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\begin{center}
\vskip 1cm{\LARGE\bf Representation of Integers of the Form \\
\vskip .1in
$x^2+ky^2-lz^2$}
\vskip 1cm
\large
Supawadee Prugsapitak and Nattaporn Thongngam\\
Algebra and Applications Research Unit \\
Division of Computational Science \\
Faculty of Science \\
Prince of Songkla University\\
Hatyai, Songkhla 90110\\
Thailand \\
\href{mailto:supawadee.p@psu.ac.th}{\tt supawadee.p@psu.ac.th} \\
\href{mailto:6310220031@psu.ac.th}{\tt 6310220031@psu.ac.th}\\
\end{center}

\vskip .2in

\begin{abstract}
A positive integer $l$ is $k$-special if for every integer $n$ there exist
non-zero integers $x, y, $ and $z$ such that $n = x^2+ky^2-lz^2$. For
all odd integers $k$, we find infinite classes of $k$-special numbers
and $2k$-special numbers.  \end{abstract}

\section{Introduction}
\label{sec:introduction}

The representations of a natural number as a sum of squares have been widely studied by many mathematicians.  Nowicki \cite{SM1} and  Lam \cite{SM2} proved that all natural numbers can be written in the form $x^2+y^2-cz^2$ where $xyz \neq 0$ if and only if $c$ is of the form $q$ or $2q$, where either $q=1$ or $q$ is a product of primes of the form $4m+1$. Later, Prugsapitak and Thongngam \cite{PT} defined  a $k$-special number. A positive integer $l$ is $k$-special if every integer $n$ can be expressed as $n=x^2+ky^2-lz^2$ for some non-zero integers $x, y, $ and $z$. They proved that $1$ is $k$-special if and only if $k$ is not divisible by $4$.

In this article, we will be interested in representing natural numbers in the form $x^2+ky^2-lz^2$ where $xyz \neq 0$ for given positive integers $k$ and $l$. For all odd integers $k$, we apply Lam's method \cite{SM2} to identify $k$-special numbers and $2k$-special numbers and additionally we show that there are infinitely many $k$-special numbers and $2k$-special numbers.

\section{Main results}
\label{sec:Main results}

In this section, we present $k$-special and $2k$-special number for an odd integer $k$.  We can show that $x^2+2ky^2$ where $\gcd(x, 2ky)=1$ and $x^2+ky^2$ where $\gcd(x, ky)=1$ are $2k$-special and $k$-special respectively by modifying Lam's approach \cite{SM2}. 

\begin{theorem}\label{2k-special}
Let $k$ and $l$ be odd integers. If $l$ can be written as $x^2+2ky^2$ for some positive integers $x$ and $y$ where $\gcd(x, 2ky)=1$, then $l$ is $2k$-special.
\end{theorem}

