Norms in quadratic fields

This is false. The smallest counterexample is $d = 34$. Let $K = \mathbb{Q}(\sqrt{34})$. The fundamental unit in $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{34}]$ is $35 + 6 \sqrt{34}$, which has norm $1$, and therefore, there is no element in $\mathcal{O}_{K}$ with norm $-1$.

However, $\frac{3}{5} + \frac{1}{5} \sqrt{34}$ has norm $-1$, so there is an element of norm $-1$ in $K$.

Jeremy Rouse already gave a counterexample, but let me expand on that somewhat. The question of whether $\mathbb{Q}(\sqrt{d})$ contains an element of norm $-1$ is purely local: this happens if and only if $d$ is a sum of two squares of rational integers. Indeed if we assume $d$ is square-free, this is saying that all odd primes dividing $d$ are congruent to $1$ modulo $4$.

The question of whether the ring of integers of $\mathbb{Q}(\sqrt{d})$ contains an element of norm $-1$ is much more subtle, and is really a question about the class group of $\mathbb{Q}(\sqrt{d})$ and the narrow class group. If we put $K_d = \mathbb{Q}(\sqrt{d})$ and $\text{CL}(K_d), \text{CL}^\sharp(K_d)$ to be the class group and narrow class group of $K_d$ respectively, then the existence of an element of norm $-1$ in $\mathcal{O}_{K_d}$ is equivalent to $\text{CL}(K_d) \cong \text{CL}^\sharp(K_d)$. This is a subtle condition. One can simplify the criterion somewhat, since really only $2^\infty$-torsion matters. The simplified criterion is the assertion that $\text{CL}(K_d)[2^k] \cong \text{CL}^\sharp(K_d)[2^k]$ for all $k \geq 1$. The condition that $\text{CL}^\sharp(K_d)[2] \cong \text{CL}(K_d)[2]$ is equivalent to the field $K_d$ containing an element of norm $-1$, and is of course a necessary condition for the ring of integers to contain an element of norm $-1$.

Edit: I should emphasize that asymptotically the sets $$S_1 = \{d : K_d \text{ contains an element of norm } -1\}$$ and $$S_2 = \{d : \mathcal{O}_{K_d} \text{ contains an element of norm } -1\}$$ do not have the same density, hence there are infinitely many counterexamples. This is proved by Fouvry and Kluners in this paper. In the same paper they also mention that one expects an asymptotic formula for the density of $S_2$, given by Stevenhagen.

Dirichlet's version of Gauss composition is in the book by Cox, (page 49 in first) with a small typo corrected in the second edition. For our purpose, duplication, it has a better look to equate $a=a'$ from the start, with $\gcd(a,b) = 1$ sufficing, $$ \left( ax^2 +bxy+ acy^2 \right) \left( aw^2 +bwz+ acz^2 \right) = c X^2 + b XY + a^2 Y^2 $$ where $$ X = axz + ayw+byz \; \; , \; \; \; Y = xw - c yz $$ so that the square of $\langle a,b,ac \rangle$ is $\langle c,b,a^2 \rangle.$

Today's question concerns $c=-1$

$$ \left( ax^2 +bxy -ay^2 \right) \left( aw^2 +bwz -az^2 \right) = - X^2 + b XY + a^2 Y^2 $$ where $$ X = axz + ayw+byz \; \; , \; \; \; Y = xw + yz $$ so that $$\langle a,b,-a \rangle^2 = \langle -1,b,a^2 \rangle.$$ We also see Stanley's fact that the discriminant is the sum of two squares, $b^2 + 4 a^2$ the way I wrote things.

By the Gauss theorem on duplication, $ \langle -1,b,a^2 \rangle$ is in the principal genus

Furthermore, we now know that the principal form is $SL_z \mathbb Z$ equivalent to $$ \langle 1,b,-a^2 \rangle $$ The principal form may not integrally represent $-1$ but does so rationally.

As to being in the same genus, we can use Siegel's definition of rational equivalence without essential denominator.

$$ \left( \begin{array}{rr} 0 & 1 \\ -a^2 & -b \\ \end{array} \right) \left( \begin{array}{rr} 1 & \frac{b}{2} \\ \frac{b}{2} & -a^2 \\ \end{array} \right) \left( \begin{array}{rr} 0 & -a^2 \\ 1 & -b \\ \end{array} \right) = \; a^2 \; \left( \begin{array}{rr} -1 & \frac{b}{2} \\ \frac{b}{2} & a^2 \\ \end{array} \right) $$

$$ \left( \begin{array}{rr} b & 1 \\ -a^2 & 0 \\ \end{array} \right) \left( \begin{array}{rr} -1 & \frac{b}{2} \\ \frac{b}{2} & a^2 \\ \end{array} \right) \left( \begin{array}{rr} b & -a^2 \\ 1 & 0 \\ \end{array} \right) = \; a^2 \; \left( \begin{array}{rr} 1 & \frac{b}{2} \\ \frac{b}{2} & -a^2 \\ \end{array} \right) $$