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'Prime Sequences' printed from http://nrich.maths.org/
In 2004 an exciting new result was proved in Number Theory by two young mathematicians Ben Green and Terence Tao. They proved that if you look in a long enough list of the prime numbers then you will be able to find numbers which form an arithmetic progression containing as many numbers as you choose! In this question we explore some of the interesting
issues surrounding arithmetic progressions of prime numbers.
An $APk$ sequence is $k\geq 3$ primes in arithmetic progression. See examples
A simple arithmetic progression of three primes starts at $3$ with common difference $4$, giving rise to the progression of prime numbers
$$
3, 7,11
$$
This is an example of $AP3$. Note that the sequence stops here because $11+4=15$, which is not a prime number. Another short arithmetic progression starts at $7$ with common difference $6$
$$
7, 13, 19
$$ 
This problem involves several linked parts leading up to a final challenge. Try some of the earlier questions to gain insights into the final challenge. These can be attempted in any order. You might find that you naturally ask yourself questions which are found later in the list of questions and you might find that one part helps in the consideration of another part. Of course, you are welcome
to go straight to the final challenge. However, you might also wish to start with one of the earlier challenges and see how many of the other challenges you naturally discover whilst exploring the underlying mathematical structure.
Consider some of these three questions first: 
Question A 
Can you find an arithmetic progression of four primes?

Question B 
How many prime APs beginning with $2$ can you find?

Question C 
How many other arithmetic progressions of prime numbers from the list of primes below can you find?

Next consider some of these three questions: 
Question A 
Why is $3, 5, 7$ the only prime AP with common difference $2$?

Question B 
What is the maximum length of a prime AP with common difference of $6$?

Question C 
If the common difference of a prime AP is $N$ then the maximum length of the prime AP is $N1$.

Now consider some of these three questions: 
Question A 
What is the maxiumum length of a prime AP with common difference $10$?

Question B 
What is the max length of a prime AP with common difference $100, 1000, 10000$ ?

Question C 
What are the possible lengths of prime APs with common difference $2p$, where $p$ is prime? Consider $p=3$ and $p> 3$ separately.

When you have thought about some of the previous problems you might like to try the
final challenge
Prove that if an AP$k$ does not begin with the prime $k$, then the common difference is a multiple of the primorial $k$#$ = 2\cdot 3\cdot 5 \cdot \dots \cdot j$, where $j$ is the largest prime not greater than $k$. 
Once you have solved this, why not try to think of some other questions about prime APs to ask?
In doing these problems you might like to see this
list of primes
2 
3 
5 
7 
11 
13 
17 
19 
23 
29 
31 
37 
41 
43 
47 
53 
59 
61 
67 
71 

73 
79 
83 
89 
97 
101 
103 
107 
109 
113 
127 
131 
137 
139 
149 
151 
157 
163 
167 
173 

179 
181 
191 
193 
197 
199 
211 
223 
227 
229 
233 
239 
241 
251 
257 
263 
269 
271 
277 
281 

283 
293 
307 
311 
313 
317 
331 
337 
347 
349 
353 
359 
367 
373 
379 
383 
389 
397 
401 
409 

419 
421 
431 
433 
439 
443 
449 
457 
461 
463 
467 
479 
487 
491 
499 
503 
509 
521 
523 
541 

547 
557 
563 
569 
571 
577 
587 
593 
599 
601 
607 
613 
617 
619 
631 
641 
643 
647 
653 
659 

661 
673 
677 
683 
691 
701 
709 
719 
727 
733 
739 
743 
751 
757 
761 
769 
773 
787 
797 
809 

811 
821 
823 
827 
829 
839 
853 
857 
859 
863 
877 
881 
883 
887 
907 
911 
919 
929 
937 
941 

947 
953 
967 
971 
977 
983 
991 
997 
1009 
1013 
1019 
1021 
1031 
1033 
1039 
1049 
1051 
1061 
1063 
1069 