Arithmetics in Other Quadratic Rings

Every quadratic ring belongs to one of the two classes:

\(1^{\circ}\) Extensions of the form \(K=\mathbb{Z}[\sqrt{d}]\), where \(d\neq1\) is a squarefree integer. The conjugation and norm are given by the formulas \(\overline{x+y\sqrt{d}}=x-y\sqrt{d}\) and \(N(x+y\sqrt{d})=x^2-dy^2\), where \(x,y\in\mathbb{Z}\).
\(2^{\circ}\) Extensions of the form \(K=\mathbb{Z}[\alpha]\) for \(\alpha=\frac{-1+\sqrt{d}}2\), where \(d=4k+1\) (\(k\in\mathbb{Z}\)) is a squarefree integer with \(d\neq1\) (then \(\alpha\) is an algebraic integer: \(\alpha^2+\alpha-k=0\)). The conjugation and norm are given by \(\overline{x+y\alpha}= x-y-y\alpha\) and \(N(x+y\alpha)=x^2-xy-ky^2\), where \(x,y\in\mathbb{Z}\).

Some of these rings are Euclidean, such as \(\mathbb{Z}[\sqrt{d}]\) for \(d=-2,-1,2,3,6,7\) and \(\mathbb{Z}\left[\frac{-1+\sqrt{d}}2 \right]\) for \(d=-7,-3,5\).

Determining all quadratic unique factorization rings (including the non-Euclidean ones) is extremely serious. Among the rings of the type \(1^{\circ}\) and \(2^{\circ}\) with \(d < 0\), apart from the ones mentioned already, the FTA holds in only five other rings: namely, the rings of the type \(2^{\circ}\) for \(d=-11,-19,-43,-67,-163\). Gauss’ conjecture that the FTA holds in infinitely many quadratic rings with a positive \(d\) has not been proved nor disproved until today.

Problem 6 Find all integer solutions of the equation \(x^2+2=y^3\).

Problem 7 Consider the sequence \(a_0,a_1,a_2,\dots\) given by \(a_0=2\) and \(a_{k+1}=2a_k^2-1\) for \(k\geq0\). Prove that if an odd prime number \(p\) divides \(a_n\), then \(p\equiv\pm1\) (mod \(2^{n+2}\)).

Consider the sequence \(x_k\) of positive numbers given by \(a_k=\cosh x_k\) (\(\cosh\) is the hyperbolic cosine, defined by \(\cosh t=\frac{e^t+e^{-t}}2\)). It is easily verified that \(\cosh(2x_k)=2a_k^2-1=\cosh x_{k+1}\), so \(x_{k+1}=2x_k\), i.e. \(x_k=\lambda\cdot 2^k\) for some \(\lambda > 0\). The condition \(a_0=2\) gives us \(\lambda=\log(2+\sqrt3)\). Therefore \[a_n=\frac{(2+\sqrt3)^{2^n}+(2-\sqrt3)^{2^n}}2.\] Let \(p > 2\) be a prime number such \(p\mid a_n\). We distinguish two cases.

\(1^{\circ}\) \(m^2\equiv 3\) (mod \(p\)) for some \(m\in\mathbb{Z}\). Then \[(2+m)^{2^n}+(2-m)^{2^n}\equiv 0\;(\mbox{mod }p).\quad\quad\quad\quad\quad (1)\] Since \((2+m)(2-m)=4-m^2\equiv 1\) (mod \(p\)), multiplying both sides of (1) by \((2+m)^{2^n}\) yields \((2+m)^{2^{n+1}}\equiv-1\) (mod \(p\)). It follows that the order of number \((2+m)\) modulo \(p\) equals \(2^{n+2}\), from which we conclude \(2^{n+2}\mid p-1\).

\(2^{\circ}\) The congruence \(m^2\equiv 3\) (mod \(p\)) has no solutions. We work in the quadratic extension \(\mathbb{Z}_p(\sqrt3)\) of the field \(\mathbb{Z}_p\) in which number \(\sqrt3\) actually plays the role of the number \(m\) from case (1). As in case (1) we have \((2+\sqrt3)^{2^{n+1}}=-1\), which means that the order of \(2+\sqrt3\) in the multiplicative group \(\mathbb{Z}_p(\sqrt3)^\ast\) equals \(2^{n+2}\). This is not enough to finish the proof as in case (1), as the group \(\mathbb{Z}_p(\sqrt3) ^\ast\) has \(p^2-1\) elements; instead, we only get that \(2^{n+2}\mid p^2-1\). However, we shall be done if we find \(u\in\mathbb{Z}_p(\sqrt3)\) for which \(u^2=2+\sqrt3\): indeed, then the order of \(u\) is \(2^{n+3}\), so \(2^{n+3}\mid p^2-1\) and therefore \(2^{n+2}\mid p-1\), since \(4\nmid p+1\).

Note that \((1+\sqrt3)^2=2(2+\sqrt3)\). Now it is enough to show that \(1/2\) is a perfect square in the field \(\mathbb{Z}_p(\sqrt3)\). This immediately follows from the relation \(a_n=0=2a_{n-1}^2-1\), as \(1/2=a_{n-1}^2\). This finishes the proof.

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