\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Mersenne Primes in Real Quadratic Fields} \vskip 1cm \large Sushma Palimar and B. R. Shankar\\ Department of Mathematical and Computational Sciences \\ National Institute of Technology Karnataka, Surathkal\\ Mangalore\\ India\\ \href{mailto:sushmapalimar@gmail.com}{\tt sushmapalimar@gmail.com}\\ \href{mailto:shankarbr@gmail.com}{\tt shankarbr@gmail.com}\\ \end{center} \vskip .2in \begin{abstract} The concept of Mersenne primes is studied in real quadratic fields with class number one. Computational results are given. The field $\mathbb{Q}(\sqrt{2})$ is studied in detail with a focus on representing Mersenne primes in the form $x^{2}+7y^{2}$. It is also proved that $x$ is divisible by $8$ and $y\equiv \pm 3$ (mod ${8}$), generalizing a result of F. Lemmermeyer, first proved by H. W. Lenstra and P. Stevenhagen using Artin's reciprocity law. \end{abstract} \section{Introduction} It is well known that $a^{d}-1$ divides $a^{n}-1$ for each divisor $d$ of $n,$ and if $n=p$, a prime, then \begin{equation*} a^{p}-1=(a-1)(1+a+a^{2}+\cdots+a^{p-1}) \end{equation*} and if $a^{p}-1$ is a prime, then $a=2$. Number theorists of all persuasions have been fascinated by prime numbers of the form $2^{p}-1$ ever since {Euclid} used them for the construction of perfect numbers. In modern times, they are named after {Marin Mersenne} (1588-1648). A well known result due to Euclid is that, if $2^{p}-1$ is a prime then $2^{p-1}(2^{p}-1)$ is perfect. Much later Euler proved the converse, every even-perfect number has this form. Mersenne primes have been studied by amateurs as well as specialists. Mersenne primes are used in cryptography too in generating pseudorandom numbers. By far, the most widely used technique for pseudorandom number generation is an algorithm first proposed by Lehmer, known as the linear congruential method. It is generated by the recursion $X_{n+1} \equiv aX_{n}$ (mod $M_{31}$), where $M_{31}$ is the Mersenne prime $2^{31}-1.$ Of the more than two billion choices for $a$, only a handful of multipliers are useful. One such value is $a=7^{5}=16807,$ which was originally designed for use in the IBM 360 family of computers, Stallings~\cite{WS}. On March 3, 1998, the birth centenary of {Emil Artin } was celebrated at the Universiteit van Amsterdam. The paper Lenstra and Stevenhagen~\cite{LS} is based on two lectures given on the occasion. We quote from Lenstra and Stevenhagen~\cite{LS}: ``Artin's reciprocity law is one of the cornerstones of \textit{class field theory.} To illustrate its usefulness in elementary number theory, we shall apply it to prove a recently observed property of Mersenne primes.'' The property of Mersenne primes referred to is the following: if $M_{p}=2^{p}-1$ is prime and $p\equiv 1$ (mod ${3}$), then $M_{p}=x^{2}+7y^{2}$ for some integers $x,y$ and one always has $x\equiv 0$ (mod ${8}$) and $y\equiv \pm 3$ (mod ${8}$). This was first observed by {Franz Lemmermeyer}. Many have attempted to generalize the notion of Mersenne primes and even-perfect numbers to complex quadratic fields with class number 1. One reason is that they have only finitely many units. Indeed, with the exception of $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-3})$, the other seven complex quadratic fields with class number 1 have only two units, $\pm1$. Spira~\cite{RS} defined Mersenne primes over $\mathbb{Q}(\sqrt{-1})$ to give a useful definition of even-perfect numbers over $\mathbb{Z}[i]$, the ring of Gaussian integers. His work was continued later by McDaniel~\cite{WM,WL} to give an analogue of {Euclid-Euler} theorem over $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-3})$. In both the papers the concept of Mersenne primes is used to give a valid definition of even-perfect numbers. Recently Berrizbeitia and Iskra~\cite{BI} studied Mersenne primes over Gaussian integers and Eisenstein integers. The primality of Gaussian Mersenne numbers and Eisenstein Mersenne