# The progression $4n+3$ and primes.

Consider an arithmetic sequence $4n+3$. This sequence contains infinitely many primes and infinitely many composites. It is clear that there cannot be $3$ consecutive primes in the sequence as every third number in the sequence is divisible by $3$.

Two consecutive primes, on the other hand, appear quite often in the sequence. My first question is:

Are there infinitely many pairs of consecutive members of the sequence that are both prime?

Also, it appears to me that composite numbers in the sequence tend to stack up quite a lot. Is there any limit to this? In particular, is it true that for any $n$, we can find $n$ consecutive members of the sequence such that they are all composite?

• ad edit: there are two questions in the post. why highlight only one of them? – Adam Dec 19 '13 at 4:31
• For $2$ consecutive primes in the sequence, it is an unsolved problem, just like the Twin Prime Conjecture. – André Nicolas Dec 19 '13 at 4:36
• If you were to change the question to to the arithmetic sequence $2n+1$, you’d be asking the Twin Prime Problem. In effect, you are asking the Twin Prime Congruent to $3$ modulo $4$ Problem, and I doubt very much that it would be answerable by elementary means. – Lubin Dec 19 '13 at 4:37

For the answer of finding $n$ consecutive members of the sequence that are composite, just take any n numbers of the sequence in the range between $4n!$ and $4n! +4n$. Say the numbers are$(4n)!+3, (4n)!+7,......,(4n)!+3+4n$ ; these are all composite. This is because the term added to 4n! is a common factor. This is a generalization for finding n consecutive composites in the naturals $1,2,.,n..$. Just use $(n+1)!+2, (n+1)!+3,...(n+1)!+n+1$. You use the same idea for primes $4n+3$ ; you just make the interval longer. NOTE that this is not necessarily the list of the smallest n consecutive composites, but it works.