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\begin{center}
\vskip 1cm{\LARGE\bf 
The Numbers $a^2 + b^2 - dc^2$
}
\vskip 1cm
\large
Andrzej Nowicki\\
Nicolaus Copernicus University\\
Faculty of Mathematics and Computer Science\\
87-100 Toru\'n \\
Poland \\
\href{mailto:anow@mat.uni.torun.pl}{\tt anow@mat.uni.torun.pl} \\
\end{center}

\vskip .2 in

\begin{abstract} 
We say that a positive integer $d$ is {\it special} if for every
integer $m$ there exist nonzero integers $a,b,c$ such that
$m=a^2+b^2-dc^2$.  In this note we present examples and some properties
of special numbers.  Moreover, we present an infinite sequence of
special numbers.
\end{abstract}

\section{Introduction}
Let $d$ be a positive integer.
If $a,b,c$ are integers, then let  $[a,b,c]_d$  denote
the number $a^2+b^2-dc^2$.
We  say that $d$ is {\it special} if for every integer $m$
there exist nonzero integers $a,b,c$ such that $m=[a,b,c]_d$.

We present examples and some properties of special numbers.
Moreover,  we present an infinite sequence of special numbers.

\section{The numbers $a^2 + b^2 - c^2$}
Observe that $0=[3,4,5]_1$  and
\begin{align*}
 -1 = [2, 2, 3]_1,  &amp; \ \  &amp;    1 = [1, 1, 1]_1,&amp; \ \ \  &amp;   -6 = [3, 1, 4]_1,&amp; \ \ &amp;  6 = [3, 1, 2]_1,\\
  -2 = [1, 1, 2]_1,&amp;&amp; 2 = [3, 3, 4]_1,&amp;&amp;   -7 = [1, 1, 3]_1,&amp;&amp; 7 = [2, 2, 1]_1,\\
  -3 = [3, 2, 4]_1,&amp;&amp; 3 = [6, 4, 7]_1,&amp;&amp;  -8 = [2, 2, 4]_1,&amp;&amp; 8 = [4, 1, 3]_1,\\
  -4 = [2, 1, 3]_1,&amp;&amp; 4 = [2, 1, 1]_1,&amp;&amp;  -9 = [6, 2, 7]_1,&amp;&amp; 9 = [3, 1, 1]_1,\\
  -5 = [4, 2, 5]_1,&amp;&amp; 5 = [5, 4, 6]_1,&amp;&amp;    -10 = [5, 1, 6]_1,&amp;&amp; 10 = [5, 1, 4]_1.
\end{align*}

One of the problems presented in \cite[Problem L25]{LeH}  states
that  every integer is of the form $[a,b,c]_1$, where $a,b,c$ are integers.
We will show that every integer is of the form $[a,b,c]_1$ where $a,b,c$
are nonzero integers.

\begin{proposition}\label{PropA}
The number $1$ is special, that is,
for every integer $m$ there exist nonzero integers $a,b,c$ such that
$
m=a^2+b^2-c^2.
$
\end{proposition}

\begin{proof}
It follows from the following equalities:
$$2k-1=[2,\ k-2, \ k-3]_1, \quad
2k=[k, \ 1, \ k-1]_1
$$
for $k\in\Bbb Z$, and $3=[6, 4, 7]_1$, $5 = [5, 4, 6]_1$,
$2 = [3, 3, 4]_1$.
\end{proof}
It is known \cite[p.\ 38]{AK}
that the equation
$x^2+y^2-z^2=3$
has infinitely many solutions in positive integers.
The  equation
$x^2+y^2-z^2=1997$
has also infinitely many  solutions in positive integers
\cite[p.\ 9]{St}.
In the next proposition we  show that
the same is true for every integer.

\begin{proposition}\label{PropB}
For every integer $m$ there are infinitely many triples $(a,b,c)$
of nonzero integers such that
$
m=a^2+b^2-c^2.
$
\end{proposition}

\begin{proof}
This is a consequence of the following two equalities.
\begin{eqnarray*}
2k-1&amp;=&amp;(2t)^2+(2t^2-k)^2-(2t^2-k+1)^2,\\
2k&amp;=&amp;(2t^2-2t-k)^2+(2t-1)^2-(2t^2-2t-k+1)^2,
\end{eqnarray*}
where $k,t$ are integers.
\end{proof}