numbers are tested using biquadratic reciprocity and cubic reciprocity laws respectively. In this paper the concept of Mersenne primes is studied in real quadratic fields $\mathbb{K}=\mathbb{Q}(\sqrt{d})$ with class number 1, so that unique factorization holds and irreducibles are always prime. We denote the ring of integers of $\mathbb{K}$ by ${\cal O_{K}}$, \begin{equation*} {\cal O_{K}} = \begin{cases} \mathbb{Z}[\sqrt{d}], & \text{ if } d \equiv 2,3\ ({\rm mod}\ 4); \\ \mathbb{Z}[\frac{1+\sqrt{d}}{2}], & \text{ if } d\equiv 1\ ({\rm mod}\ 4). \end{cases} \end{equation*} Since $\mathbb{K}$ is a unique factorization domain, irreducibles are primes in these domains. Hence for any $\eta$ $\in$ $\mathbb{K}$ the two factorings \begin{center} $\eta = \pi_1^{k_1}\pi_2^{k_2}\cdots \pi_r^{k_r} $ and $\eta = \epsilon_1 \pi_1^{k_1}\epsilon_2 \pi_2^{k_2} \cdots \epsilon_r \pi_r^{k_r}$ \end{center} are considered to be one and the same, where $\epsilon_i$ are units and $\pi_i$ are irreducibles. We define $M_{p,\alpha}=\frac{\alpha^{p}-1}{\alpha-1}\,$ such that $\alpha \in \mathcal{O_{K}}$ is irreducible and $u=\alpha-1$ is a unit other than $\pm 1$. Then $M_{p,\alpha}$ may be called as an analog of Mersenne prime if the norm of $M_{p,\alpha}$ namely $N(M_{p,\alpha})=N(\frac{\alpha^{p}-1}{a-1})$ is a rational prime. Condition for the irreducibility of $ \alpha=1+u \in {\cal O_{K}}$ such that $\alpha-1$ is a unit (other than $\pm 1$) is derived in the next section. For this we study the case $N(\alpha-1)=N(u)=\pm 1$ separately. We also give a list of such quadratic fields and a few Mersenne primes in those fields. Computational results show that, among real quadratic fields, Mersenne primes in $\mathbb{Q}(\sqrt{2})$ have a definite structure. The special property of the usual Mersenne primes observed by {Franz Lemmermeyer} and proved in Lenstra and Stevenhagen~\cite{LS} admits a generalization to Mersenne primes over $\mathbb{Q}(\sqrt{2})$. This property appears to be special only to $\mathbb{Q}(\sqrt{2})$. Some interesting properties of Mersenne primes and recent primality tests to check the primality of Mersenne numbers in $\mathbb{Q}(\sqrt{2})$ are given. Also, the usual Mersenne primes given by $M_{p}=2^{p}-1,$ can be obtained from the field $\mathbb{K}\,=\,\mathbb{Q}(\sqrt{2})$ without altering the conditions on $M_{p,\alpha}$. \section{Basic Results} Below we consider various cases under which $\alpha$ is irreducible. \begin{theorem}\label{th1} Let $d\equiv 2,3\ ({\rm mod}\ 4)$ and $N(\alpha-1)=-1$. Then $\alpha$ is irreducible if and only if $d=2$ and $u\in \{1+\sqrt{2},\,1-\sqrt{2},\,-1+\sqrt{2},\,-1-\sqrt{2} \}$. \end{theorem} \begin{proof} Let $\alpha$ be irreducible and $u=a+b\sqrt{d}$. Then $\alpha=(a+1)+b\sqrt{d}$. Hence $N(\alpha)= (a+1)^{2}-2b^{2}=N(u)+2a+1= 2a$. Since $\alpha$ is irreducible, $2a$ should be a rational prime. Hence $a=\pm1$. With $a=1$, $u=1+b\sqrt{d}$ and $ N(u)=-1= 1-b^{2}d$. i.e., $b^{2}d=2$. Since $d$ is square-free, $d=2$ and $b=\pm1$. Similarly with $a=-1$, we get $b=\pm1$ and $d=2$. Conversely let $d=2$ and $u=a+b\sqrt{2}$ be any unit in $\mathbb{Q}(\sqrt{2})$. Then $\alpha=(a+1)+b\sqrt{2}$ and $N(\alpha)=(a+1)^{2}-2b^{2}=a^{2}-2b^{2}+2a+1=N(u)+2a+1=2a$, is a rational prime, if and only if $a=\pm1$. As before, we get $b=\pm 1.$ Hence different choices of $u$ for which $\alpha$ is irreducible are respectively, $1+\sqrt{2},\,1-\sqrt{2},\,-1+\sqrt{2}\,$ and $-1-\sqrt{2}$. As $1+\sqrt{2}$ is the fundamental unit, these values are, $u,-u^{-1},u^{-1},-u$. Corresponding $\alpha$ values are, $2+\sqrt{2},\,2-\sqrt{2},\,\sqrt{2}$ and $-\sqrt{2}$. \end{proof} Since $2-\sqrt{2}$ and $-\sqrt{2} $ are the conjugates of $2+\sqrt{2}$ and $\sqrt{2}$ respectively, we compute $M_{p,\alpha}$ with $\alpha=2+\sqrt{2}$ and $\sqrt{2}$. For $\alpha=2+\sqrt{2}$ a few Mersenne primes in $\mathbb{Q}(\sqrt{2})$ are given below: \begin{table}[H] \centering \begin{tabular}{|l|l|l|l|} \hline $p$ &$M_{p,\alpha}$ &$N(M_{p,\alpha})$ \\ \hline $2$ &$3+\sqrt{2}$ &$7$ \\ \hline $3$ &$9+5\sqrt{2} $ &$31$ \\ \hline $5$ &$97+67\sqrt{2}$ &$431$ \\ \hline $7$ &$1121+791\sqrt{2}$ &$5279$ \\ \hline $11$ &$152193+107615\sqrt{2}$ &$732799$ \\ \hline \end{tabular} \caption{Mersenne primes in $\mathbb{Q}(\sqrt{2}).