\begin{proof} 
Let $l$ be an odd integer and $l=x^2+2ky^2$ for some positive integers $x$ and $y$ where $\gcd(x, 2ky)=1$. We first find the representation of odd numbers of the form $a^2+2kb^2-lc^2$ where $abc \neq 0$. Since $\gcd(x, 2ky)=1$, there exist integers $\alpha_0$ and $\beta_0$ such that 
\begin{align}
x\alpha_0+2ky\beta_0=1. \nonumber
\end{align}
For a positive integer $n$, let $\alpha_n=\alpha_0+2nky$ and $\beta_n=\beta_0-nx$. It is easy to see that 
\begin{align}
x\alpha_n+2ky\beta_n=x(\alpha_0+2nky)+2ky(\beta_0-nx)=x\alpha_0+2ky\beta_0=1.\nonumber
\end{align}
Next, let $a_n=xj+\alpha_n$, $b_n=yj+\beta_n$ and $c_n=j$, where $j$ is an integer which will be chosen later.
Thus
\begin{align}
a_n^2+2kb_n^2-lc_n^2&amp;=(xj+\alpha_n)^2+2k(yj+\beta_n)^2-(x^2+2ky^2)j^2 \nonumber\\
&amp;=x^2j^2+2xj\alpha_n+\alpha_n^2+2ky^2j^2+4ykj\beta_n+2k\beta_n^2-x^2j^2-2ky^2j^2 \nonumber\\
&amp;=2xj\alpha_n+4ykj\beta_n+\alpha_n^2+2k\beta_n^2 \nonumber\\
&amp;=2j(x\alpha_n+2ky\beta_n)+\alpha_n^2+2k\beta_n^2. \nonumber
\end{align}
Since $x$ is odd and $x\alpha_n +2ky\beta_n=1$, we can see that $\alpha_n$ is odd. Thus
\begin{align}
a_i^2+2kb_i^2-lc_i^2=2j +\alpha_i^2+2k\beta_i^2 \nonumber
\end{align}
for non-negative integers $i$.
Let $m$ be an odd integer. For a non-negative integer $n$, we choose a suitable value of an integer $j_n$ such that 
\begin{align}m=2j_n+\alpha_n^2+2k\beta_n^2=a_n^2+2kb_n^2-lc_n^2 \nonumber
\end{align}
where $a_n=xj_n+\alpha_n$, $b_n=yj_n+\beta_n$ and $c_n=j_n$.
We can see that $a_nb_nc_n=0$ if and only if $m$ is one of the following values; $\alpha_n^2+2k\beta_n^2, \alpha_n^2+2k\beta_n^2-{2\alpha_n}/{x}$ or $\alpha_n^2+2k\beta_n^2-{2\beta_n}/{y}$.
Since 
\begin{align}
    \lim_{n \to \infty}\alpha_n  =\lim_{n \to \infty} -\beta_n=\infty, \nonumber
\end{align}
there exists a non-negative integer $n$ such that $\alpha_n^2+2k\beta_n^2-{ 2\alpha_n}/{x}&gt;m$ and $\beta_n&lt;0$.
Therefore we obtain a representation for $m$, namely $m=a_n^2+2kb_n^2-lc_n^2$ where $a_nb_nc_n \neq 0$.


We next find the representation of even integers.
Since $\gcd(x, 2ky)=1$, we have $\gcd(x, ky)=1$. Then there exist integers $\alpha_0$ and $\beta_0$ such that 
\begin{align}
x\alpha_0+ky\beta_0=1. \nonumber
\end{align}
For a positive integer $n$, let $\alpha_n=\alpha_0+kny$ and $\beta_n=\beta_0-nx$. Then
\begin{align}
x\alpha_n+ky\beta_n&amp;=x(\alpha_0+kny)+ky(\beta_0-nx)=x\alpha_0+kxny+ky\beta_0-kynx
=x\alpha_0+ky\beta_0=1 \nonumber.
\end{align}
Now let $a_n=xj+2\alpha_n$, $b_n=yj+\beta_n$ and $c_n=j$ where $j$ is an integer which will be determined later. 
Then
\begin{align}
a_n^2+2kb_n^2-lc_n^2 \nonumber
&amp;=(xj+2\alpha_n)^2+2k(yj+\beta_n)^2-(x^2+2ky^2)j^2 \nonumber\\
&amp;=x^2j^2+4xj\alpha_n+4\alpha_n^2+2ky^2j^2+4ykj\beta_n+2k\beta_n^2-x^2j^2-2ky^2j^2 \nonumber\\
&amp;=4j(x\alpha_n+ky\beta_n)+4\alpha_n^2+2k\beta_n^2. \nonumber
\end{align}
Since $x$ is odd, we have
\begin{align}
4\alpha_n^2+2k\beta_n^2
&amp;=4(\alpha_0+kny)^2+2k(\beta_0-nx)^2 \nonumber\\
&amp;=4\alpha_0^2+8kny\alpha_0+4k^2n^2y^2+2k\beta_0^2-4knx\beta_0+2kn^2x^2 \nonumber\\
&amp;\equiv 2k\beta_0^2+2kn^2x^2 \pmod 4 \nonumber\\
&amp; \equiv 2k\beta_0^2+2kn^2 \pmod 4 \nonumber.
\end{align}
Thus,
\begin{displaymath}
4\alpha_n^2+2k\beta_n^2 \equiv
\begin{cases} 
2k\beta_0^2 \pmod 4, &amp;\text{if $n$ is even;} \\
2k(\beta_0^2+1) \pmod 4, &amp;\text{if $n$ is odd.}
\end{cases}
\end{displaymath}

