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For any arbitrary integer $a$, prove that $2 \mid a \left( a+1 \right)$ and $3 \mid a \left( a+1 \right) \left( a+2 \right)$

I know that if $2 \mid a \left( a+1 \right)$ then $a \left( a+1 \right)=2q$; $q \in \mathbb{Z}$.

And if $3 \mid a \left( a+1 \right) \left( a+2 \right)$, then $a \left( a+1 \right) \left( a+2 \right) = 3r$; $r \in \mathbb{Z}$.

But how do I prove that both are true?

pabhp
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  • Also: https://math.stackexchange.com/q/497859/42969. – Martin R Aug 24 '19 at 14:06
  • Normally I prefer to use the close-as-duplicate option over close-for-lack-of-context, but here the first proposed duplicate uses the "base case" that here is the first part of the problem, and the second linked "duplicate" only addresses the second part. Despite the abundance of rapidly posted Answers, this Question does not meet my (rather de minimus) standard for context, so I'm voting to close. Perhaps the OP could show their attempt? – hardmath Aug 24 '19 at 17:00

6 Answers6

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If $a$ is odd then $a+1$ is even or if $a$ is even then $a+1$ is odd. In both cases the product $a(a+1)$ is even so divisible by $2$. In the sequence $a$, $a+1$, $a+2$ there is always a multiple of $3$ because there is a multiple of $3$ every $3$ number. So: $a(a+1)(a+2)$ is divisible by $3$.

Matteo
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The product of two successive natural numbers are always divisible by $2$ and the product of three successive natural numbers is always divisible by $3$

Dr. Sonnhard Graubner
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Or saying it different, one of the two numbers $a,a+1$ is divisible by 2 and so $a(a+1)$ is divisible by 2.

Similarly, one of the three numbers $a,a+1,a+2$ is divisible by 3 and so $a(a+1)(a+2)$ is divisible by 3.

Wuestenfux
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$$a(a+1)(a+2)$$ represents the number of strings of length $3$ using $a+2$ different letters w/o repetition.
Dividing that by $3!$ gives the number of ways of choosing $3$ letters from $a+2$ ignoring order. This has to be an integer.

AgentS
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Any proof will somewhere hide an induction or inductive definition.

Here $n(n+1)-(n-1)n=2n$ is even and $n(n+1)(n+2)-(n-1)n(n+1)=3n(n+1)$ is divisible by $3$ and this is sufficient to provide the inductive step.

Mark Bennet
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I think OP understands each statement is separately true, but is uncertain if that implies they are simultaneously true for any $a,(a+1),(a+2)$, or perhaps how to prove that.

I would explain it this way: Since you know that $2\mid a(a+1)$ for any natural number $a$, you can be certain that $2\mid a(a+1)\cdot m$ for any natural numbers $\{a,m\}$, and in particular when $m=(a+2)$. Thus you know that $2\mid a(a+1)(a+2)$ for any $a$. You also already know that $3\mid a(a+1)(a+2)$ for any $a$, so the two statements are simultaneously true for any $a$.

Keith Backman
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