$} \end{table} The next three Mersenne primes are found at $p=73$, with \[N(M_{p,\alpha}) = 851569055172258793218602741480913108991,\] $p=89$ with \[N(M_{p,\alpha})=290315886781191681464330388772329064268797313023,\] and at $p=233$ with \[N(M_{p,\alpha}) = 18060475427282023033368001231166441784737806891537 \] \[806547065314167911959518498581747712829157156517940837234519177963497324543.\] With $\alpha = \sqrt{2}$, $M_{p,\alpha}=\frac{(\sqrt{2})^{p}-1}{\sqrt{2}-1}$. Thus $N(M_{p,\alpha})=2^{p}-1$, giving all the usual Mersenne numbers. \begin{theorem}\label{th2} Let $d \equiv 1\ ({\rm mod}\ 4) $ and $\alpha-1=u=\frac{a+b\sqrt{d}}{2}$ be a unit such that, $N(u)=N(\alpha-1)=-1$. Then $\alpha$ is irreducible if and only if $a$ is a rational prime and $b$ is some odd integer. \end{theorem} \begin{proof} By hypothesis, $ N(\alpha)= \frac{(a+2)^{2}-db^{2}}{4} = a$, since $N(u)=-1$. For $\alpha$ to be irreducible $a$ should be an odd rational prime. Indeed if $a=2$ then $u=\frac{2+b\sqrt{d}}{2}$ and $N(u)=\frac{4-db^{2}}{4}= -1 $ $\Rightarrow b^{2}d=8.$ This is impossible since $d \equiv 1$ (mod ${4}$). Hence it is clear that $b$ is some odd integer. Thus the analogs of Mersenne primes are defined for $d \equiv 1$ (mod ${4}$) whenever units are of the form $u= \frac{p+(2n+1)\sqrt{d}}{2}$, where $n \in \mathbb{Z} $ and $p$ is an odd rational prime. The converse is straightforward since the norm of $\alpha$ is $a=p$, a rational prime by assumption. \end{proof} Table~\ref{table2} shows the values of $d \equiv 1$ (mod ${4}$), $d<500$ for which the class number is 1, $N(u)=-1$ and $\alpha$ is irreducible. \begin{table}[H] \centering \begin{tabular}{|l|l|l|l|} \hline $\mathbb{Q}(\sqrt{d})$ &$u$ &$\alpha$ & $N(\alpha)$ \\ \hline $\mathbb{Q}(\sqrt{13})$ &$(\frac{3+\sqrt{13}}{2})$ &$(\frac{5+\sqrt{13}}{2})$ &$3$ \\ \hline $\mathbb{Q}(\sqrt{29})$ &$(\frac{5+\sqrt{29}}{2})$ &$(\frac{7+\sqrt{29}}{2})$ &$5$ \\ \hline $\mathbb{Q}(\sqrt{53})$ &$(\frac{7+\sqrt{53}}{2})$ &$(\frac{9+\sqrt{53}}{2})$ &$7$ \\ \hline $\mathbb{Q}(\sqrt{149})$ &$(\frac{61+5\sqrt{149}}{2})$ &$(\frac{63+5\sqrt{149}}{2})$ &$61$ \\ \hline $\mathbb{Q}(\sqrt{173})$ &$(\frac{13+\sqrt{173}}{2})$ &$(\frac{15+\sqrt{173}}{2})$ &$13$ \\ \hline $\mathbb{Q}(\sqrt{293})$ &$(\frac{17+\sqrt{293}}{2})$ &$(\frac{19+\sqrt{293}}{2})$ &$17$ \\ \hline \end{tabular} \caption{ $d \equiv 1$ (mod ${4}$) $N(u)=-1$ ; $\alpha$ is irreducible. } \label{table2} \end{table} \begin{theorem}\label{th3} Let $ d \equiv 2,3\ ({\rm mod}\ 4)$ and $u=a+b\sqrt{d}$ be a unit, such that $N(u)=1$. Then $\alpha$ is always reducible. \end{theorem} \begin{proof} By hypothesis, $\alpha=(a+1)+b\sqrt{d}$ and $ N(\alpha)= 2(1+a)$, which is prime only if $a=0$, which contradicts $N(u)=1$. Hence $\alpha$ is not irreducible. \end{proof} \begin{theorem}\label{th4}Let $d \equiv\ 1\ ({\rm mod}\ 4)$ and $u=\frac{a+b\sqrt{d}}{2}$ be a unit such that $N(u)=1$. Then, $\alpha$ is irreducible if and only if $a+2$ is a rational prime and $b$ is some odd integer. \end{theorem} \begin{proof} By hypothesis, $ N(\alpha)= \frac{(a+2)^{2}-db^{2}}{4} = a+2$, since $N(u)=1$. For $\alpha$ to be irreducible $a+2$ should be a rational prime. Clearly $a \neq 0$ and $a^{2} \equiv 1$ (mod ${4}$). Since $d \equiv 1$ (mod ${4}$) it is clear that $b$ is some odd integer. Thus the analogs of Mersenne primes are defined for $d \equiv 1$ (mod ${4}$) whenever units are of the form $u= \frac{a+(2n+1)\sqrt{d}}{2}$, where $n \in \mathbb{Z} $ and $a+2$ is an odd rational prime. Converse is straightforward as in theorem \ref{th2}. \end{proof} Table~\ref{table4} below shows the values of $d \equiv 1$ (mod ${4}$), $d<500$ for which the class number is 1, $N(u)=1$ and $\alpha$ is