\section{Properties of special numbers}
In this section we present some elementary properties of special numbers.
The following, well known lemma (see, for example, \cite{Sierpinski88}),
will play an important role.

\begin{lemma}\label{LemDwaKw}
A positive integer $m$ is a sum of two integer squares if and only if
all prime factors of $m$ of the form $4k+3$
have even exponent in the prime factorization of $m$.
\end{lemma}

Now we prove

\begin{proposition}\label{SumDkwa}
Every special number is a sum of two integer squares.
If a non-square positive integer $d$ is special, then
$d$ is a sum of two nonzero integer squares.
\end{proposition}

\begin{proof}
Let $d$ be a special number.
There exist nonzero integers  $a,b,c$ such that  $[a,b,c]_d=d$.
Thus, we have the equality
$$a^2+b^2=d(c^2+1),$$
which says that  $d(c^2+1)$ ia a sum of two squares.
Hence, by Lemma \ref{LemDwaKw},
all prime factors of $d(c^2+1)$ of the form $4k+3$
have even exponent in the prime factorization of $d(c^2+1)$.
Since $c^2+1$ is also a sum of two squares,
all prime factors of $d$ of the form $4k+3$
have even exponent in the prime factorization of $d$.
Hence, again by Lemma \ref{LemDwaKw},
$d$ is a sum of two integer squares.
Now it is also  clear that if additionally $d$ is non-square,
then $d$ is a sum of two nonzero integer squares.
\end{proof}

Note that $4=2^2+0^2$ is a sum of two integer squares and the number $4$
is not special.
The number $8=2^2+2^2$ is a sum of two nonzero squares
and $8$ is not special.
In general we have


\begin{proposition}\label{PodzCztery}
If a positive integer $d$ is divisible by $4$, then $d$ is not special.
\end{proposition}

\begin{proof}
Let $d=4k$ where $k$ is a positive integer, and assume that $d$ is special.
Then $a^2+b^2-dc^2=3$ for some nonzero integers $a,b,c$.
This implies that the number $a^2+b^2$ is of the form $4k+3$.
But  integers of the form $4k+3$ are not
sums of two squares. 
Thus the assumption that $d$ is special leads to a
contradiction.
\end{proof}

\begin{proposition}\label{PodzPrim}
If a positive integer $d$ is divisible by a prime number of the form $4k+3$,
then $d$ is not special.
\end{proposition}

\begin{proof}
Let $p$ be a prime number of the form $4k+3$.  Assume that $p\mid d$
and $d$ is special.
Then $d$ is a sum of two squares (by Proposition \ref{SumDkwa})
and this implies (by Lemma \ref{LemDwaKw}) that $p^2\mid d$.
Moreover, there exist nonzero integers $a,b,c$ such that $a^2+b^2-dc^2=p$.
In this case $p$ divides the sum of two squares $a^2+b^2$ and so,
again by Lemma \ref{LemDwaKw}, the integer  $a^2+b^2$ is divisible by $p^2$.
Hence, $p^2$ divides $p$. Thus the assumption that $d$ is special leads to a
contradiction.
\end{proof}

As a consequence of the above propositions we obtain the following theorem.

\begin{theorem}\label{TwSp}
Every special number is of the form $q$ or $2q$, where either $q=1$ 
or $q$ is a product of prime numbers of the form $4k+1$.
\end{theorem}

\begin{question}\label{Pyt}
Let $d=q$ or $d=2q$, where $q$ is a product of prime numbers of the form
$4k+1$. Is it true that $d$ is a special number?
\end{question}

We do not know the answer to  the above question.