For every non-negative integer $r$,
 we obtain two identities for representing even integers given by 
\begin{align}
a_{2r}^2+2kb_{2r}^2-lc_{2r}^2=(xj+2\alpha_{2r})^2+2k(yj+\beta_{2r})^2-(x^2+2y^2)j^2=4j+4\alpha_{2r}^2+2k\beta_{2r}^2 \nonumber \end{align}
and
\begin{align}
a_{2r+1}^2+2kb_{2r+1}^2-lc_{2r+1}^2&amp;=(xj+2\alpha_{2r+1})^2+2(yj+\beta_{2r+1})^2-(x^2+2y^2)j^2 \nonumber \\&amp;=4j+4\alpha_{2r+1}^2+2k\beta_{2r+1}^2. \nonumber
\end{align}

Let $m$ be an even integer. There are four cases to consider; namely,
\begin{itemize}
\item {\it Case 1:} $m \equiv 0 \pmod {4}$ and $\beta_0^2 \equiv 0$ (mod $4$),

\item {\it Case 2}: $m \equiv 2 \pmod {4}$ and $\beta_0^2 \equiv 0$ (mod $4$),

\item {\it Case 3}: $m \equiv 0\pmod {4}$ and $\beta_0^2 \equiv 1$ (mod $4$),

\item {\it Case 4}: $m \equiv 2 \pmod {4}$ and $\beta_0^2 \equiv 1$ (mod $4$).
\end{itemize}


Since all cases can be handled similarly, we provide the proof for Case 1 as follows.
We choose a suitable value of an integer $j_{2r}$ such that 
\begin{align}
    m=4j_{2r}+4\alpha_{2r}^2+2k\beta_{2r}^2=a_{2r}^2+2kb_{2r}^2-lc_{2r}^2 \nonumber
\end{align}
where $a_{2r}=xj_{2r}+2\alpha_{2r}$, $b_{2r}=yj_{2r}+\beta_{2r}$ and $c_{2r}=j_{2r}$. We can see that $a_{2r}b_{2r}c_{2r} =0$ if and only if $m$ is one of the following values: $4\alpha_{2r}^2+2k\beta_{2r}^2, 4\alpha_{2r}^2+2k\beta_{2r}^2-{ 8\alpha_{2r}}/{x}$ or $4\alpha_{2r}^2+2k\beta_{2r}^2-{4\beta_{2r}}/{y}$. Since \begin{align}
\lim_{r \to \infty}\alpha_{2r}  =\lim_{r \to \infty} -\beta_{2r}=\infty, \nonumber
\end{align}
there exists a non-negative integer $r$ such that $4\alpha_{2r}^2+2k\beta_{2r}^2-{ 8\alpha_{2r}}/{x}&gt;m$ and $\beta_{2r}&lt;0$. Therefore we obtain a representation for $m$, namely $m=a_{2r}^2+2kb_{2r}^2-lc_{2r}^2$ where $a_{2r}b_{2r}c_{2r} \neq 0$.
\end{proof}

For an odd integer $k$, we obtain a sufficient condition for an odd integer to be $2k$-special. The necessary condition is unknown to us. However we can provide a necessary condition for the $2$-special number. That is, if $d$ is $2$-special, then $d=m^2+2n^2$ for some integers $m$ and $n$. We first recall a well-known lemma (see, for example, \cite{AW}) before proving our mentioned conclusion.

\begin{lemma}\label{lem3}
Let $p$ be a prime. Then $p=x^2+2y^2$ for some integers $x$ and $y$ if and only if $p=2$ or $p \equiv \modd{1,3} {8}$.
\end{lemma}