irreducible. \begin{table}[H] \centering \begin{tabular}{|l|l|l|l|} \hline $\mathbb{Q}(\sqrt{d})$ & $u$ &$\alpha$ & $N(\alpha)$ \\ \hline $\mathbb{Q}(\sqrt{21})$ &$\frac{5+\sqrt{21}}{2}$ &$\frac{7+\sqrt{21}}{2}$ &$7$ \\ \hline $\mathbb{Q}(\sqrt{77})$ &$\frac{9+\sqrt{77}}{2}$ &$\frac{11+\sqrt{77}}{2}$ &$11$ \\ \hline $\mathbb{Q}(\sqrt{93})$ &$\frac{29+3\sqrt{93}}{2}$ &$\frac{31+3\sqrt{93}}{2}$ &$31$ \\ \hline $\mathbb{Q}(\sqrt{237})$ &$\frac{77+5\sqrt{237}}{2}$ &$\frac{79+5\sqrt{237}}{2}$ &$79$ \\ \hline $\mathbb{Q}(\sqrt{437})$ &$\frac{21+\sqrt{437}}{2}$ &$\frac{23+\sqrt{437}}{2}$ &$23$ \\ \hline $\mathbb{Q}(\sqrt{453})$ &$\frac{149+7\sqrt{453}}{2}$ &$\frac {151+7\sqrt{453}}{2}$ &$151$ \\ \hline \end{tabular} \caption{ $d \equiv 1$ (mod ${4}$); $N(u)=1$; $\alpha$ is irreducible. } \label{table4} \end{table} As an illustration we consider the following table. \begin{table}[H] \centering \begin{tabular} {|l|l|l|l|} \hline $p$ &$N(M_{p,\alpha})$ \\ \hline $17$ &$223358425353211 $ \\ \hline \end{tabular} \caption{ Mersenne primes in $ \mathbb{Q}(\sqrt{21}),\,u=\frac{5+\sqrt{21}}{2}$.} \end{table} The next Mersenne prime is found at $p=47$. Similar calculations are obtained for $\mathbb{Q}(\sqrt{77})$, the fundamental unit is $u=\frac{9+\sqrt{77}}{2}$ and $\alpha=\frac{11+\sqrt{77}}{2}.$ \begin{table}[H] \centering \begin{tabular}{|l|l|l|l|} \hline $p$ &$N(M_{p,\alpha})$ \\ \hline $2$ &$23 $ \\ \hline $7$ &$10248701 $ \\ \hline \end{tabular} \caption{Mersenne primes in $ \mathbb{Q}(\sqrt{77}), \,u=\frac{9+\sqrt{77}}{2}$.} \end{table} The next Mersenne prime is found at $p=71$. The values of $u$ for Tables~\ref{table2} and \ref{table4} are taken from Cohen~\cite{cohen}. \subsection{Observations} \begin{enumerate} \item In Tables~\ref{table2} and \ref{table4} above, we have chosen only the fundamental unit $u$ in $\mathbb{Q}(\sqrt{d})$. However it is possible that $\alpha=1+u$ is not irreducible with $u$ as fundamental unit and yet $\alpha^{\prime}=1+u^{\prime}$ is irreducible for some other unit $u^{\prime}$ in $\mathbb{Q}(\sqrt{d})$. As an illustration we consider $\mathbb{Q}(\sqrt{5})$. Here $u=\frac{1+\sqrt{5}}{2}$ is the fundamental unit. But, $\alpha=1+u=\frac{3+\sqrt{5}}{2}=u^{2}$ is again a unit! However, with $u^{\prime}=u^{2}=\frac{3+\sqrt{5}}{2}$, we get $\alpha^{\prime}=1+u^{\prime}=\frac{5+\sqrt{5}}{2}$ and $N(\alpha^{\prime})=5$, so $\alpha^{\prime}$ is irreducible. Another choice is $u^{5}=u^{''}=\frac{11+5\sqrt{5}}{2}$ and $\alpha^{''}=1+u^{''}=\frac{13+5\sqrt{5}}{2}$ is irreducible since $N(\alpha^{''})=11$. \item Theorems \ref{th1} and \ref{th3} imply the following: Among all fields $\mathbb{Q}(\sqrt{d})$, $d\equiv 2,3$ (mod ${4}$) $\mathbb{Q}(\sqrt{2})$ is the only field where $ \alpha=1+u$ is irreducible. There are essentially only two choices for $\alpha$, namely $\sqrt{2}$ and $2+\sqrt{2}$. \end{enumerate} Similar to usual Mersenne primes in $\mathbb{Z}$, quadratic Mersenne norms have the following properties: \subsection{Properties of \texorpdfstring{$N(M_{p,\alpha})$}{N(Mp,a)}} \begin{enumerate} \item If $N(M_{n,\alpha})$ is prime, then $n$ is prime. \item The sequence $\{N(M_{n,\alpha})\}_{n=1}^{\infty}$ is an increasing sequence of integers that starts at $1.$ \item If $d$ divides $n$ then $M_{d,\alpha}$ divides $M_{n,\alpha}$ in $\mathbb{Q}(\sqrt{d})$ and $N(M_{d,\alpha})$ divides $N(M_{n,\alpha})$. \item If $d$ and $n$ are relatively prime then $M_{d,\alpha}$ is relatively prime to $M_{n,\alpha}$ in $\mathbb{Q}({\sqrt{d}})$ and $N(M_{n,\alpha})$ is relatively prime to $N(M_{d,\alpha})$. \end{enumerate} Experimental evidence shows that Mersenne primes are sparse in $\mathbb{Q}(\sqrt{d})$ for $d \equiv 1$ (mod ${4}$). Some interesting properties of Mersenne primes in $\mathbb{Q}(\sqrt{2})$ are given below. \subsection{Properties of \texorpdfstring{ $N(M_{p,\alpha})$ in $\mathbb{Q}(\sqrt{2})$} {N(Mp,a) in Q({2}}} \begin{enumerate} \item Since $\alpha=1+u\,=\,2+\sqrt{2}\,=\,u\sqrt{2}$, where $u$ is the fundamental unit, we have $ \alpha^{n}=a_{n}+b_{n}\sqrt{2}=u^{n}(\sqrt{2})^{n},$ for any integer $n>0$ and $a_{n},b_{n}\in \mathbb{Z}$. A small calculation also reveals that \begin{equation*} \alpha^{n}= \begin{cases} (2^{\frac{n-1}{2}}\sqrt{2})u^{n}, & \text{ if } n \text{ is odd};\\ 2^{\frac{n}{2}}u^{n}, & \text{ if } n \text{ is even}.