1

Looking at David M. Burton's Elementary Number Theory the last few days and I've found an exercise where you are to show:

  • $2$ divides $n(n+1)$
  • $6$ divides $n(n+1)(n+2)$
  • $24$ divides $n(n+1)(n+2)(n+3)$
  • $120$ divides $n(n+1)(n+2)(n+3)(n+4)$

but the book does not go on to investigate the general case.

I've done some thinking about it and for coprime integers it's well-known you can divide coprime product by each of the factors: $2 \times 3 \times 5$ obviously divides numbers of the form $n(n+1)(n+2)(n+3)(n+4)$ because there's always a number in there divisible by $2$, one divisible by $3$, and one divisible by $5$. Not so obvious for four. You notice that either $n+2$ and $n+4$ or $n+1$ and $n+3$ have both a power of $2$ and a power of $4$ in them. So all in all you have tha $2 \times 3 \times 4 \times 5$ divides $n(n+1)(n+2)(n+3)(n+4)$. You can assemble special cases piecemeal with relative ease, but I can't see how it is at all clear that it is always the case that $n!$ will always divide $n$ consecutive integers.

We take it for granted that a sequence of $n$ consecutive integers always contains at least one which is divisible by each number up to $n$.

I just can't quite get around whether I can justify then jumping to the conclusion that $n!$ divides $n(n+1)(n+2)...(n+(n-1))$.

RobPratt
  • 45,619
Prime Mover
  • 5,005
  • 1
    Good oberservation. Yes, this is true. The quotients are the binomial coefficients $$\binom{n+1}2,\binom{n+2}3,\dots,\binom{n+k-1}k,\dots$$ where $\binom ab$ is the number of subsets of size $b$ in a set of size $a.$ https://en.wikipedia.org/wiki/Binomial_coefficient#:~:text=In%20mathematics%2C%20the%20binomial%20coefficients,%E2%89%A5%200%20and%20is%20written – Thomas Andrews Sep 16 '22 at 19:03
  • Any $n$ consecutive numbers exhaust all the residue classes $\mod{n}$. – Anthony Sep 16 '22 at 19:12
  • 1
    Please search for answers before asking questions. – Bill Dubuque Sep 16 '22 at 19:14
  • That a sequence of $n$ consecutive integers contains a multiple of $n$ has an easy proof. But that alone does not suffice to deduce the sought result - for which see the linked dupes. – Bill Dubuque Sep 16 '22 at 19:34

0 Answers0