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Suppose we have a prime $p\equiv 3\mod 4$ and $p>3$ with the property that for all primes $q<p/4$, we have that $\left(\frac{q}{p}\right)=-1$. I believe that in this case it is true that the class number of $\mathbb{Q}[\sqrt{-p}]$ is 1, but I am not sure how to prove this. Any ideas?

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  • $\begingroup$ I presume you mean all non-square $q < p/4$? Otherwise, the hypotheses are never met since $q=1$ is always a residue. $\endgroup$ – Barry Smith Apr 3 '15 at 23:11
  • $\begingroup$ Actually, its worse than that. Since, for instance, if $2$ and $3$ are both nonresidues, then $6$ will be a residue. I guess you're assuming $q$ is prime? $\endgroup$ – Barry Smith Apr 3 '15 at 23:29
  • $\begingroup$ Yeah sorry, I forgot to state $q$ was prime. $\endgroup$ – ruadath Apr 4 '15 at 1:20
  • $\begingroup$ The class number of $\mathbb{Q}[\sqrt{-p}]$ is one only for a finite number of primes, see en.wikipedia.org/wiki/… , imaginary quadratic fields. $\endgroup$ – Jack D'Aurizio Apr 4 '15 at 1:25
  • $\begingroup$ Moreover, your hypothesis is very unlikey to hold, since by Burgess bound and Vinogradov's amplification trick it is known that the size of the least quadratic non-residue $\pmod{p}$ is $\ll p^{\frac{1}{4\sqrt{e}}}$. $\endgroup$ – Jack D'Aurizio Apr 4 '15 at 2:57
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Hint: Minkowski's bound shows that every ideal class contains an element of norm at most $$\sqrt{|-p|} \left( \frac{4}{\pi} \right) \frac{2!}{2^2} = \frac{2\sqrt{p}}{\pi}.$$ Using your hypotheses, can you find a way to show that all ideals of small norm are principal?

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  • $\begingroup$ I made that observation already, and I was able to show that all prime ideals of such small norm are non-split, but I was not able to show that this implies they were principal. $\endgroup$ – ruadath Apr 3 '15 at 22:55
  • $\begingroup$ @ruadan What does an inert prime look like? $\endgroup$ – Brandon Carter Apr 3 '15 at 22:58

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