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 $AP-k$ 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 $AP-3$. 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 $N-1$.
|
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 |