# prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$

## prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$

Let $P(n)$ be the proposition we want to prov, ie: $P(n):=24 \mid(n)(n+1)(n+2)(n+3)$

For $P(1)$ we have: $24 \mid(1)(1+1)(1+2)(1+3)\implies6 \mid(1)(2)(3)(4)\implies24 \mid24$, so $P(1)$is true.

For $P(2)$ we have: $24 \mid(2)(2+1)(2+2)(2+3)\implies6 \mid(2)(3)(4)(5)\implies24 \mid120$, so $P(1)$is true

Inductive Hypothesis: Let $n=k$ and we assume that $P(k):=24\mid k(k+1)(k+2)(k+3)$ is true.

Inductive Step: $$(k+1)(k+2)(k+3)(k+4)$$ $$k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$ Using the assumption of $P(k) \implies \exists a\in \mathbb Z$, such that, $(k+1)(k+2)(k+3)=24\cdot a$

so: $$=24\cdot a +4(k+1)(k+2)(k+3)$$

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For the obvious induction to work, we need the fact that the product of $3$ consecutive integers is divisible by $6$. If we are going to do that by induction, we need the fact that the product of $2$ consecutive integers is divisible by $2$. But we can also give non-induction arguments for these facts, (and also for the fact about $4$ consecutives). One can also do a double induction and prove that the product of $m$ consecutives is divisible by $m!$. – André Nicolas Jan 9 '14 at 7:31

Continuing from where you left, we just need to prove that for $n=k+1$, $P(n)$ is an integer multiple of $24$.
$$P(k+1) = (k+1)(k+2)(k+3)(k+4)$$ $$P(k+1) = k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$ $1st$ term on right hand side is $P(k)$ which is an integer multiple of $24$ from your inductive hypothesis.
$2nd$ term on the right hand side has a product of $3$ consecutive integers and hence divisible by $6$. So on a whole divisible by $4*6=24$.
On a whole the right hand side is divisible by $24$. Hence $P(k+1)$ is an integral multiple of $24$.
thank you for the last comment, i didn´t had in mind that $6\mid n(n+1)(n+2)$ – Victor Francisco Salazar Garci Jan 9 '14 at 7:27