Are most primes in a prime arithmetic progression of length at least 3? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:26:14Z http://mathoverflow.net/feeds/question/57377 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57377/are-most-primes-in-a-prime-arithmetic-progression-of-length-at-least-3 Are most primes in a prime arithmetic progression of length at least 3? Stanley Yao Xiao 2011-03-04T17:30:29Z 2011-03-07T05:11:32Z <p>Following the following two previous questions on mathoverflow:</p> <p><a href="http://mathoverflow.net/questions/34197/are-all-primes-in-a-pap-3/34298#34298" rel="nofollow">http://mathoverflow.net/questions/34197/are-all-primes-in-a-pap-3/34298#34298</a></p> <p>and</p> <p><a href="http://mathoverflow.net/questions/2214/covering-the-primes-by-3-term-aps" rel="nofollow">http://mathoverflow.net/questions/2214/covering-the-primes-by-3-term-aps</a></p> <p>I have attempted to show that infinitely many primes are in an arithmetic progression of length 3 in the primes following Ben Green's comment that one can do this using the circle method; but I have not found any success. Can anyone suggest (with more detail perhaps) a way to show that infinitely many primes are in an arithmetic progression of length at least 3?</p> <p>Edit: In view of the comments, I have rephrased the question: My intention was to ask what can one do (Ben Green suggested the circle method, but gave no details) to show that 'most' primes (that is, the exceptional set has upper density 0) are in an arithmetic progression of at least 3 in the primes.</p> http://mathoverflow.net/questions/57377/are-most-primes-in-a-prime-arithmetic-progression-of-length-at-least-3/57635#57635 Answer by Terry Tao for Are most primes in a prime arithmetic progression of length at least 3? Terry Tao 2011-03-07T05:03:57Z 2011-03-07T05:11:32Z <p>It is a <a href="http://www.ams.org/mathscinet-getitem?mr=2180408" rel="nofollow">theorem of Ben Green</a> that every subset of the primes of positive relative density contains a progression of length three. As an immediate consequence, the set of primes \$A\$ which are not the first term in a progression of primes of length three has density zero (otherwise \$A\$ would contain a length three progression, a contradiction).</p> <p>Ben's proof is, strictly speaking, an application of the circle method, but is probably overkill for your problem (roughly speaking, Ben wants to find a length three progression with all three elements in an arbitrary dense subset \$A\$ of the primes; for your problem, one only needs to study the simpler problem where the smallest term of the progression needs to be in \$A\$ but the other two elements lie in the set \$P\$ of primes). So a simpler proof (still by the circle method) is likely to exist. (Note, by the way, that a <a href="http://www.ams.org/mathscinet-getitem?mr=2245880" rel="nofollow">later paper of Ben and myself</a> gives a slightly simpler proof of Ben's theorem (and, of course, we also have <a href="http://www.ams.org/mathscinet-getitem?mr=2415379" rel="nofollow">a more complicated proof</a> as well).)</p> <p>Another approach is to modify <a href="http://www.ams.org/mathscinet-getitem?mr=374063" rel="nofollow">the argument of Montgomery and Vaughan</a> (also, ultimately, based on the circle method) that shows that the number of even numbers that are not the sum \$p_1+p_2\$ of two primes is very low in density (much lower than the density of the primes, in particular). (Actually, <a href="http://www.ams.org/mathscinet-getitem?mr=327703" rel="nofollow">an older and somewhat simpler paper of Vaughan</a> already suffices for this.) The same argument should also show that the odd integers \$n\$ that are not the first term \$2p_1-p_2\$ of an 3-term AP whose other two terms \$p_1,p_2\$ are primes larger than \$n\$, also has density much smaller than that of the primes.</p>