$-1$ is not a sum of two squares in $\mathbb{Q}(\sqrt{-7})$

Question

The question is the following: Prove that $-1$ is not the sum of two squares in $\mathbb{Q}(\sqrt{-7})$.

I have a solution which makes use of Algebraic Number Theory, namely the Dirichlet Unit Theorem about units in a number field. Hopefully this solution is correct.

Suppose that $x^2 + y^2 = -1$ with $x, y \in \mathbb{Q}(\sqrt{-7})$. By working in $\mathbb{Q}(\sqrt{\pm 7})$, we can factor this as $(x+iy)(x-iy) = -1$, where $x, y \in \mathbb{Q}(\sqrt{-7})$. It follows that $x+iy$ is an element with complex absolute value $1$, for any embedding $\mathbb{Q}(\sqrt{\pm 7}) \hookrightarrow \mathbb{C}$, so by the Dirichlet Unit Theorem it follows that $x+iy$ is a root of unity, which must be inside $\mathbb{Q}(\sqrt{\pm 7})$. However it is not too difficult to find that those are $\pm 1, \pm i$ and those do not give solutions to the equation.

However I am wondering if there is a more elementary solution to this problem, especially one that does not require the use of the Dirichlet Unit Theorem.

Answer 1

An elementary solution that shows that $-1$ is not a sum of squares from $\mathbb{Z}[\sqrt{-7}]$. Assume $\sum_k (a_k + b_k \sqrt{-7})^2 = -1$ that is $\sum_k (a_k^2 - 7 b_k^2) =-1$ and $\sum_k 2 a_k b_k = 0$, adding the last two equalities we get $$\sum_k (a_k + b_k)^2 =-1 +8 \sum_k ( b_k)^2$$

so a sum of three squares $\equiv -1 \pmod 8$, not possible.

Answer 2

COMMENT.- With abstraction about whether the unit theorem is correctly applied above, a more elementary demonstration matching rational and non-rational parts seems to give something unlikely. In fact from $$\begin{cases}x=a+b\sqrt{-7}\\y=c+d\sqrt{-7}\end{cases}$$ one has $$x^2+y^2=-1\iff a^2+c^2-7(b^2+d^2)=-1 \text{ and }2(ab+cd)=0$$ from which the system in $a^2$ and $b^2$ $$\begin{cases}a^2-7b^2=7d^2-c^2-1\\a^2b^2=(cd)^2\end{cases}$$ whose solution is $$2a^2=7d^2-c^2-1\pm\sqrt{(7d^2-c^2-1)^2-28cd}\\14b^2=-(7d^2-c^2-1)\pm\sqrt{(7d^2-c^2-1)^2-28cd}$$ This is very restrictive for two reasons: $a$ and $b$ must be rational and exchanging the unknowns for $c$ and $d$ the solution in a and b must have the same form.

Answer 3

Your question could be put under the general headline "elements of a field $K$ of characteristic $\neq 2$ which are represented by a non-degenerate quadratic form with coefficients in $K$", see e.g. Cassels-Fhröhlich, ANT, exercise 4 at the end of the book. In the simple case of 2 or 3 variables, it is more convenient to adopt the normic point of view.

So fix $K$ and introduce the biquadratic extension $L=K(\sqrt a, \sqrt b)$, whose 3 quadratic subextensions are $K(\sqrt a), K(\sqrt b), K(\sqrt ab)$. Here $K=\mathbf Q, a=-1, b=7, ab=-7$, and your question is equivalent to a NSC for $c=-1\in \mathbf Q(\sqrt {-7})$ to be a norm from $L$. The general normic criterion is that a given $c\in K\subset K(\sqrt ab)$ is a norm from $L$ iff $c$ is a product of a norm from $K(\sqrt a)$ and a norm from $K(\sqrt b)$. A detailed proof is given in exercise 4.3 op. cit. Applying this criterion, we get that $-1$ is a sum of two squares in $\mathbf Q(\sqrt {-7})$ iff $-1\in \mathbf Q$ is a product $(x^2+y^2)(u^2-7v^2)$, with $x,y,u,v \in \mathbf Q$. If the coefficients were in $\mathbf Z$ (but a priori they are not), one could conclude by comparing modules and using the Pell-Fermat equation (or Dirichlet's unit theorem). Isn't some hypothesis missing ?

Because the 4-variable equation above, say (E), admits rational solutions, as can be shown by "elementary" ways. First pick up a rational solution of the equation $x^2+y^2=r^2$ (a pythagorean triangle). Eq. (E) then reduces to $(u^2-7v^2)=-r^{-2}$, or, changing variables but keeping the same notations, $u^2-7v^2=-1$, which admits rational solutions (Pell-Fermat).

Answer 4

This problem can be restated in terms of quadratic forms representing $0$. Let $K=\mathbb Q(\sqrt{-7})$.

Here, $F= X^2 + Y^2 + Z^2$ is the quadratic form to consider. It arises, naturally when considering $a,b,c \in \mathcal O_{\mathbb Q(\sqrt{-7})}=\mathbb Z[(1+\sqrt{-7})/2]=:R$ such that $$ \left(\dfrac{a}{b}\right)^2 + \left(\dfrac{c}{b}\right)^2 = -1. $$

It is true that $F=0$ has no solution in $R$ other than $(0,0,0)$. A proof follows by the Infinite Descent Argument below.

Let $(a,b,c)\in R^3\setminus\{0,0,0\}$ be a putative solution to $F=0$ such that $gcd(a,b,c)=1$, here we use that $R$ is a PID. Let $\mathfrak p_2$ be any prime ideal of $R$ containing $2$.

Then $R/\mathfrak p_2^2\simeq \mathbb Z/4\mathbb Z$ and the points $(0,0,0), (0,0,2),(0,2,2),(2,2,2)$ are the solutions of $F=0$ in $\mathbb Z/4\mathbb Z$ (up to change of coordinates). Thus, $a,b,c\in \mathfrak p_2$ i.e., $\mathfrak p_2 \mid gcd(a,b,c)$. This contradicts the assumptions above.

Remarks:

1. In $2$-adic terms one deduces that $F=0$ has no non-zero solution in $\mathbb Z_2$ (or $\mathbb Q_2)$.
2. Local-global principles (Hasse-Minkowski theorem) apply to the quadratic form $F$. As a consequence, if $F=0$ has no solution in $K$ then there must be a prime ideal $\mathfrak p$ in $K$ such that $F=0$ has no non-zero solution over the localization (or completion) $K_{\mathfrak p}$ of $K$ at $\mathfrak p$.