Is there any way to determine whether the fundamental unit of a quadratic field has negative or positive norm, except by actually computing the unit to all of its (many) digits? And, similarly, when $d$ is 1 mod 4, the ring of integers will have a basis with denominators, but sometimes the unit does belong to ${\mathbb Z} [\sqrt d]$—is there any way to tell "in advance" if that will happen, i.e., any way other than computing the unit explicitly? (I can't find any such methods in books, but of course that doesn't prove they don't exist.)
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1$\begingroup$ on the first bit: for a prime $p \equiv 1 \pmod 4$ there is always a solution to $x^2 - p y^2 = -1.$ With $p \equiv q \equiv 1 \pmod 4$ and Legendre $(p,q) = -1,$ there is a solution to $x^2 - pq y^2 = -1.$ No guarantee if mutual residues, the first failures are $x^2 - 205 y^2 \neq -1 $ and $x^2 - 221 y^2 \neq -1 $ $\endgroup$– Will JagyCommented Jun 12, 2022 at 21:33
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1$\begingroup$ That is interesting; I did not know that. One can pretty easily show that if the class number is 1 and the norm of the unit is negative then d is prime. By your observation, if the class number is 1, then the norm is negative if and only if d is prime. $\endgroup$– Michael BeesonCommented Jun 13, 2022 at 1:28
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1$\begingroup$ The second question is discussed here: matwbn.icm.edu.pl/ksiazki/aa/aa74/aa7435.pdf $\endgroup$– Franz LemmermeyerCommented Jun 13, 2022 at 6:23
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$\begingroup$ Michael, I looked at Franz's link; I think you would enjoy D. A. Buell, Binary Quadratic Forms. The stuff about discriminants $5 \pmod 8$ is collected in Theorem 7.5 on page 118. $\endgroup$– Will JagyCommented Jun 13, 2022 at 22:02
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This might be useful:
Stevenhagen, Peter, The number of real quadratic fields having units of negative norm, Exp. Math. 2, No. 2, 121-136 (1993). ZBL0792.11041.
As Stevenhagen explains, if the discriminant $D$ has a prime factor that is 3 mod 4 then there can't be any units with norm $-1$; and if this criterion doesn't apply, then it seems to be random but with a slight bias towards $-1$ rather than $+1$ (and he gives a conjectural exact formula for the probability).
Stevenhagen doesn't explicitly discuss the probability of the fundamental unit being in $\mathbf{Z}[\sqrt{d}]$, but it seems the pattern here is similar: there are congruence criteria which handle some cases (if $d = 1 \bmod 8$ then $u$ is always in $\mathbf{Z}[\sqrt{d}]$), and the behaviour is random otherwise.
answered Jun 12, 2022 at 21:20
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15$\begingroup$ Note that Stevenhagen's conjecture has now been proved by Kuymans-Pagano, arXiv 2201.13424. $\endgroup$ Commented Jun 12, 2022 at 21:52