfect $n$-th power.

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.

• 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. May 10, 2020 at 20:43
• Sorry but I already tried a lot of analytical stuff but I couldn't proceed. May 10, 2020 at 20:46
• Can you please explain the analytical stuff that you tried? May 10, 2020 at 20:48
• Can something be done from my progress? May 10, 2020 at 20:51
• 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 $K$? May 10, 2020 at 21:09

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 $$\dfrac{n^2-1}{n} on the LHS, so for sufficiently large $$k$$, all these numbers fall into the interval.

• I actually think the method is fine. You simply need a different estimate. Well done +1 May 11, 2020 at 7:17
• Thanks, that is corrected now. May 11, 2020 at 7:23
• Perhaps some modification will work for even $n$ as well, which is still pending. May 11, 2020 at 7:24
• 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. 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 $$k^n for all $$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.