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 \( x^2+2=y^3 \).

Show solution

Let us write the equation as \( (x+\sqrt{-2}) (x-\sqrt{-2})=y^3 \). For \( x \) even we have \( y^3\equiv2 \) (mod 4), which is impossible; therefore \( x \) is odd. Then \( x+\sqrt{-2} \) and \( x-\sqrt{-2} \) are coprime elements of \( \mathbb{Z}[\sqrt{-2}] \) whose product is a perfect cube. Using the FTA in \( \mathbb{Z}[\sqrt{-2}] \) we conclude that \( x+\sqrt{-2} \) and \( x-\sqrt{-2} \) are both perfect cubes. Hence there exist \( a,b\in\mathbb{Z} \) such that \( (a+b\sqrt{-2})^3=x+\sqrt{-2} \). Comparing the coefficients at \( \sqrt{-2} \) yields \( b(3a^2-2b^2)=1 \); therefore \( b=1 \) and \( a=\pm1 \). Now we easily obtain that \( x=\pm5 \) and \( y=3 \) is the only integral solution of the equation.

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 \( 2^{n+2} \)).

Show solution

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.