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A result I would like to know is if there are infinitely primes congruent to $1 \pmod 4$, with fractional part in an interval strictly contained in $\left(0, \dfrac 1 4 \right)$. The title question seems to be true, and seems to be a natural way to prove the above.

Here the case for all primes is proven. However, as I am not a number theorist, I do not follow the argument too well. For my case, I would like a reference to an appropriate source, and maybe a proof which can be easily understood after finding the source.

Another way to look at it would be a generalisation of the result by Glyn Harman in On the distribution of $\sqrt p$ modulo one to this set of $p$. Unfortunately, I have not been able to find this either.

Ѕᴀᴀᴅ
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Rhys Evans
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    See this answer. The proof works the same way as for the primes, showing that for $h \ne 0$ : $$\sum_{n < x} \Lambda(n) e(h \sqrt{n}) = o(x), \qquad \sum_{n < x} \chi_4(n)\Lambda(n) e(h \sqrt{n}) = o(x)$$ (where $\chi_4(2n+1) = (-1)^n$ is the non-trivial character modulo $4$). Adding the two estimates, we have the result. – reuns May 31 '17 at 02:56
  • @user1952009 Do you have a reference for the second sum? – Sungjin Kim May 31 '17 at 03:12
  • @i707107 No need, everything works the same way – reuns May 31 '17 at 03:19
  • @user1952009 I tried it, but didn't work nicely. Can you please post it as an answer? – Sungjin Kim May 31 '17 at 03:32
  • @i707107 Where are you stuck ? Did you look at the linked answer and Iwianec's book ? – reuns May 31 '17 at 03:35
  • @user1952009 Yes, I tried it, I was struggling with an analogue of 13.39 for the twisted sum. But, I think I am slowly getting your hint. – Sungjin Kim May 31 '17 at 03:44
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    @i707107 I can't write directly the equivalent formula for $\chi_4(n) \Lambda(n)$, but it will come in the same way from $$L(s,\chi_4) = \sum_{n=1}^\infty \chi_4(n) n^{-s},\qquad L'(s,\chi_4) = -\sum_{n=1}^\infty \chi_4(n) n^{-s}\ln n, \qquad \frac{1}{L(s,\chi_4)} = \sum_{n=1}^\infty \chi_4(n) \mu(n) n^{-s} ,\qquad\frac{L'}{L}(s,\chi_4) = -\sum_{n=1}^\infty \chi_4(n) \Lambda(n) n^{-s}$$ – reuns May 31 '17 at 04:00
  • I know that it is true now, which is the biggest relief. Unfortunately, I need this result to be clear to people (including me!) that do not have any knowledge of what you are using. That is why I was hoping for a paper with the result to reference. I'll continue the search, but thanks anyway! – Rhys Evans May 31 '17 at 12:35

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