Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(I know that there's an infinity of primes congruent to 5 mod 6, but I don't know if there is an infinity of primes congruent to 1 mod 6.)

But what I'd really like to know is whether or not the number of primes ≡1 mod 6 less than a given n can be said to be asymptotically equal to the number of primes ≡5 mod 6 less than that n. So the "total" number of primes ≡1 mod 6 would be equal to the number of primes ≡5 mod 6.

(I hope this is understandable. I'd also like to know how to phrase this inquiry in a more standard fashion, if someone would tell me how to fix it.)

share|improve this question
2  
You might ask like this: let $\pi_1(n)$ be the number of primes of the form $6k+1$ that are less than $n$, and $\pi_5(n)$ similarly. Then you can ask whether $\pi_1(n)/\pi_5(n)$ approaches a limit as $n$ increases without bound, and if so, whether that limit is 1. –  MJD Aug 22 '12 at 1:13
    
That's precisely what I mean; thank you. –  Annick Aug 22 '12 at 1:14
1  
(You didn't ask for this, but for an elementary proof that there are infinitely many primes congruent to $1 \bmod 6$, consider the possible primes dividing a number of the form $k^2 - k + 1$.) –  Qiaochu Yuan Aug 22 '12 at 1:28
    
Got it, thanks. Great to know. –  Annick Aug 22 '12 at 2:35
add comment

2 Answers

up vote 7 down vote accepted

Yes, this is true. A more general statement follows from a suitably strong form of Dirichlet's theorem on arithmetic progressions, namely that asymptotically the proportion of primes which are congruent to $a \bmod n$ (for $\gcd(a, n) = 1$) is $\frac{1}{\varphi(n)}$.

In the particular case that $n = 6$ it is possible to give a more elementary proof of a weaker result. Define the Dirichlet L-function

$$L(s, \chi_6) = \sum_{n=1}^{\infty} \frac{\chi_6(n)}{n^s}$$

where $\chi_6$ is the unique nontrivial Dirichlet character $\bmod 6$. This takes the form $\chi_6(n) = 1$ if $n \equiv 1 \bmod 6$, $\chi_6(n) = -1$ if $n \equiv 5 \bmod 6$, and $\chi_6(n) = 0$ otherwise. The Euler product of this L-function is

$$L(s, \chi_6) = \prod_p \left( \frac{1}{1 - \chi_6(p) p^{-s}} \right) = \prod_{p \equiv 1 \bmod 6} \left( \frac{1}{1 - p^{-s}} \right) \prod_{p \equiv 5 \bmod 6} \left( \frac{1}{1 + p^{-s}} \right).$$

It is possible to explicitly evaluate $L(1, \chi_6)$ and in particular to show that it is not zero; in fact,

$$L(1, \chi_6) = \int_0^1 \frac{1 - x^5}{1 - x^6} \, dx$$

and this can be evaluated using partial fractions (but note that the integrand is always positive so this number is definitely positive). So we conclude that

$$-\log L(s, \chi_6) = \sum_{p \equiv 1 \bmod 6} \log (1 - p^{-s}) + \sum_{p \equiv 5 \bmod 6} \log (1 + p^{-s})$$

approaches a nonzero constant as $s \to 1$ (if summed in the appropriate order) even though the first and second terms separately approach $\mp \infty$. So the contributions coming from primes in each residue class cancel out asymptotically. This is not quite as strong as the desired statement, though; if you fill in all the details in what I've said you'll show that the Dirichlet density of the primes congruent to $\pm 1 \bmod 6$ are the same but this should still be true for the natural density and this requires a further argument (I am not sure how much further, though).

For more details see any book on analytic number theory, e.g. Apostol.

share|improve this answer
    
Thanks so much for this highly detailed answer and explanatory links. –  Annick Aug 22 '12 at 1:27
add comment

Look up "prime races" at http://www.dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf.

Their conclusion: "It does seem that “typically” qn + a has fewer primes than qn + b if a is a square modulo q while b is not."

So, since 1 is a quadratic residue mod 6, while 5 is not, there will typically be more primes of the form 6n+5 than 6n+1 (though their ratio does tend to 1).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.