How do I prove that $(a+1)(a+2)$ is divisible by $3$ given $3\nmid a$?

Question

So I'm pretty new to this and I was wondering how I prove that

prove that $(a+1)(a+2)$ is divisible by $3$, given that $a$ is an integer that is not divisible by $3$?

I know that if I set $a = 1$ I get

$$(1+1)(1+2) = 6$$

also that I'll get

$$a^2 + 3a + 2$$

and $6$ is divisible by $3$, but I'm not quite sure how to prove it...

Answer 1

Every integer that is not divisible by $3$ can be written either as $3k+1$ or $3k+2$ for some integer $k$.

In the first case $(a+1)(a+2)=(3k+2)(3k+3)=3(3k+2)(k+1)$, while in the second case $(a+1)(a+2)=(3k+3)(3k+4)=3(k+1)(3k+4)$.

In both cases you get $3$ multiplied by some integer.

Answer 2

Hint:

Use congruences. If $\;a\not\equiv 0\mod 3$, either $a\equiv -1$ or $a\equiv 1\mod 3$. What can you say of $(a+1)(a+2)$ in each case?

Answer 3

Every third integer is divisible by three. We know $3\nmid a$. Hence either $3\mid a+1$ xor $3\mid a+2$ since $a, a+1, a+2$ are three consecutive integers.