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The problem is

Prove that for any positive integers $m$ and $n$, there exist a set of $n$ consecutive positive integers each of which is divisible by a number of the form $a^m$, where $a$ is some integer in $\Bbb{N}$.

When I first saw this question, it seemed quite trivial to start with $a=1$. I don't think that there can be any other possibility of $a$. Kindly help me if I'm going wrong.

Any hint would be appreciated. If possible, a hint/solution involving basic theorems of integers (preferably not modular arithmetic) would be highly helpful.

Blue
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Aarush Mishra
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    Presumably the problem meant to exclude $a=1$. Hint: consider divisibility by $p_1^m, \dots,p_n^m$ where the $p_j$ are distinct primes. And I suggest embracing modular arithmetic—trying to avoid it is just putting obstacles in your own way. – Greg Martin May 15 '22 at 18:57
  • @GregMartin Why did you consider exactly n distinct primes? – Aarush Mishra May 15 '22 at 19:04
  • @GregMartin I completely agree that modular arithmetic might be the way to go. But this problem was included as a part of arithmetic of integers chapter of the textbook 'Challenge and Thrill of Pre-College Mathematics', so I felt going with the essence of the chapter, where modular arithmetic isn't introduced :) – Aarush Mishra May 15 '22 at 19:11

2 Answers2

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I've seen this problem appear in exercise lists when introducing Chinese Remainder Theorem (in my case for math olympiads).

The solution is usually to take a positive integer $k$ that is congruent to $-i \bmod p_i^m$ where $p_i$ is the $i$'th prime. If we do this we can note in the consecutive integers $k+1,\dots,k+n$ the $i$'th integer is divisible by $p_i^m$.

Update: I think this result is very closely related to chinese remainder theorem. Note that for large vlaues of $n$ you are going to require arbitrarily many different primes to divide the different terms, I don't believe there is going to be a nice shortcut that allows you to ignore CRT like arguments.

Asinomás
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  • Is there any other way than CRT to approach this problem? – Aarush Mishra May 15 '22 at 22:05
  • Sorry I don't know. – Asinomás May 15 '22 at 22:08
  • This question is related: https://math.stackexchange.com/questions/160547/consecutive-non-square-free-numbers#:~:text=For%20the%20info%2C%20the%20first,and%20(44%2C45), also this one: https://mathoverflow.net/questions/151210/runs-of-consecutive-non-squarefree-integers – Asinomás May 15 '22 at 22:10
  • @AarushMishra I have been thinking about the problem and expanded my solution, what do you think? – Asinomás May 16 '22 at 14:41
  • Yep, now I have started reading the basics of modular arithmetic. I am still on the basic properties though. Would surely give a try to this problem again after understanding CRT. Thanks for your help – Aarush Mishra May 16 '22 at 18:20
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 09 '23 at 22:33
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We need to show that $\exists a_i$ s.t. $a_i^m | x_i$ where $x_1, x_2 \ldots, x_n$ are $n$ consecutive natural numbers and $a_i \neq 1$, for $1\le i\le n$. Thus, $a_i$ is divisible by at least one prime, let $p_i$ be the smallest prime dividing $a_i$. Notice that $p_i^m|x_i$. So, we will concern ourselves with $x_i$ divisible by $m$th power of a prime.

Define: $$p = \prod_{i = 1}^n p_i^m$$ Given any integer $m$, we prove the claim by inducting on $n$. (The proof does not specifically use $m$, hence there is no need to induct on it.)

For $n = 1$, the problem is trivial: take $x_1 = 2^m$. Suppose the claim is true for $n = k$, i.e. $\exists$ prime $p_1,\ldots p_k$ s.t. $p_i^m | x_i$ for $1\le i\le k$, where $x_1< \ldots< x_k$ are consecutive positive integers.
Let $y_i = x_i + zp$, and $y_{k+1} = y_k + 1$. We will find an algorithm for finding a non-negative integer $z$ s.t. $y_{k+1}$ is a multiple of $m$th power of a prime.
Let $q$ be the smallest prime larger than $x_k$. Singe $\gcd(p,q^m) = 1$, by Bezout's Lemma we conclude that $\exists u,v \in \Bbb Z$ s.t : $$up + vq^m = 1$$WLOG assume $u\le 0$, then we have $$u(-1-x_i)p \equiv -1 - x_i \pmod{q^m}$$Setting $z = u(-1-x_i)\ge 0$ we observe that: $$\begin{aligned}y_{n+1} &= x_n + u(-1-a_i)p + 1\\ &\equiv x_n -x_n - 1+1 \pmod{q^m}\\ &\equiv 0 \pmod{q^m}\end{aligned}$$ Thus, $q^m|y_{n+1}$ which completes our proof.

D S
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  • Although CRT is most suitable for this question, the OP had asked for an answer without it. Also, we can avoid the mod by some elementary application of division lemma – D S Sep 08 '23 at 12:25
  • This answer is essentially the same as Asinomás's answer, except instead of invoking CRT by name, you have repeated inline a proof of Easy CRT to solve $, x\equiv 0\pmod{p},, x\equiv -1-a_i\pmod{q^m}.,$ But this is no simpler than the general proof of CRT, so it is better to simply invoke CRT and let the reader learn the general proof - whose idea is obfuscated when inlined in the above special case; i.e. don't work against abstraction when it is no more difficult than the special case at hand. – Bill Dubuque Sep 08 '23 at 15:40
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Sep 09 '23 at 22:33