# Thread: 3 consecutive prime numbers.

1. ## 3 consecutive prime numbers.

Of the form n, n+2, n+4

We have 3, 5, 7

29, 31 ,33 etc

I guess they must be infinite? How is it proved.

I realise the final digit has to be in the pattern 9,1,3 or 7,9,1 but what next?

2. 33 is not prime i guess

3. i don't think any other numbers exist

4. It can easily be proved that if you have 3 integers of the form n, n+2, n+4 then one of them must be divisible by 3 (try it yourself). Thus no such prime triplets exist, other than 3,5,7.

5. yea at least one should be divisible by 3

i guess this thread is closed

6. If you assume n=3k then case closed
n=3k+1 then n+2=3(k+1)
n=3k+2 then n+4=3(k+2)

So, that proves it?

How would you formally write this assuming nothing excepting basic axioms or addition etc.

Are there any other proofs?
Is there any way of obtaining every single proof of a question like this in general? Suppose you found 10 proofs out of 10, how would you prove or ascertain that there is no eleventh proof? Or is this idiotic speculation?

If you add the three you get 3(n+2), is it true that the sum of 3 different integers, if divisible by three, then at least one of the three must also be divisible by three? How is this proved, if so?

7. Originally Posted by berachia
If you assume n=3k then case closed
n=3k+1 then n+2=3(k+1)
n=3k+2 then n+4=3(k+2)

So, that proves it?

How would you formally write this assuming nothing excepting basic axioms or addition etc.

Are there any other proofs?
Is there any way of obtaining every single proof of a question like this in general? Suppose you found 10 proofs out of 10, how would you prove or ascertain that there is no eleventh proof? Or is this idiotic speculation?

If you add the three you get 3(n+2), is it true that the sum of 3 different integers, if divisible by three, then at least one of the three must also be divisible by three? How is this proved, if so?
You did formally write the proof. You split it into all possible cases and showed that for each case, one of those integers is divisible by 3.