\\ \end{cases} \end{equation*} Rewriting this, \begin{equation*} \alpha^{n}= \begin{cases} (2^{\frac{n-1}{2}}\sqrt{2}) (v_{n}+w_{n}\sqrt{2}), & \text{ if } n \text{ is odd}, \; w_{n}, v_{n}\in \mathbb{Z}; \\ 2^{\frac{n}{2}}(v_{n}^{\prime}+w_{n}^{\prime}\sqrt{2}), & \text{ if } n \text{ is even},\; v_{n}^{\prime},w_{n}^{\prime}\in \mathbb{Z}. \end{cases} \end{equation*} It can be noted that $w_{n}$, the coefficient of $\sqrt{2}$ in $u^{n}$ is odd if $n$ is odd. Further, $\,2^{\frac{n+1}{2}}w_{n}=a_{n}$ and $2^{\frac{n-1}{2}}v_{n}=b_{n}$. And $w_{n}^{\prime}$, the coefficient of $\sqrt{2}$ in $u^{n}$ is even if $n$ is even. Further, $2^{\frac{n}{2}}v_{n}^{\prime}=a_{n}\;$ and $\;2^{\frac{n}{2}}w_{n}^{\prime}=b_{n}$. For $n$ odd, we have $ N(u)^{n}=-1$, so \[N(\alpha^{n})=N(2^{\frac{n-1}{2}}\sqrt{2})N(u)^{n}= N(2^{\frac{n-1}{2}})N(\sqrt{2}) (-1)^{n}=2^{n-1}(-2)(-1)\,=\,2^{n}.\] For $n$ even, $ N(u)^{n}=1$, and \[N(\alpha^{n})=N(2^{\frac{n}{2}})N(u)^{n}=N(2^{\frac{n}{2}})( 1)=2^{n}. \] \item For any odd prime $p,$ let $\alpha^{p}= (2^{\frac{p-1}{2}}\sqrt{2}) (v_{p}+w_{p}\sqrt{2})$. Then \begin{eqnarray*} N(\alpha^{p}-1) &=& (2^{\frac{p+1}{2}}w_{p}-1)^{2}-2(2^{\frac{p-1}{2}}{v_{p}})^{2} \\ &=& (2^{p+1}{w_{p}}^{2}+1-2^{\frac{p+3}{2}}w_p)-2^{p}{v_p}^{2} \\ &=& 2^{p}(2{w_{p}}^{2}-{v_p}^{2})-2^{\frac{p+3}{2}}w_p +1 \\ &=& 2^{p} -2^{\frac{p+3}{2}}w_p +1. \end{eqnarray*} Thus, \[N(M_{p,\alpha})=\, 2^{\frac{p+3}{2}}w_{p}-2^{p}-1. \] \item As already pointed out, $a_{p}$ has a factor of $2^{\frac{p+1}{2}}$. Hence $2a_{p}\equiv 0$ (mod ${4}$). This further implies that, $N(M_{p,\alpha})\equiv -1$ (mod ${4}$) for $p\geq2$ and $N(M_{p,\alpha})\equiv -1$ (mod ${8}$) for $p>2.$ The next three properties are consequences of quadratic reciprocity, and $\left(\frac{\cdot}{\cdot}\right)$ denotes the $Legendre$ symbol. \item Let $p$ be an odd prime and $p\equiv \pm 1$ (mod ${8}$. Then \begin{displaymath}2^{\frac{p+3}{2}}=2^{2}2^{\frac{p-1}{2}}\equiv 4\pmod{p}.\end{displaymath} If $p\equiv \pm 3$ (mod ${8}$), then \begin{displaymath} 2^{\frac{p+3}{2}}=2^{2}2^{\frac{p-1}{2}}\equiv -4\pmod{p}.\end{displaymath} Combining the above we get \begin{equation*} N(M_{p,\alpha}) \equiv \begin{cases} 4w_{p}-3 \ {\rm (mod\ } p), & \text{ if } p \equiv \pm 1\ ({\rm mod} \ 8); \\ -4w_{p}-3 \ {\rm (mod\ } p) , & \text{ if } p \equiv \pm 3\ ({\rm mod } \ 8). \end{cases} \end{equation*} \item If $N(M_{p,\alpha})$ is a rational prime and $q$ is any other prime then \begin{equation*} \left(\frac{N(M_{p,\alpha})}{q}\right)\left(\frac{q}{N(M_{p,\alpha})}\right)= \begin{cases} 1, & \text{ if } q\equiv 1\ ({\rm mod } \ 4); \\ -1, & \text{ if } q\equiv 3\ ({\rm mod } \ 4). \end{cases} \end{equation*} \item If $N(M_{p,\alpha})$ is a rational prime then $\left( \frac{2}{N(M_{p,\alpha})}\right)=1$ since $N(M_{p,\alpha})\equiv -1$ (mod ${8}$). Hence \mbox{$\sqrt{2} \in \mathbb{F}_{N(M_{p,\alpha})}$} the finite field with \mbox{$N(M_{p,\alpha})$} elements. \end{enumerate} \section{Testing for primality} Several primality tests are available and some are specially designed for special numbers, an example being the famous Lucas-Lehmer test for the usual Mersenne primes. We show that the generalized Mersenne numbers of $\mathbb{Q}(\sqrt{2})$ can be put in a special form, so that, recent primality tests can be used to determine whether they are prime. Now, \[N(M_{p,\alpha})= \,2^{\frac{p+3}{2}}w_{p}-2^{p}-1,\] or \[N(M_{p,\alpha})=2^{\frac{p+3}{2}}(w_{p}-2^{\frac{p-3}{2}})-1.\] Since $w_{p}$ is odd, $(w_{p}-2^{\frac{p-3}{2}})$ is odd for $p>3.