-1

Prove that for any positive integers $m$ and $n$, there exists set on $n$ consecutive positive integers each of which is divisible by a number of the form $a^m$ where $a$ is any integer.

To be honest, I'm not even sure what the question is asking. I mean should I write $a=1$ and move on, or is there an actual non trivial solution.

Any hint/ help is appreciated!

amWhy
  • 210,739
Aditya_math
  • 1,887

1 Answers1

1

It's a Chinese Remainder Theorem problem. Take the system

$$x\equiv 0 \pmod{2^m}$$

$$x+1 \equiv 0\pmod{3^m}$$

etc.

$$x+n-1 \equiv 0 \pmod{p_k^m}$$

where $p_k$ is the $k$th prime.

B. Goddard
  • 33,728
  • So is it essentially asking for a proof of CRT? – Aditya_math Nov 06 '20 at 15:19
  • It's not obvious to me that the moduli have to occur in the numerical order as you present them. Is it not possible to construct $k!$ such solutions by taking the $p_k^m$ in each possible order? For $k=3$ e.g., can't we take $2^m$ second or third rather than first (using $3^m$ or $5^m$ as the first modulus)? – Keith Backman Nov 06 '20 at 15:46
  • 1
    @KeithBackman Nobody said that you have to take the moduli in numerical order. The guy asked for a hint, I gave him a hint. It's a homework problem posted with no effort shown, so a hint is all he gets. – B. Goddard Nov 06 '20 at 16:24
  • @B. Goddard Right. I just wanted to be certain I wasn't missing something subtle with regard to the specifics of your hint. – Keith Backman Nov 06 '20 at 16:59
  • @B.Goddard i just read your comment and appreciate your answer but just wanted to let you know this wasn't a homework problem, as a matter of fact my school doesn't even teach number theory. This was a problem in a book i am solving. In this chapter, CRT had not been introduced yet so i was not sure what to do which is what i mentioned in the OP. – Aditya_math Nov 11 '20 at 21:19
  • 1
    The point is that it's a "textbook" problem. If CRT hasn't been introduced yet, then the purpose of the problem is to lead you to the CRT. – B. Goddard Nov 11 '20 at 22:25
  • It suffices for the moduli to be powers of (pairwise) coprimes, which can be generated by the method used in Euclid's proof of infinitely many primes – Bill Dubuque Nov 13 '20 at 09:29