\begin{proposition}\label{PellNiesk}
Let $d$ be a non-square positive integer and let
$m$ be an integer.
Assume that there exists a triple  $(a,b,c)$ of positive integers
such that $[a,b,c]_d=m$.
Then such triples $(a,b,c)$ are infinitely many.
\end{proposition}

\begin{proof}
Let $[a,b,c]_d=m$ for some positive integers $a,b,c$.
Then the Pell equation
$$ x^2-dz^2=m-b^2$$
has a solution in positive integers $(x,z)=(a,c)$.
It follows from the theory of Pell equations
\cite{Sierpinski88, Barb, NowPEL}
that
then this equation has infinitely many positive solutions.
Let   $(u,v)$ be such a solution.
Then the triple  $(u,b,v)$ is a  solution in positive integers
of the equation  $x^2+y^2-dz^2=m$.
\end{proof}

\section{Examples}
We already know that the number $1$ is special.
In this section we present the all special numbers smaller than $50$.

Consider the case $d=2$.
Let us recall that $[a,b,c]_2=a^2+b^2-2c^2$.
Observe that  $0=[1,1,1]_2$  and we have
\begin{align*}
  -1 = [4, 1, 3]_2 ,&amp; \ \ &amp; 1 = [8, 3, 6]_2,&amp; \ \ &amp;   -6 = [1, 1, 2]_2,&amp; \ \ &amp;6 = [2, 2, 1]_2, \\
 -2 = [12, 4, 9]_2,&amp;&amp; 2 = [3, 1, 2]_2,&amp;&amp;    -7 = [4, 3, 4]_2,&amp;&amp; 7 = [4, 3, 3]_2, \\
  -3 = [2, 1, 2]_2,&amp;&amp; 3 = [2, 1, 1]_2,&amp;&amp;    -8 = [3, 1, 3]_2,&amp;&amp; 8 = [3, 1, 1]_2, \\
  -4 = [8, 2, 6]_2,&amp;&amp; 4 = [16, 6, 12]_2,&amp;&amp; -9 = [5, 4, 5]_2,&amp;&amp; 9 = [4, 1, 2]_2, \\
  -5 = [3, 2, 3]_2,&amp;&amp; 5 = [3, 2, 2]_2,&amp;&amp;   -10 = [2, 2, 3]_2,&amp;&amp; 10 = [3, 3, 2]_2.
\end{align*}

\begin{proposition}\label{PropAdwa}
The number $2$ is special, that is,
for every integer $m$ there exist nonzero integers $a,b,c$ such that
$m=a^2+b^2-2c^2$.
\end{proposition}

\begin{proof}
This is a consequence of the  equalities
$
2k-1=[k-1, \ k, \ k-1]_2,\quad\\
4k=[k-1, \ k+1, \ k-1]_2,\quad
4k+2=[k-3, \ k+1, \ k-2]_2$
(where $k$ is an integer),
and
$1=[8,3,6]_2$,  $-1=[4,1,3]_2$,
$-4=[9,2,6]_2$. \ $4=[16,6,12]$,
 \ $-2=[12,4,9]_2$, \ $10= [3,3,2]_2$,  \  $14=[4,4,3]_2$.
\end{proof}

Note the following consequence of Propositions \ref{PropAdwa} and
\ref{PellNiesk}.

\begin{proposition}\label{DwaNiesk}
For every integer $m$ there are infinitely many triples $(a,b,c)$,
of nonzero integers such that
$
m=a^2+b^2-2c^2.
$
\end{proposition}

\begin{example}\label{Exa21}
Some solutions $(x,y,z)$ of the equation
$ x^2+y^2-2z^2=1:$
\begin{align*}
(8, 3, 6),&amp; \ &amp; (15, 8, 12),&amp; \ &amp; (24, 15, 20),&amp; \ &amp; (33, 8, 24),&amp; \ &amp; (35, 24, 30),\\
(48, 3, 34),&amp;&amp;  (48, 17, 36),&amp;&amp; (48, 35, 42),&amp;&amp;(63, 48, 56),&amp;&amp;(72, 33, 56),\\
(72, 15, 52),&amp;&amp; (80, 63, 72),&amp;&amp;(93, 8, 66),&amp;&amp; (93, 48, 74),&amp;&amp; (99, 80, 90).
\end{align*}
\end{example}

\begin{example}\label{exa22}
For every integer $a$ we have $[a+2,a,a+1]_2=2$.
\end{example}

