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\begin{document}
\begin{center}
\epsfxsize=4in \leavevmode\epsffile{logo129.eps}
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\begin{center}
\vskip 1cm{\LARGE\bf Sums of Squares
and Products \\
\vskip .1in of Jacobsthal Numbers} \vskip 1cm
\large
Zvonko \v Cerin \\
Department of Mathematics \\
University of Zagreb\\
Bijeni\v{c}ka 30 \\
Zagreb 10 000 \\
Croatia \\
\href{mailto:cerin@math.hr}{\tt cerin@math.hr} \\
\end{center}


\vskip .2in

\begin{abstract}
We consider sums of squares of odd and even
terms of the Jacobsthal sequence and sums of
their products. We also study the analogous alternating sums.
These sums are related to 
products of appropriate Jacobsthal numbers and
several integer sequences. The formulas that
we discover show that a certain translation
property for these sums holds, so that in
practice, only sums of initial values 
and the information where the
summation begins are necessary.
\end{abstract}


\vskip .2in


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%               %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%   Section 1   %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%               %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%\keywords {Jacobsthal
%numbers, Jacobsthal-Lucas numbers, integer
%sequences, sum of squares, combinatorial
%identity, Maple V} \subjclass {Primary 11B39,
%11Y55, 05A19}

\section{Introduction}
The Jacobsthal and Jacobsthal-Lucas sequences
\z {J_n} and \z {j_n} are defined by the
recurrence relations \q {J_0=0,\qquad
J_1=1,\qquad J_n=J_{n-1}+2\,J_{n-2}\,\,\text{
for \z {n\geqslant 2},}} and \q {j_0=2,\qquad
j_1=1,\qquad j_n=j_{n-1}+2\,j_{n-2}\,\,\text{
for \z {n\geqslant 2}.}}

In Sections 2--4 we consider sums of
squares of odd and even terms of the
Jacobsthal sequence and sums of their
products. These sums have nice representations
as products of appropriate Jacobsthal and
Jacobsthal-Lucas numbers.


The numbers \z {J_k} appear as the integer
sequence \z {A001045} from \re {sloane} while
the numbers \z {j_k} is \z {A014551}. The
properties of these numbers are summarized in
\re {hor}. For the convenience of the reader
we shall now explicitly define these
sequences.

The first eleven terms of the sequence \z
{J_k} are \z {0}, \z {1}, \z {1}, \z {3}, \z
{5}, \z {11}, \z {21}, \z {43}, \z {85}, \z
{171} and \z {341}. It is given by the formula
\z {J_n={\frac{2^n-(-1)^n}{3}}}.

The first eleven terms of the sequence \z
{j_k} are \z {2}, \z {1}, \z {5}, \z {7}, \z
{17}, \z {31}, \z {65}, \z {127}, \z {257}, \z
{511} and \z {1025}. It is given by the
formula \z {j_n=2^n+(-1)^n}.

In the last three sections we look into the
alternating sums of squares of odd and even
terms of the Jacobsthal sequence and the
alternating sums of products of two
consecutive Jacobsthal numbers. These sums
also have nice representations as products of
appropriate Jacobsthal and Jacobsthal-Lucas
numbers.

These formulas for ordinary sums and for
alternating sums have been discovered with the
help of a PC computer and all algebraic
identities needed for the verification of our
theorems can be easily checked in either
Derive, Mathematica or Maple V. Running times
of all these calculations are in the range of
a few seconds.

Similar results for Fibonacci, Lucas, Pell,
and Pell-Lucas numbers have recently been
discovered by G. M. Gianella and the author in
papers \cite{c1-D,c2-B,c3-P,c4-M,c5-PL,c6-P}. They
improved some results in \cite{r&amp;l}.

\section{Jacobsthal even squares}

The following lemma is needed to accomplish
the inductive step in the proof of the first
part of our first theorem.

For any \z {n=0,\,1,\,2,\dots} let \z
{{\beta}_n={\frac{2^{4n+2}-1}{3}}=J_{4n+2}},
\z
{{\gamma}_n={\frac{2^{4n+2}+1}{5}}={\frac{j_{4n+2}}{5}}}
and \z
{{\tau}_n={\frac{2^{4n}-1}{15}}={\frac{J_{4n}}{5}}}.


