# Relationship between class number and Legendre symbol

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?

• I presume you mean all non-square $q < p/4$? Otherwise, the hypotheses are never met since $q=1$ is always a residue. – Barry Smith Apr 3 '15 at 23:11
• 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? – Barry Smith Apr 3 '15 at 23:29
• Yeah sorry, I forgot to state $q$ was prime. – ruadath Apr 4 '15 at 1:20
• 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. – Jack D'Aurizio Apr 4 '15 at 1:25
• 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 $\ll p^{\frac{1}{4\sqrt{e}}}$. – Jack D'Aurizio Apr 4 '15 at 2:57

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?