Consider now the case $d=5$.
Let us recall that $[a,b,c]_5=a^2+b^2-5c^2$.
Observe that  $0=[1,2,1]_5$  and we have
\begin{align*}
   -1 = [12, 10, 7]_5,&amp;   \ &amp; 1 = [10, 9, 6]_5,&amp; \    &amp;
  -2 = [3, 3, 2]_5,&amp;&amp; 2 = [9, 1, 4]_5,    \\
  -3 = [1, 1, 1]_5,&amp;&amp; 3 = [2, 2, 1]_5,&amp;   &amp;
  -4 = [5, 4, 3]_5,&amp;&amp; 4 = [20, 3, 9]_5, \\
  -5 = [6, 2, 3]_5,&amp;&amp; 5 = [3, 1, 1]_5,&amp;  &amp;
  -6 = [7, 5, 4]_5,&amp;&amp; 6 = [5, 1, 2]_5,  \\
 -7 = [3, 2, 2]_5,&amp;&amp; 7 = [6, 4, 3]_5,&amp;  &amp;
  -8 = [6, 1, 3]_5,&amp;&amp; 8 = [3, 2, 1]_5, \\
  -9 = [10, 4, 5]_5,&amp;&amp; 9 = [5, 2, 2]_5,&amp; &amp;
  -10 = [7, 11,6]_5,&amp;&amp; 10 = [3,9, 4]_5.
\end{align*}

\begin{proposition}\label{PropApiec}
The number $5$ is special, that is,
for every integer $m$ there exist
nonzero integers $a,b,c$ such that
$
m=a^2+b^2-5c^2.
$
\end{proposition}

\begin{proof}
It follows from the equalities
$$
k^2+(2k-2)^2-5(k-1)^2=2k-1, \quad (k-2)^2+(2k-1)^2-5(k-1)^2=2k,
$$
and
$=-1=[12,10,7]_5$, \ $1=[10,9,6]_5$, \
$2=[9,1,4]_5$, \ $4=[20,3,9]_5$.
\end{proof}

Note the following consequence of Propositions \ref{PropApiec} and
\ref{PellNiesk}.

\begin{proposition}\label{PiecNiesk}
For every integer $m$ there are infinitely many triples $(a,b,c)$,
of positive integers such that
$
m=a^2+b^2-5c^2.
$
\end{proposition}

\begin{proposition}\label{PropExa}
Let $d=q$ or $d=2q$, where $q$ is a product of prime numbers of the form
$4k+1$. If $d\leqslant 50$, then  $d$ is special.
\end{proposition}

