Primes in arithmetic progression

How many primes can you find in arithmetic progression (prime AP for short)? Recall that in an AP, the differences between successive members are the same, so the AP has the form $a$, $a+d$, $a+2d$, $a+3d$, $a+4d$, .... Here $a$ is called the "first term" while $d$ is the "common difference". For example, $1$, $4$, $7$, $10$ is a four term AP with common difference $d = 3$.

Here are some examples of prime APs:

• {$3$, $5$, $7$}, with length $3$ and common difference $2$;
• {$5$, $11$, $17$, $23$, $29$}, with length $5$ and common difference $6$;
• {$7$, $37$, $67$, $97$, $127$, $157$}, with length $6$ and common difference $30$.
• {$7$, $157$, $307$, $457$, $607$, $757$, $907$}, with length $7$ and common difference $150$.

It is easy to check that the prime AP {$3$, $5$, $7$}, with length $3$ and common difference $2$, is the only such prime AP, and a prime AP with common difference $2$ cannot exceed a length of $3$. To see why, we only need to examine the remainders left by these numbers on division by $3$.

It is also easy to see the following:

• If a prime AP has length greater than $2$, then its common difference must be an even number, and it must feature only odd primes (because $2$ is the only even prime).
• If a prime AP has length greater than $3$, then its common difference must be a multiple of $3$, and therefore a multiple of $6$ as well (since it also must be an even number). To see why, we only need to examine the remainders left by these numbers on division by $3$.
• If a prime AP has length greater than $5$, then its common difference must be a multiple of $5$, and therefore a multiple of $30$ as well (since it also must be a multiple of $6$). To see why, we only need to examine the remainders left by these numbers on division by $5$.
• If a prime AP has length greater than $7$, then its common difference must be a multiple of $7$, and therefore a multiple of $210$ as well (since it also must be a multiple of $30$). To see why, we only need to examine the remainders left by these numbers on division by $7$.

This chain of results may clearly be extended indefinitely.

By these means we can list innumerable necessary conditions for the existence of long prime APs.

But finding such APs is quite a different matter. As of now, it is largely based on trial-and-error, but guided by clever heuristics. In other words: try, try, and try again! To get a sense of the difficulties involved in this, try finding a prime AP of length $10$ or more - on your own!

Recently (earlier this month, in fact), researchers found a prime AP of length $26$. Here is a link to the site where the discovery was announced: AP26 discovery. The first prime of this AP is
\[43142746595714191,\]
and its common difference is this number:
\[23681770 \times (2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23).\]
In other words, this AP consists of the numbers
\[43142746595714191 + 5283234035979900 \times n\]
for $n = 0$, $1$, $2$, ..., $25$.

Note that the number
\[2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 = 223092870,\]
which is a factor of the common difference, is simply the product of the first $9$ primes. (You will find numbers of this kind repeatedly featuring in such searches.) You can check that the last member of this prime AP is the number
\[175223597495211691.\]
To verify that all these $26$ numbers are really prime would require a good computer algebra system, like Derive or Mathematica or the free open source package, Maxima. Here is what Maxima tells you when you ask it whether the $27$-th term of the AP is prime or not - that is, the number
\[43142746595714191 + 5283234035979900 \times 26;\]
Maxima returns the factorization
\[29 \times 31 \times 41 \times 59 \times 85433250011.\]
So this particular AP cannot be extended beyond $26$ terms.

As of now, this is the "champion" - the holder of the "world record"! No prime AP of length greater than $26$ is currently known.

But just a few years back, Terence Tao and Ben Green proved that there exist prime APs of any desired length! (We had featured Terence Tao in a previous posting.) In other words, there must exist a prime AP of length exceeding one billion; only, we may never ever find it, for it may well involve stupendously large numbers ...

Here are two other sites which will tell you a lot about this topic: Prime AP records and Wikipedia-primes in AP.
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