$ For $ p>3,$ \[ N(M_{p,\alpha})=h.2^{\frac{p+3}{2}}-1, \quad where \quad h=(w_{p}-2^{\frac{p-3}{2}}), \quad odd. \] An algorithm to test the primality of numbers of the form $h\cdot2^{n}\pm1$, for any odd integer $h$ such that, $h\neq 4^{m}-1$ for any $m$ is described in Bosma~\cite{BW}. It can be noted that, $h$ is not equal to $4^{m}-1$ in $M_{p,\alpha}$ for any $m$. Hence this algorithm can be used to test the primality of $M_{p,\alpha}$. \section{Primes of the form \texorpdfstring{$x^{2}+7y^{2}$}{x2+7y2}} The problem of representing a prime number by the form $x^{2}+ny^{2}$, where $n $ is any fixed positive integer dates back to Fermat. This question was best answered by $Euler$ who spent $40$ years in proving Fermat's theorem and thinking about how they can be generalized, he proposed some conjectures concerning $p=x^{2}+ny^{2}$, for $n>3$. These remarkable conjectures, among other things, touch on quadratic forms and their composition, genus theory, cubic and biquadratic reciprocity. Refer Cox~\cite{DC} for a thorough treatment. Euler became intensely interested in this question in the early 1740's and he mentions numerous examples in his letters to Goldbach. One among several of his conjectures stated in modern notation is \begin{displaymath} \left(\frac{-7}{p}\right)=1\Longleftrightarrow p\equiv 1,9,11,15,23,25 \pmod{28}. \end{displaymath} The following lemma gives necessary and sufficient condition for a number $m$ to be represented by a form of discriminant $D$. \begin{lemma} Let $D\equiv 0,1$ (mod ${4}$) and $m$ be an integer relatively prime to $D.$ Then $m$ is properly represented by a primitive form of discriminant $D$ if and only if $D$ is a quadratic residue modulo $m$. \end{lemma} As a corollary, we have the following: \begin{corollary} Let $n$ be an integer and $p$ be an odd prime not dividing $n.$ Then \(\left(\frac{-n}{p}\right)=1\) if and only if $p$ is represented by a primitive form of discriminant $-4n.$ \end{corollary} In 1903, Landau proved a conjecture of Gauss (theorem \ref{th7} below). Let $h(D)$ denote the number of classes of primitive positive definite forms of discriminant $D$, i.e., $h(D)$ is equal to the number of reduced forms of discriminant $D$. \begin{theorem}\label{th7} Let $n$ be a positive integer. Then \begin{center} $h(-4n) =1 \Leftrightarrow n= 1,2,3,4\,\,or\, 7$. \end{center} \end{theorem} One may note that, $x^{2}+ny^{2}$ is always a reduced form with discriminant $-4n.$ In this paper we consider the case $n=7$ and represent $N(M_{p,\alpha})$ in the form $x^{2}+7y^{2}$ whenever $M_{p,\alpha}$ is a Mersenne prime in $\mathbb{Q}(\sqrt{2})$. $x^{2}+7y^{2}$ is the only reduced form of discriminant $-28$, and it follows that \begin{displaymath} p\,= \, x^{2}+7y^{2} \Longleftrightarrow p\equiv 1,9,11,15,23,25\pmod{28}. \end{displaymath} for primes $p\neq 7$. The special property of the usual Mersenne primes over $\mathbb{Z}$ referred to in the beginning, Lenstra and Stevenhagen~\cite{LS}, has the following generalization over $\mathbb{Q}(\sqrt{2})$: \begin{theorem}\label{main-th} If $N(M_{p,\alpha})$ is a rational prime, with $\alpha=2+\sqrt{2}$, then $N(M_{p,\alpha})$ is always a quadratic residue $\pmod{7}$, and hence it can be written as $x^{2}+7y^{2}$. Also, $x$ is divisible by $8$, and $y \equiv \pm 3 \pmod{8}$. \end{theorem} The detailed proof is given in Palimar~\cite{thes}, using Artin's reciprocity law. Here we prove the theorem in two stages: first we show that $N(M_{p,\alpha})$ is always a quadratic residue (mod ${7}$). Next, we give an outline of the proof that $x$ is divisible by $8$, and $y \equiv \pm 3$ (mod ${8}$). The first few Mersenne primes in $\mathbb{Q}(\sqrt{2})$ with $\alpha=2+\sqrt{2},$ as well as the representations of their norms as $x^{2}+7y^{2}$ is given in Table~\ref{table7}. \begin{table}[H] \centering \begin{tabular}{|l|l|l|l|} \hline $p$ &$M_{p,\alpha}$ &$x^{2}+7y^{2}$ \\ \hline $5$ &$431$ &$16^{2}+7\cdot5^{2}$ \\ \hline $7$ &$5279$ &$64^{2}+7\cdot13^{2}$ \\ \hline $11$ &$732799$ &$856^{2}+7\cdot3^{2}$ \\ \hline \end{tabular} \caption{Norms of Mersenne primes in $\mathbb{Q}(\sqrt{2})$ in the form $x^{2}+7y^{2}$. } \label{table7} \end{table} For $p=73$, \[ N(M_{p,\alpha})=851569055172258793218602741480913108991 =\] \[(28615996544447548272)^{2}+7\cdot(2161143775888286749)^{2}\] For $p=\,89$, \[N(M_{p,\alpha}) = 290315886781191681464330388772329064268797313023 =\] \[(363706809248848497658560)^{2}+7\cdot(150253711001099458172317)^{2}\] For $p=\,233$, \[ N(M_{p,\alpha})= 1806047542728202303336800123116644178473780689153780654706531416\] \[7911959518498581747712829157156517940837234519177963497324543.