\begin{proof}
If $d&lt;10$, then $d=1,2$ or $5$, and we already know that in this case
$d$ is special. If $d\geqslant 10$, then we have the following equalities:
\begin{alignat*}{3}
\ [k, 3 k - 3, k - 1]_{10} \quad &amp;=&amp;&amp; \quad [k - 5, 3 k - 8, k - 3]_{10} \quad &amp;&amp;= \quad 2k-1,\\
\  [k + 1, 3 k - 3, k - 1]_{10}\quad &amp;=&amp;&amp; \quad [k - 9, 3 k - 13, k - 5]_{10}&amp;&amp;= \quad 4k,\\
\ [k - 1, 3 k + 1, k]_{10} \quad &amp;=&amp;&amp; \quad [k - 21, 3 k - 39, k - 14]_{10} &amp;&amp;= \quad 4k+2.\\ \\
\ [2 k - 4, 3 k - 10, k - 3]_{13}\quad &amp;=&amp;&amp; \quad [2 k - 30, 3 k - 36, k - 13]_{13}&amp;&amp;= \quad 2k-1,\\
\ [2 k - 3, 3 k - 2, k - 1]_{13}\quad &amp;=&amp;&amp; \quad [2 k - 29, 3 k - 54, k - 17]_{13}&amp;&amp;= \quad 2k. \\ \\
\ [k, 4 k - 4, k - 1]_{17}\quad &amp;=&amp;&amp; \quad [k - 34, 4 k - 106, k - 27]_{17}&amp;&amp;= \quad 2k-1,\\
 \ [k - 8, 4 k - 19, k - 5]_{17}\quad &amp;=&amp;&amp; \quad [k - 76, 4 k - 357, k - 65]_{17}&amp;&amp;= \quad 2k. \\ \\
\ [3 k - 18, 4 k - 30, k - 7]_{25}\quad &amp;=&amp;&amp; \quad [3 k - 68, 4 k - 80, k - 21]_{25} \quad &amp;&amp;= \quad 2k-1,\\
\ [3 k - 4, 4 k - 3, k - 1]_{25}\quad &amp;=&amp;&amp; \quad [3 k - 104, 4 k - 153, k - 37]_{25} \quad &amp;&amp;=  \quad 2k. \\ \\
\ [k, 5 k - 5, k - 1]_{26}\quad &amp;=&amp;&amp; \quad [k - 13, 5 k - 44, k - 9]_{26}&amp;&amp;= \quad 2k-1.\\
\ [k + 1, 5 k - 5, k - 1]_{26}\quad &amp;=&amp;&amp; \quad [k - 25, 5 k - 83, k - 17]_{26}&amp;&amp;= \quad 4k,\\
\ [k - 5, 5 k - 9, k - 2]_{26}\quad &amp;=&amp;&amp; \quad [k - 57, 5 k - 217, k - 44]_{26}&amp;&amp;= \quad 4k+2. \\ \\
\ [2 k - 8, 5 k - 14, k - 3]_{29}\quad &amp;=&amp;&amp; \quad [2 k - 66, 5 k - 188, k - 37]_{29}&amp;&amp;=  \quad 2k-1,\\
\ [2 k - 7, 5 k - 26, k - 5]_{29}\quad &amp;=&amp;&amp; \quad [2 k - 65, 5 k - 142, k - 29]_{29}&amp;&amp;=  \quad 2k. \\  \\
\end{alignat*}
\begin{alignat*}{3}
\ [3 k - 7, 5 k - 16, k - 3]_{34}&amp;=&amp;&amp; \quad [3 k - 24, 5 k - 33, k - 7]_{34}\quad &amp;&amp;=2k-1,\\
\ [3 k - 11, 5 k - 27, k - 5]_{34} \quad&amp;=&amp;&amp; \quad [3 k - 45, 5 k - 61, k - 13]_{34}\quad &amp;&amp;=4k,\\
\ [3 k - 1, 5 k + 1, k]_{34} \quad&amp;=&amp;&amp; \quad [3 k - 69, 5 k - 135, k - 26]_{34}\quad &amp;&amp;=4k+2. \\  \\
\ [k, 6 k - 6, k - 1]_{37} \quad&amp;=&amp;&amp; \quad [k - 74, 6 k - 376, k - 63]_{37}\quad &amp;&amp;=2k-1,\\
 \ [k - 18, 6 k - 77, k - 13]_{37} \quad&amp;=&amp;&amp; \quad [k - 166, 6 k - 891, k - 149]_{37}\quad &amp;&amp;=2k. \\  \\
\ [4 k - 48, 5 k - 68, k - 13]_{41} \quad&amp;=&amp;&amp; \quad [4 k - 130, 5 k - 150, k - 31]_{41}\quad &amp;&amp;=2k-1,\\
\ [4 k - 5, 5 k - 4, k - 1]_{41} \quad&amp;=&amp;&amp; \quad [4 k - 251, 5 k - 332, k - 65]_{41}\quad &amp;&amp;=2k.\\  \\
\ [k, 7 k - 7, k - 1]_{50} \quad&amp;=&amp;&amp; \quad [k - 25, 7 k - 132, k - 19]_{50}\quad &amp;&amp;=2k-1,\\
\ [k + 1, 7 k - 7, k - 1]_{50} \quad&amp;=&amp;&amp; \quad [k - 49, 7 k - 257, k - 37]_{50}\quad &amp;&amp;=4k,\\
  \ [k - 11, 7 k - 41, k - 6]_{50} \quad&amp;=&amp;&amp; \quad [k - 111, 7 k - 641, k - 92]_{50}\quad &amp;&amp;=4k+2.
\end{alignat*}
\end{proof}
By similar methods we are ready to prove, using a computer,  that the
same is true for $d&lt;1000$.
Hence, we know that if $d&lt;1000$, then the answer to Question \ref{Pyt} is affirmative.


