3, 5, 7 are the only three consecutive odd natural numbers that are prime. Can someone please help me proof this theorem. So far I assumed towards contradiction that there exist three consecutive odd natural numbers that are primes P, P+2, and P+4. 

Theorem: 3, 5, 7 are the only three consecutive odd natural numbers that are prime.
All odd primes, except for $3$, are of the form $6n \pm 1$.

So, if two odd primes bigger than $3$ exists and differ by $2$,  they must be of the form

$6n-1, 6n+1$ for some integer $n$.

The number that are $2$ before and $2$ after will show up as

$6n-3, 6n-1, 6n+1, 6n+3$

If $n=1$, you get $3,5,7,9$ and the first three numbers are prime.

For any $n&gt;1$, only the middle two numbers might be prime.

A simpler proof.

Let $(x, x+2, x+4)$ be three consecutive odd numbers. 

Modulo 3, those three numbers will have to look like one of

(0, 2, 1)  or  (1, 0, 2)  or (2, 1, 0)

Hence at least one of them will be a multiple of three.

Since the number 3 is a prime number, we need to look at those cases where one of those three numbers is 3.

(-1, 1, 3)  or
(1, 3, 5)  or
(3, 5, 7)

So (3, 5, 7) is the only sequence of three consecutive odd prime numbers.
Here $nk$ represents $10n +k$


We start by saying ,$$$$leaving 2 all prime numbers have odd units place , $$$$possible combinations for 3 consecutive prime numbers will be.

$$n1,n3,n5$$
$$n3,n5,n7$$
$$n5,n7,n9$$
$$n7,n9,(n+1)1$$
$$n9,(n+1)1,(n+1)3$$
As all the numbers $\gt 5$ with units place 5 are divisible by 5, so by eliminating cases with n5 we are left with,

$$n7,n9,(n+1)1$$
$$n9,(n+1)1,(n+1)3$$


  Now if we see 3's table 
  $$9,39,69,\cdots,(3n)9$$
  Are divisible by 3,and
  $$21,51,81,\cdots,(3n+2)1$$
  Are divisible by 3,
  $$27,57,87,\cdots,(3k+2)7$$
  Are divisible by 3


So on our possible cases if $n=3k$


  $$(3k)7,(3k)9,(3k+1)1$$
  $$(3k)9,(3k+1)1,(3k+1)3$$
  As  $(3n)9$ is divisible by 3, $n\not= 3k$ for three consecutive prime numbers to exist.


For $n=3k +1$


  $$(3k+1)7,(3k+1)9,(3k+2)1$$
  $$(3k+1)9,(3k+2)1,(3k+2)3$$
  As (3k+2)1 is divisible by 3,
  $$n\not= 3k+1$$ for three consecutive prime no to exist


For $n=3k+2$


  $$(3k+2)7,(3k+2)9,(3(k+1))1$$
  $$(3k+2)9,(3(k+1))1,(3(k+1))3$$
  As (3k+2)7 is divisible by 3 ,
  $$n\not= 3k+2$$for three consecutive prime numbers.


We all cases are disapproved we thus say there are no three consecutive prime numbers $\gt 5$.