MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the answer to this question, engelbret mentions "a conjecture of H. L. Montgomery (not the one on pair correlations, another one), which implies both the GRH and the Elliott-Halberstam conjecture.". My question is : what does this conjecture says, and where is it stated ? I am also interested in any references that would discuss that conjecture.

share|cite|improve this question
I just saw the following question… by you discussing this very conjecture. – Lucia Sep 2 '13 at 16:02
yes, I just noticed it now as well, strange... – Carlo Beenakker Sep 2 '13 at 17:49
Lucia, Carlo, somehow I didn't realize that the conjecture you mention, and that indeed I knew, was implying both GRH and Elliot-Halberstam, hence was the conjecture engelbert was talking about. This was stupid of me. I kind of imagined that there was a deeper conjecture of Montgomery, on the distribution of zero's on the critical line, which would imply and provide an explanation for the you guys mention. Somehow I am not completely convinced by the answer to my question 115438. – Joël Nov 6 '13 at 2:15

I think the conjecture you have in mind is that for an arithmetic progression $a\pmod q$ with $(a,q)=1$ and $x\ge q$ $$ \Psi(x;q,a) = \frac{x}{\phi(q)} + O\Big(\frac{x^{\frac 12+\epsilon}}{\sqrt{q}}\Big). $$ This is stronger than GRH which has the error $O(x^{\frac 12+\epsilon})$, and just summing the errors over all $q\le x^{1-\epsilon}$ you get the Elliott-Halberstam conjecture as well.

share|cite|improve this answer

discussion: Limitations to the equidistribution of primes (Friedlander and Granville, 1989) and Small gaps between prime numbers (Soundararajan, 2007) (page 16).

the sources to Montgomery are

Topics in multiplicative number theory (Springer, 1971)

Proc. Symp. Pure Math. 28, 307 (1976)

share|cite|improve this answer

Your Answer


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.