\section{An infinite sequence of special numbers}

In this section we prove that the set of special numbers
is infinite.
In our proof we  use the following well known lemma \cite{Sierpinski88, Barb, NowPEL}
concerned with the sequence  \cite[A001110]{Sloane}.
Let us recall  that every number of the form
$
t_n=\frac{n(n+1)}{2}=1+2+\dots+n
$
is called {\it triangular} .

\begin{lemma}\label{lemPel}
There are infinitely many  square triangular numbers.
Examples:
$$
t_1=1^2, \quad t_8=6^2, \quad t_{49}=35^2, \quad t_{288}=204^2, \quad t_{1681}=1189^2.
$$
\end{lemma}

\begin{proof}
The Pell equation $x^2-8y^2=1$ has infinitely many solutions in positive
integers.
Let $(x,y)$ be  one of such solutions.
Then $x$ is odd. Let $x=2n+1$ where $n$ is a positive integer.
Then we have $t_n=\frac{n(n+1)}{2}=y^2$.
\end{proof}

\begin{theorem}\label{NSp}
There are infinitely many special numbers.
\end{theorem}

\begin{proof}
We know from the previous lemma that there are
infinitely many  positive integers $u$
such that $u^2=\frac{k(k+1)}{2}$ for some positive integer $k$.
Let $d=(2u)^2+1$ with $u\geqslant2$.
Observe that $d=k^2+(k+1)^2$.
We will show that the number $d$ is special. Let $m$ be an integer.

First assume that $m$ is even. Let $m=2s$, where $s$ is an integer.
We have the equality
$$
\Big((k+1)(s-1)+1\Big)^2+\Big(k(s-1)-1\Big)^2-d\big(s-1\Big)^2=2s.
$$
Thus, if $m=2s$ with $s\neq1$, then there exist nonzero integers $a,b,c$
such that $[a,b,c]_d=m$.
Consider the case $s=1$, that is, $m=2$.
Since $d$ is non-square, the Pell equation $x^2-dz^2=1$ has  a
solution $(x,z)$ such that $x,z$ are positive integers.
Then we have $[x,1,z]_d=2$. Therefore, every even integer $m$ is of the form
$[a,b,c]_d$ with nonzero integers $a,b,c$.

Now assume that $m$ is odd. Let $m=2s-1$ where $s$ is an integer.
We have the equality
$$
s^2+(2us-2u)^2-d(s-1)^2=2s-1.
$$
Thus, if $m=2s-1$ with $s\neq1$, then there exist positive integers $a,b,c$
such that $[a,b,c]_d=m$.
Consider the case $s=1$, that is, $m=1$.
Since $d-4$ is non-square (because $d=4u^2+1$ with $u\geqslant2$),
the Pell equation $x^2-(d-4)z^2=1$ has  a
solution $(x,z)$ such that $x,z$ are positive integers.
Then we have $[x,2z,z]_d=1$. Therefore, every odd integer $m$ is also
of the form
$[a,b,c]_d$ with nonzero integers $a,b,c$.
\end{proof}

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\bibitem{Barb}
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A.~Nowicki,
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 \bibitem{Sloane}
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\bibitem{St}
A. M. Storozhev,
{\it International Mathematics Tournament of Towns 1997--2002},
Book~5, AMT Publishing, 2006.
\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11D09; Secondary 11B99.

\noindent \emph{Keywords: } 
Pell's equation, sum of squares.
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\noindent (Concerned with sequences
\seqnum{A000217} and
\seqnum{A001110}.)

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\noindent
Received November 18 2014;
revised versions received  December 16 2014; December 18 2014.
Published in {\it Journal of Integer Sequences}, January 24 2015.

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