Using the above lemma, we have the following.
\begin{lemma}\label{lem4}
A positive integer $n$ can be written as $x^2+2y^2$ for some integers $x$ and $y$ if and only if all primes of the form $8k+5$ or $8k+7$ have an even exponent in the prime factorization of $n$.
\end{lemma}
\begin{proof}
If $n=x^2+2y^2$ for some integers $x$ and $y$, then \begin{align}
n=x^2+2y^2=(x+y\sqrt{-2})(x-y\sqrt{-2}). \nonumber
\end{align}
Let $p$ be a prime of the form $8k+5$ or $8k+7$ and $p$ divides $n$. Since
$-2$ is a quadratic non-residue modulo $p$, we see that $p$ is a prime in $\mathbb{Z}[\sqrt{-2}]$. Thus $p$ divides $x+y\sqrt{-2}$ or $p$ divides $x-y\sqrt{-2}$. If $p$ divides $x+y\sqrt{-2}$, then $p$ divides $x-y\sqrt{-2}$. Hence $p$ divides $2x$ and $p$ divides $2y$. Since $p$ is odd, we have that $p$ divides both $x$ and $y$. Similarly, we can show that if $p$ divides $x-y\sqrt{-2}$, then $p$ divides both $x$ and $y$. Write $x=px_1$ and $y=py_1$ for some integers $x_1$ and $y_1$. Therefore $n=p^2x_1^2+2p^2y_1^2=p^2(x_1^2+2y_1^2)$. If $p\nmid x_1^2+2y_1^2$, then $p^2||n$. If $p$ divides $x_1^2+2y_1^2$, then $p^2$ divides $x_1^2+2y_1^2$. We can repeat the process until we arrive at the conclusion that $p$ has even multiplicity in the prime factorization of $n$.
Conversely, we know that $2$ and all primes $p$ where $p \equiv 1, 3$ (mod $8$) can be written as $a^2+2b^2$ for some integers $a$ and $b$ and a product of integers of the form $a^2+2b^2$ is still an integer of the form $a^2+2b^2$. Therefore the result follows.
\end{proof} 


\begin{theorem}
If $d$ is $2$-special, then $d=m^2+2n^2$ for some integers $m$ and $n$.
\end{theorem}
\begin{proof}
Let $d$ be $2$-special. For all integers $c$ there exist non-zero integers $x, y, $ and $z$ such that $x^2+2y^2-dz^2=2dc^2$. So $x^2+2y^2=d(z^2+2c^2)$. By Lemma \ref{lem4}, all primes of the form $8k+5$ or $8k+7$ have even exponent in the prime factorization of $x^2+2y^2$ and $2c^2+z^2$. Thus all primes of the form $8k+5$ or $8k+7$ have even exponent in the prime factorization of $d$. Hence again by Lemma \ref{lem4}, we have that $d$ is of the form $m^2+2n^2$ for some integers $m$ and $n$.
\end{proof}


The converse of the above theorem is not true, as we will discover in the next theorem that $8$ is not $2$-special.

\begin{theorem}
Let $k$ be an odd integer. If $d$ is divisible by $8$, then $d$ is not $2k$-special.
\end{theorem}
\begin{proof}
Let $d$ be divisible by 8. Suppose on the contrary that $d$ is $2k$-special. 
Then
\begin{align}
 x^2+2ky^2-dz^2=5 \nonumber
\end{align}
 for some non-zero integers $x, y, $ and $z$. So $x^2+2ky^2 \equiv 5$ (mod $8$). 
Since $x^2 \equiv 0, 1, 4 $ (mod $8$) and $2ky^2 \equiv 0, 2k $ (mod $8$), it is easy to see that $x^2+2ky^2\equiv 0, 1, 4, 2k, 2k+1, 2k+4 $ (mod $8$). Since $k$ is odd, we can see that $x^2+2ky^2 \not\equiv 5$ (mod $8$). This is a contradiction.
\end{proof}
Next, we consider $k$-special numbers where $k$ is odd. 

\begin{theorem}\label{k-special}
Let $l$ and $k$ be odd integers. If $l=x^2+ky^2$ for some positive integers $x$ and $y$ and $\gcd(x, ky)=1$, then $l$ is $k$-special.
\end{theorem} 
\begin{proof}
Suppose $l=x^2+ky^2$ for some positive integers $x$ and $y$ where $\gcd(x,ky)=1.$ Since $\gcd(x,ky)=1$, there exist integers $\alpha_0$ and $\beta_0$ such that $x\alpha_0+ky\beta_0 =1$.

