(i) Prove that the product of three consecutive odd numbers is divisible by 3.
(ii) Prove that $p= 3$ is the only prime such that $p$, $p + 2$, and $p + 4$ are all prime.
(i) For any integer $n$ we can use the Division Algorithm to write it as
$$ n = 3m + r$$
for some integer $m$, and $0 \le r < 3$.
If $n$ is odd, the product of three consecutive odd numbers is
$$ (3m + r)(3m + r + 2)(3m + r + 4) $$
There are 3 cases to consider, $r=0, r=1, r=2$.
- $r=0$, the product is $(3m)(3m + 2)(3m + 4)$ which is divisible by 3.
- $r=1$, the product is $(3m + 1)(3m + 3)(3m + 5) =3(3m + 1)(m + 1)(3m + 5) $ which is divisible by 3.
- $r=3$, the product is $(3m + 2)(3m + 4)(3m + 6) =3(3m + 2)(3m + 4)(m + 2) $ which is divisible by 3.
In all cases, the product $n(n+2)(n+4)$ is divisible by 3.
So the product of 3 consecutive odd numbers is divisible by 3.
(In fact, the result holds for 3 consecutive even numbers).
(b) Lets split the primes into three cases
- primes greater than 3, $p>3$.
- prime 3, $p=3$
- prime 2, $p=2$
Let's consider each case in turn.
The first case, all primes $p>3$, are odd, and we have shown that the product $p(p+2)(p+4)$ is divisible by 3. Since $p$ is not divisible by 3, it must be that one of $(p+2)$ and $(p+4)$ is divisible by 3, and therefore not prime.
For the second case, we only consider the prime $p=3$. Here $p+2=5$ and $p+4=7$, all of which are prime.
For the third case, we only consider the prime $p=2$. Here $p+2=4$ and $p+4=6$, both of which are not prime.
We have shown that 3 is the only prime $p$ such that $p$, $p+2$ and $p+4$ are all prime.