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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
up vote 2 down vote accepted

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 $24$.

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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
@VictorFranciscoSalazarGarci No Problem. If you are satisfied with any of the answers you get for your question accept the answer so that others do not waste their time by posting again. – lsp Jan 9 '14 at 7:29

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