Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

Question

Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

Since $24 = 2^3 \cdot 3$ it follows that $(n - 1)n(n + 1)(n + 2)$ is divisible by $3$ since it contains the product of three consecutive numbers.

If $n$ is odd then $n-1$ and $n+1$ are even and other one of them is divisible by $2$ and the other by $4$, thus the product is divisible by $2^3$.

If $n$ is even then $n$ and $n+2$ are both even and similar reasoning follows that other one is divisible by $2$ and the other by $4$.

Is this correct? For the odd case if I have that $n=1$ then $n-1=0$ and $n+1=2$, but neither one is divisible by $4$ here? Or does $0$ count as being divisible by $4$?

So in other words, you want an elementary proof that $\binom{n+2}{4}$ is an integer for all integers $n$? — Daniel Schepler, Nov 10 '20 at 21:12
What Daniel said. For any $k$ we have that the "descending factorial" $n(n-1)(n-2)\cdots (n-k+1)$ is divisible by $k!$ for all $n$. — Jyrki Lahtonen, Nov 10 '20 at 21:16
Yes, that a standard method. See the linked dupes for this and other well-known methods. — Bill Dubuque, Nov 10 '20 at 22:25

David G. Stork · Answer 1 · 2020-11-10T21:54:15.017

1

You have four consecutive terms, so exactly one of them is divisible by $4$, another is divisible by $2$, and at least one other is divisible by $3$.

QED.

edited Nov 10 '20 at 21:54

answered Nov 10 '20 at 20:57

David G. Stork

29,774

score 0 · Answer 2 · answered Nov 10 '20 at 20:53

hint

Other approach, by induction.

For $ n=1 $, it is true.

Let $ n\ge 1 $ such that $$24|\color{red}{(n-1)n(n+1)(n+2)}.$$

Replacing $ n $, by $ n+1$,

$$n(n+1)(n+2)(n+3)=\color{red}{(n-1)n(n+1)(n+2)}+4n(n+1)(n+2)$$

So, you just need to prove that $$6|n(n+1)(n+2)$$

Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.

2 Answers2