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
\( x^2+2=y^3 \).
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
\( 2^{n+2} \)).
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.