# Prime Numbers and Progressions

Is it possible to find prime numbers separated by $2$ of any length? For example, $3,5$ is a sequence of prime numbers separated by $2$ and of length $2$. But could we find $10^{100}$ prime numbers separated by $2$? If so, why doesn't this imply the Twin Prime Conjecture? Is it because we are looking at twin prime pairs like $(2,3), (3,5), (5,7), \dots$ instead of one long sequence? That is, an infinite sequence of prime numbers separated by $2$ doesn't imply an infinite number of twin prime pairs?

• The only case where $n,n+2,n+4$ are all prime is when $n=3$. Note that exactly one of $n, n+2, n+4$ is a multiple of $3$. – Hagen von Eitzen Jun 6 '13 at 17:32
• Green-Tao guarantees that there are arbitrarily long arithmetic progressions in the primes, but you don't get to control what the common differences are. In fact, as the progressions get longer, the common difference necessarily also increases (exercise). – Qiaochu Yuan Jun 6 '13 at 18:07
• If you really want to explore the patterns of differences between primes which may (and are conjectured to) repeat infinitely often, the following blog-post is a good place to start, though a demanding read: terrytao.wordpress.com/2013/06/03/… – Mark Bennet Jun 6 '13 at 18:15

Hint $\$ mod $\,3\!:\ \{0, 2, 4\}\equiv \{0,1,2\}\ \Rightarrow\ \{n, n\!+\!2, n\!+\!4\}\equiv \{0,1,2\},\$ so one is divisible by $\,3.\,$

Similarly if $\,a\,$ is coprime to $\,m\,$ then $\,\{n, n\!+\!a,\ldots, n\!+\!(m\!-\!1)a\} =\,$ all residue classes mod $\,m,\,$ hence one of them is divisible by $\,m\,$ (since $\,x\mapsto n+ax\,$ is $\,1$-$1\,$ so onto, mod $\,m).$

In any three consecutive odd numbers, one will be divisible by 3. So one of $2n+1, 2n+3 \text{ or } 2n+5$ will be divisible by 3, and if $n \gt 1$ they can't all be prime. The sequence $3, 5, 7$ is the one exception to this.

The Green-Tao paper shows that there are unbounded arithmetic progressions in the primes, not infinite. Here's the difference. If you pick some integer bound, say 1000, I can (in principle) find an arithmetic progression longer than that: say, 1001 numbers each separated by some $k$ from the last such that all 1001 are prime. You can repeat this with a larger bound, say a million, if you want. But you can't choose "$\infty$", and there are no infinite arithmetic progressions in the primes.

• This isn't the issue. Green-Tao doesn't guarantee that you can get a particular common difference. – Qiaochu Yuan Jun 6 '13 at 18:06
• @QiaochuYuan: I didn't claim it does! I merely attempted to explain the paper, as asked in the question. – Charles Jun 6 '13 at 18:25

The maximum string of consecutive primes each a distance of $2$ from the preceding prime is $3$, i.e. $\{3,5,7\}$. An intuitive reason for this is as follows:

Consider the base 10 "decimal" system that we use to count in. In this system we have certain special numbers that we can look at and immediately determine whether or not they are prime; i.e. any number that ends with an even number (apart from the number $2$) and any number that ends in $5$ (apart from the number $5$) must not be a prime. The reason for this is because we know immediately that they are divisible by $2$ or $5$ respectively. The actual reason we can make such an instantaneous assumption is due to the fact that the base we are working in is divisible by both $2$ and $5$. Numbers ending in $1,3,7,9$ are not so simple.

If we now analyse our numbers using a different base i.e. base $6$ then any number ending in either $0,2,3,4$ must not be prime (apart from the individual numbers $2,3$). Again the reason we can make this assumption is because, in base $6$, if our number ends in $0$ it must be divisible by $6$, if it ends in $2$ it must be divisible by $2$, if it ends in $3$ it must be divisible by $3$ and if it ends in $4$ it must be divisible by $2$. So the numbers that can contend as possible primes, in base $6$, look like this (bracketed): $$(2),(3),4,(5),10,(11),12,13,14,(15),20,(21),22,23,24,(25),30,(31),32,33,34,(35),40,(41),42,43,44,(45),50,(51),52,53,54,(55),\dots$$

Or in base 10:

$$(2),(3),4,(5),6,(7),8,9,10,(11),12,(13),14,15,16,(17),18,(19),20,21,22,(23),24,(25),26,27,28,(29),30,(31),32,33,34,(35),\dots$$

Notice the repeated skip a number skip three numbers pattern that we can observe. This is the reason why we cannot have strings longer than $3$.

Again I stress the above bold words. I am not claiming numbers in brackets are prime, just that numbers not in bracket must, necessarily, not be primes.

I hope this provides some intuition.