1. ## Proof by induction

Can anyone help me get started with this proof using induction? Thank a lot everyone.

Prove that the product of any three consecutive positive integers is divisible by 6.

2. Originally Posted by jzellt
Can anyone help me get started with this proof using induction? Thank a lot everyone.

Prove that the product of any three consecutive positive integers is divisible by 6.

Base case $1 \times 2 \times 3 = 6$ so the proposition is true for the consecutive numbers $1,\ 2,\ 3$.

Suppose the proposition is true for three consecutive numbers starting from $k$.

Then $6$ divides $[k \times (k+1) \times (k+2)]$, now consider the product of three consecutive integers starting for $k+1$:

$(k+1) \times (k+2) \times (k+3) = k \times (k+1) \times (k+2) + 3 \times (k+1) \times (k+2)$

but $6$ divides $[k \times (k+1) \times (k+2)]$ and as one of $(k+1)$ and $(k+2)$ must be even $6$ divides $[3 \times (k+1) \times (k+2)]$, and so we conclude that $6$ divides $[(k+1) \times (k+2) \times (k+3)]$.

Now we have proven the base case and that if the proposition is true for three consecutive integers starting from $k$ it is true for three consecutive integers starting from $k+1$, and so by induction it is true for every set of
three consective positive integers.

RonL

3. Hello, jzellt!

A slight variation of CaptainBlack's proof.

Prove that the product of any three consecutive positive integers is divisible by 6.
We want to prove

. . $\text{For all }n\in I^+,\;n(n+1)(n+2) \:=\:6a\;\;\text{for some integer }a.$

Verify $S(1)\!:\;\;1\cdot2\cdot3\:=\:6$. . . True!

Assume $S(k)\!:\;\;k(k+1)(k+2) \;=\;6a$ for some integer $a$

Add $3(k+1)(k+2)$ to both sides:
. . $k(k+1)(k+2) \:{\color{blue}+\:3(k+1)(k+2)} \;=\;6a \:{\color{blue}+ \:3(k+1)(k+2)}$

The left side is: . $k(k+1)(k+2)+ 3(k+1)(k+2)$
. . Factor: . $(k+1)(k+2)(k+3)\;\;{\color{red}\text{[L]}}$

The right side is: . $3[2a + (k+1)(k+2)]$
. . $(k+1)(k+2)$ is the product of two consecutive integers.
. . Hence, it is divisible by 2:. $(k+1)(k+2) \:=\:2b$
The right side is:. $3(2a+2b) \:=\:6(a+b)\;\;{\color{red}\text{[R]}}$

From [L] and [R],we have: . $(k+1)(k+2)(k+3)\:=\:6(a+b)$

We have established $S(k+1)$..The inductive proof is complete.