\] The corresponding representation is \[(86527345603258677818378326573842407929031070590321223524182584)^{2} +\] \[7\cdot(38865140256563104639356290982349294477380709218952585423373629)^{2}.\] We now show that, if $N(M_{p,\alpha})$ is a prime then, $N(M_{p,\alpha})$ can be written as $x^{2}+7y^{2}$. Since $N(M_{p,\alpha})= 2^{\frac{p+3}{2}}w_{p}-2^{p}-1,$ representing a prime in the form $x^{2}+7y^{2}$ depends on $w_{p}$. Now, we find the values of $v_{p}$ and $w_{p}$ (mod ${7}$). As we know, for any odd $n$, $N(u^{n})\,=\,v_{n}^{2}-2 w_{n}^{2}\,=-1$. If $u^{n}=v_{n} +w_{n}\sqrt{2}$, then $v_{n}$ and $w_{n}$ satisfy the following recursions: $v_{n+1}=v_{n}+2w_{n}$ and $w_{n+1}=v_{n}+w_{n}$, with initial conditions $v_{1}=1,\,w_{1}=1$. The above recursions can be used to show that $v_{n}$ and $w_{n}$ satisfy the following: \begin{equation*} v_{n+2}=3v_{n}+4w_n \, ,\, w_{n+2}=2v_n+3w_n; \end{equation*} \begin{equation*} v_{n+3}=7v_n+ 10w_n \, , \, w_{n+3}=5v_n+10w_n; \end{equation*} \begin{equation*} v_{n+4}=17v_n+24w_n \,,\,w_{n+4}=12v_n+17w_n; \end{equation*} \begin{equation*} v_{n+5}=41v_n+58w_n \,,\, w_{n+5}=29v_n+41w_n; \end{equation*} \begin{equation*} v_{n+6}=99v_n+140w_n \,,\, w_{n+6}=70v_n+99w_n; \end{equation*} From the above one may also easily obtain the following congruences: \begin{equation*} \{v_{6k+1}\}\equiv1 \pmod{7} \; ,\;\{w_{6k+1}\}\equiv1 \pmod{7}; \end{equation*} \begin{equation*} \{v_{6k+5}\}\equiv6 \pmod{7}\; ,\; \{w_{6k+5}\}\equiv1 \pmod{7}. \end{equation*} Since only odd prime powers greater than $3$ are considered, we have listed only the congruences for indices congruent to $\pm 1 \pmod{6}$. Hence \begin{equation} \label{eq2} N(M_{p,\alpha})=2^{\frac{p+3}{2}}w_{p}-2^{p}-1\equiv 2^{\frac{p+3}{2}}- 2^{p}-1 \pmod{7}. \end{equation} Let us solve equation (\ref{eq2}) for $p>3$. If $p=3k+1$ then, $2^{p}\equiv 2 \pmod{7}$ and $2^{\frac{p+3}{2}}\equiv 4 \pmod{7}$, so\begin{displaymath}N(M_{p,\alpha})\equiv 1 \pmod{7}.\end{displaymath} If $p=3k+2$ then, $2^{p}\equiv 4 \pmod{7}$ and $2^{\frac{p+3}{2}}\equiv 2 \pmod{7}$, so \begin{displaymath}N(M_{p,\alpha})\equiv 4\pmod{7}.\end{displaymath} Thus in both cases $N(M_{p,\alpha})$ can always be represented as $x^{2}+7y^{2}$. To prove theorem \ref{main-th} we need the following lemma. \begin{lemma} If $N(M_{p,\alpha})$ is a rational prime and $N(M_{p,\alpha})=x^{2}+7y^{2}$, then $x\equiv 0 \pmod {4}$, and $y\equiv \pm 3 \pmod {8}$. \end{lemma} \begin{proof} From the previous discussion, we know \begin{equation} \label{lem-eq} N(M_{p,\alpha})=x^{2}+7y^{2}. \end{equation} But $N(M_{p,\alpha})=2^{\frac{p+3}{2}}w_{p}-2^{p}-1.$ Clearly we may take $p>6$. So either $p=6k+1$ or $p=6k+5$. If $p=6k+1$, then \begin{equation} \label{lem-eq1} N(M_{p,\alpha})=2^{\frac{6k+4}{2}}w_p-2^{6k+1}-1\, \equiv -1 \equiv 7\pmod{8}. \end{equation} If $p=6k+5$, then also, \begin{equation} \label{lem-eq2}N(M_{p,\alpha})=2^{\frac{6k+5}{2}}w_p-2^{6k+5}-1\equiv 7\pmod{8}.