For a positive integer $n$, we define $\alpha_n=\alpha_{0}+nky$ and $\beta_n=\beta_{0}-nx$. 
Then
\begin{align}
x\alpha_n+ky\beta_n&amp;=x(\alpha_0+nky)+ky(\beta_0-nx) \nonumber\\
&amp;=x\alpha_0+xnky+ky\beta_0-nkyx \nonumber\\
&amp;=x\alpha_0+ky\beta_0=1.\nonumber
\end{align}

Let $a_n =xj+\alpha_n$, $b_n=yj+\beta_n$ and $c_n=j$, where $j$ is an integer which will be selected later. Thus
\begin{align*}
a_n^2+kb_n^2-lc_n^2
&amp;=(xj+\alpha_n)^2+k(yj+\beta_n)^2-(x^2+ky^2)j^2\\
&amp;=x^2j^2+2xj\alpha_n+\alpha_n^2+ky^2j^2+2kyj\beta_n+k\beta_n^2-x^2j^2-ky^2j^2\\
&amp;=2xj\alpha_n+\alpha_n^2+2kyj\beta_n+k\beta_n^2\\
&amp;=2j(x\alpha_n+ky\beta_n)+\alpha_n^2+k\beta_n^2.
\end{align*} 
We have
\begin{align}
\alpha_{n}^2+k\beta_{n}^2 \nonumber
&amp;=(\alpha_0+nky)^2+k(\beta_0-nx)^2 \nonumber\\
&amp;=\alpha_0^2+2kn \alpha_0y+n^2k^2y^2+k\beta_0^2-2nk\beta_0x+kn^2x^2 \nonumber\\
&amp;\equiv \alpha_0^2+k\beta_0^2+n^2k^2y^2+kn^2x^2 \pmod 2. \nonumber 
\end{align}
Therefore, 
\begin{displaymath}
\alpha_{n}^2+k\beta_{n}^2 \equiv
\begin{cases}
\alpha_0^2+k\beta_0^2 \pmod 2, &amp; \text{if $n$ is even;} \\
 \alpha_0^2+k\beta_0^2+y^2+x^2 \pmod 2, &amp;\text{if $n$ is odd.}
\end{cases}
\end{displaymath}
Since $l$ and $k$ are odd, we can see that $x$ and $ y$ have different parities.
Thus
\begin{displaymath}
\alpha_{n}^2+k\beta_{n}^2 
\equiv
\begin{cases}
\alpha_0^2+k\beta_0^2 \pmod 2, &amp;\text{if $n$ is even;} \\
 \alpha_0^2+k\beta_0^2+1\pmod 2, &amp;\text{if $n$ is odd.}
\end{cases}
\end{displaymath}
For all non-negative integers $r$, we obtain the following identities
\begin{align}
a_{2r}^2+kb_{2r}^2-lc_{2r}^2 &amp;=2j+\alpha_{2r}^2+k\beta_{2r}^2 \equiv
 2j+\alpha_0^2+k\beta_0^2 \pmod{2}, \nonumber\\
a_{2r+1}^2+kb_{2r+1}^2-lc_{2r+1}^2 &amp;=2j+\alpha_{2r+1}^2+k\beta_{2r+1}^2 \equiv 2j+\alpha_0^2+k\beta_0^2+1 \pmod{2}. \nonumber
\end{align}


By using both identities, we can demonstrate that all integers can be expressed in the form $a^2+kb^2-lc^2$. Next, we will illustrate how to choose $a,b,$ and $c$ so that $abc \neq 0$.

\bigskip

\noindent{\it Case 1}:  $\alpha_0^2+k\beta_0^2 \equiv 0\pmod 2$. 

