Help with mathematical induction!

Question

My teacher is asking me to: Prove that 2 is a factor of (n+1)(n+2) for all positive integers n. I need to solve this problem through mathematical induction but I am completely lost! Please help me and all other confused math students!! thanks!! And don't tell my teacher lol.

Answer 1

Hint:

$$(n+2)(n+3)=(n+1)(n+2)+2(n+1).$$

Answer 2

To show that 2 is a factor of $(n+1)(n+2)$ means to show that $(n+1)(n+2)$ is even, or there exists $k \in \mathbb{Z}$ such that $(n+1)(n+2) = 2k$. This is your induction statement.

$$P(n):(n+1)(n+2)=2k, \ k\in \mathbb{Z}$$

You must show your base case $(n=1)$ such that:

$$P(1): (1+1)(1+2)=2k$$

This is obviously true for $k=3$. Now, we assume the statement $P(n)$ to be true and must show $P(n+1)$ to be true, which states

$$P(n+1):((n+1)+1)((n+1)+2)=2k, \ k\in\mathbb{Z}$$ $$P(n+1):(n+2)(n+3)=2k, \ k\in\mathbb{Z}$$

Since $(n+3)-(n+2) = 1$, the two terms are one digit apart, implying opposite parity. Thus, one must be even and the other odd. Without loss of generality, assume $n+3$ to be even, such that there exists $j\in\mathbb{Z}$ with $n+3=2j$. By substitution,

$$(n+2)(n+3) = 2j(n+2)$$

Since $\mathbb{Z}$ is closed under multiplication, let $k\in\mathbb{Z}$ with $k=j(n+2)$ such that $$(n+2)(n+3) = 2k$$ as was desired.

EDIT: Using Yves' hint would make this cleaner and you would not have to invoke the opposite parity idea, which may not be foundational at this level.