\end{equation} But right hand side of equation (\ref{lem-eq}) is $x^{2}+7y^{2}$. We show that $x$ must be even and $y$ odd. For, if $x$ is odd and $y$ is even, then $x^{2}\equiv 1$ (mod ${8}$) and either $y^{2}\equiv 0$ (mod ${8}$) or $y^{2}\equiv 4$ (mod ${8}$). If $y^{2}\equiv 0$ (mod ${8}$), then $x^{2}+7y^{2}\equiv 1$ (mod ${8}$) contradicting equations (\ref{lem-eq1}) and (\ref{lem-eq2}); and if $y^{2}\equiv 4$ (mod ${8}$), then \begin{displaymath}x^{2}+7y^{2}\equiv 1+7.4\equiv 5\pmod{8},\end{displaymath} again contradicting equations (\ref{lem-eq1}) and (\ref{lem-eq2}). Thus $x$ is even and $y$ is odd. Hence by equation (\ref{lem-eq}) \begin{displaymath} 7\equiv x^{2}+7y^{2}\equiv x^{2}+7\pmod{8}, \end{displaymath} since $y^{2}\equiv 1$ (mod ${8}$) and so, $x^{2}\equiv 0$ (mod ${8}$) implying $x\equiv 0$ (mod ${4}$). We now prove that $y\equiv \pm3$ (mod ${8}$) Let $p\equiv 1$ (mod ${6}$). From equation (\ref{lem-eq1}) \begin{displaymath} N(M_{p,\alpha})=x^{2}+7y^{2} =2^{\frac{6k+4}{2}}w_p-2^{6k+1}-1. \end{displaymath} Reducing modulo $16$, we get $N(M_{p,\alpha})\equiv -1$ (mod ${16}$). But $N(M_{p,\alpha})=x^{2}+7y^{2}$ and $x\equiv 0$ (mod ${4}$). Hence $7y^2\equiv -1$ (mod ${16}$), yielding $y^{2}\equiv 9$ (mod ${16}$). This proves that $y\equiv \pm3 $ (mod ${8}$). The same result follows from equation(\ref{lem-eq2}) when $p\equiv 5$ (mod ${6}$). \end{proof} \begin{proof}[Outline of the proof of Theorem~\ref{main-th}] We now show that $x\equiv 0$ (mod ${8}$). Virtually, the proof given in Lenstra and Stevenhagen~\cite{LS} carries over word-for-word, and so, we merely give an outline. All details and notation are as in Lenstra and Stevenhagen~\cite{LS}. By definition, $N(M_{p,\alpha})=\frac{(2+\sqrt{2})^p-1}{1+\sqrt{2}}. \frac{(2-\sqrt{2})^p-1}{1-\sqrt{2}}.$ Denote the two factors on the right by $v_p$ and $\bar v_p$. It is easy to see that $v_p$ and $\bar v_p$ are both totally positive. We compute the Artin symbols of $v_{p}{\mathbb{Z}}_E$ and $\bar v_{p}{\mathbb{Z}}_E$, and show that they are both trivial. We need to consider only two cases: $p\equiv 1$ (mod ${6}$) and $p\equiv 5$ (mod ${6}$). Since $\sqrt{2}\equiv 3,4$ (mod ${7}$), by taking $\sqrt{2}=4$ in $v_p$ and $\sqrt{2}=3$ in $\bar v_p$, a straightforward computation shows that, $v_p\equiv 1$ (mod ${7}$) and $\bar v_p\equiv 1$ (mod ${7}$). This completes the proof. \end{proof} \section{Acknowledgments} The authors gratefully acknowledge the kind help and encouragement given by Prof. Chandan Singh Dalawat during their visit to Harish-Chandra Research Institute, Allahabad, India, in December 2011. His lucid explanation of Artin reciprocity and short introduction to Class field theory was of great help. \begin{thebibliography}{99} \bibitem{BI} Pedro Berrizbeitia and Boris Iskra, Gaussian and Eisenstein Mersenne primes, \emph{Math. Comp.} {\bf 79} (2010), 1779--1791. \bibitem{BW} Wieb Bosma, Explicit primality criteria for $h.2^{k}\pm1$, \emph{Math. Comp.} {\bf 61} (1993), 97--109. \bibitem{cohen} Henri Cohen, {\it A Course in Computational Algebraic Number Theory}, Springer-Verlag, 1993. \bibitem {DC} David A. Cox, {\it Primes of the form $x^{2}+ny^{2}$}, Wiley -Interscience, 1989. \bibitem{LS} Hendrik W. Lenstra and Peter Stevenhagen, {Artin reciprocity and Mersenne primes}, \emph{Nieuw Arch. Wiskd. (5)} {\bf 1 } (2000), 44--54. \bibitem {WM} Wayne L. McDaniel, {Perfect Gaussian integers}, \emph{Acta Arith.} {\bf 25} (1974), 137--144. \bibitem{WL} Wayne L. McDaniel, {An analogue in certain unique factorization domains of the Euclid-Euler theorem on perfect numbers}, \emph{Int. J. Math. Math. Sci.} {\bf 13} (1990), 13--24. \bibitem{thes} Sushma Palimar, {Computations in $p$-adic discrete dynamics and real quadratic fields}, Ph.D Thesis, National Institute of Technology Karnataka, Surathkal, India, 2012. \bibitem {RS} Robert Spira, {The complex sum of divisors}, \emph{Amer. Math. Monthly} {\bf 68} (1961), 120--124. \bibitem {WS} William Stallings, {\it Cryptography and Network Security: Principles and Practices}, Prentice-Hall, Inc., 2006. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11R04. \noindent \emph{Keywords: } Mersenne primes, Artin's reciprocity law. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A033207}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received May 2 2012; revised version received May 21 2012. Published in {\it Journal of Integer Sequences}, June 11 2012. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}