We first consider how to represent an even integer $m$. We select an appropriate value for $j_{2r}$ such that $$m=2j_{2r}+\alpha_{2r}^2+k\beta_{2r}^2=a_{2r}^2+kb_{2r}^2-lc_{2r}^2$$ where $a_{2r}=xj_{2r}+\alpha_{2r}$, $b_{2r}=yj_{2r}+\beta_{2r}$ and $c_{2r}=j_{2r}$. We can see that $a_{2r}b_{2r}c_{2r} = 0$ if and only if $m$ is one of the following values: $\alpha_{2r}^2+k\beta_{2r}^2, \alpha_{2r}^2+k\beta_{2r}^2-{2\alpha_{2r}}/{x}$ or $\alpha_{2r}^2+k\beta_{2r}^2-{2\beta_{2r}}/{y}$. Since \[\lim_{r \to \infty}\alpha_{2r}  =\lim_{r \to \infty} -\beta_{2r}=\infty,\] there exists a non-negative integer $r$ for which $\alpha_{2r}^2+k\beta_{2r}^2-{ 2\alpha_{2r}}/{x}&gt;m$ and $\beta_{2r}&lt;0$. As a result, we have a representation for $m$, namely $m=a_{2r}^2+kb_{2r}^2-lc_{2r}^2$ where $a_{2r}b_{2r}c_{2r} \neq 0$.


Next, we consider a representation for an odd integer $m$. Let $m$ be an odd integer. We select an appropriate value for $j_{2r+1}$ such that $$m=2j_{2r+1}+\alpha_{2r+1}^2+k\beta_{2r+1}^2=a_{2r+1}^2+kb_{2r+1}^2-lc_{2r+1}^2$$ where $a_{2r+1}=xj_{2r+1}+\alpha_{2r+1}$, $b_{2r+1}=yj_{2r+1}+\beta_{2r+1}$ and $c_{2r+1}=j_{2r+1}$. It is easy to see that $a_{2r+1}b_{2r+1}c_{2r+1} = 0$ if and only if $m$ is one of the values listed below: $\alpha_{2r+1}^2+k\beta_{2r+1}^2, \alpha_{2r+1}^2+k\beta_{2r+1}^2-{2\alpha_{2r+1}}/{x}$ or $\alpha_{2r+1}^2+k\beta_{2r+1}^2-{2\beta_{2r+1}}/{y}$. Since \[\lim_{r \to \infty}\alpha_{2r+1}=\lim_{r \to \infty} -\beta_{2r+1}=\infty,\] there exists a non-negative integer $r$ such that $\alpha_{2r+1}^2+k\beta_{2r+1}^2-{ 2\alpha_{2r+1}}/{x}&gt;m$ and $\beta_{2r+1}&lt;0$. As a consequence, a representation for $m$ is obtained as follows: $m=a_{2r+1}^2+kb_{2r+1}^2-lc_{2r+1}^2$ where $a_{2r+1}b_{2r+1}c_{2r+1} \neq 0$.

\bigskip

\noindent{\it Case 2}: $\alpha_0^2+k\beta_0^2 \equiv 1\pmod 2$. This case can be handled similarly as in Case 1. Thus $l$ is $k$-special as desired.
\end{proof}

Finally, we obtain the following consequence of Theorem \ref{2k-special} and Theorem \ref{k-special}.

\begin{corollary}
Let $k$ be an odd integer. There are infinitely many $k$-special numbers and $2k$-special numbers. 
\end{corollary}





\section{Acknowledgments}

We would like to express our gratitude to the reviewers for their comments and suggestions on how to improve our manuscript.

\begin{thebibliography}{9}

\bibitem{AW} 
S. Alaca and K. S. Williams, {\it Introductory Algebraic Number Theory}, Cambridge University Press, 2004.

\bibitem{SM2} P. C. H. Lam, Representation of integers using {$a^2 + b^2 - dc^2$}, {\it J. Integer Sequences} {\bf 18} (2015),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL18/Lam/lam2.html}{Article 15.8.6}.

\bibitem{SM1}A. Nowicki,
The numbers {$a^2+b^2-dc^2$}, {\it J. Integer Sequences} {\bf 18} (2015),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL18/Nowicki/nowicki3.html}{Article 15.2.3}.

\bibitem{PT}S. Prugsapitak and N. Thongngam,
The numbers {$a^2+mb^2-c^2$}, {\it J. Integer Sequences} {\bf 24} (2021),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL24/Prugsapitak/prug3.html}{Article 21.7.7}.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11E25.

\noindent \emph{Keywords:} sum of squares.

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received  November 10 2021; 
revised versions received December 21 2021; April 15 2022; April 24 2022; October 18 2022; October 24 2022.
Published in {\it Journal of Integer Sequences}, October 25 2022.


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