\begin{lem}\label{l1}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline}
 \hspace{2cm}   ({\beta}_{n+1}\,{\gamma}_{n+1}-
{\beta}_{n}\,{\gamma}_{n})\,J_{2k}^2+
8\,({\beta}_{n+1}\,{\tau}_{n+1}-
{\beta}_{n}\,{\tau}_{n})\,J_{2k}\\
    =J_{2k+4n+4}^2+J_{2k+4n+2}^2-
J_{4n+4}^2-J_{4n+2}^2.\hspace{2cm}
%\quad\text{\rm (2.1)}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
Let \z {A=2^{2k}}, \z {B=2^{8n}} and \z
{C=2^{4n}}. The difference of the left hand
side and the right hand side of the relation
(2.1) is equal to
\begin{multline*}
{\frac {544}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {40}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k}+  \left(
-1 \right) ^{1+4\,n} \right) (A+1)C\\+ { \frac
{272}{9}}\,\left( \left( -1 \right)
^{4\,k}+2\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n}+
 \left( -1 \right) ^{1+8\,n+4\,k}\right).
\end{multline*}
It is obvious that all coefficients in the
above expression are zero and the proof is
complete.
\end{proof}

There are essentially three types of proofs in
this paper that we indicate as {\textbf{P1}},
{\textbf{P2}} and {\textbf{P3}}.

In {\textbf{P1}} and {\textbf{P2}} we use the
substitutions \z {A=2^{2k}}, \z {B=2^{8n}}, \z
{C=2^{4n}} and \z {M=2^k}, \z {P=(-1)^k},
respectively. We prove that the difference of
the left hand side and the right hand side of
the relation from the statement is equal to
zero with algebraic simplification to an
expression that is obviously vanishing. This
part is easily done with the help of a
computer (for example in Maple V) so that we
shall only indicate the final expression.

In {\textbf{P3}} we argue by induction on \z
{n} taking care first of the initial value \z
{n=0} and then showing that it holds for \z
{n=r+1} under the assumption that it is true
for \z {n=r}.

The following lemma is needed to accomplish
the inductive step in the proof of the second
part of our first theorem.

For any \z {n=0,\,1,\,2,\dots} let \z
{{\eta}_n=2^{4n+4}+1=j_{4n+4}}.

\begin{lem}\label{l2}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline}
\hspace{2cm}({\tau}_{n+2}\,{\eta}_{n+1}-
{\tau}_{n+1}\,{\eta}_{n})\,J_{2k}^2+
8\,({\beta}_{n+1}\,{\tau}_{n+2}-
{\beta}_{n}\,{\tau}_{n+1})\,J_{2k}\\
    =J_{2k+4n+6}^2+J_{2k+4n+4}^2-
J_{4n+6}^2-J_{4n+4}^2.\hspace{2cm}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {8704}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {160}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k}+  \left(
-1 \right) ^{1+4\,n} \right) (A+1)C\\+ { \frac
{4352}{9}}\,\left( \left( -1 \right)
^{4\,k}+2\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n}+
 \left( -1 \right) ^{1+8\,n+4\,k}\right).
\end{multline*}
\end{proof}

\begin{thm}\label{t1}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq
{\sum_{i=0}^{m}\,J_{2k+2i}^2=
\sum_{i=0}^{m}\,J_{2i}^2+{\beta}_n\,J_{2\,k}
\left[{\gamma}_n\,J_{2k}+ 8\,{\tau}_n\right],}
if \z {m=2n} and \z {n=0,\,1,\,2,\dots}  and
\qq { \sum_{i=0}^{m}\,J_{2k+2i}^2=
\sum_{i=0}^{m}\,J_{2i}^2+{\tau}_{n+1}\,J_{2\,k}
\left[{\eta}_n\,J_{2k}+ 8\,{\beta}_n\right],}
if \z {m=2n+1} and \z {n=0,\,1,\,2,\dots}.
\end{thm}
\begin{proof}[Proof of (2.3)] ({\textbf{P3}}).
The proof is by induction on \z {n}. When \z
{n=0} we obtain \q {
J_{2k}^2=J_{0}^2+{\beta}_0\,J_{2k}\,
[{\gamma}_0\,J_{2k}+8\,{\tau}_0]=J_{2k}\,J_{2k}=
J_{2k}^2,} because \z {J_0=0}, \z
{{\beta}_0=1}, \z {{\gamma}_0=1} and \z
{{\tau}_0=0}.

Assume that the relation (2.3) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)}\,J_{2k+2i}^2=
J_{2k+4r+2}^2+J_{2k+4r+4}^2+
\sum_{i=0}^{2r}\,J_{2k+2i}^2=
J_{2k+4r+2}^2+J_{2k+4r+4}^2+\\
\sum_{i=0}^{2r}\,J_{2i}^2+{\beta}_r\,J_{2\,k}
\left[{\gamma}_r\,J_{2k}+ 8\,{\tau}_r\right]
=\sum_{i=0}^{2(r+1)}\,J_{2i}^2+{\beta}_{r+1}\,J_{2\,k}
\left[{\gamma}_{r+1}\,J_{2k}+
8\,{\tau}_{r+1}\right],
\end{multline*}
where the last step uses Lemma 1 for \z
{n=r+1}. Hence, (2.3) is true for \z {n=r+1}
and the proof is completed.
\end{proof}
\begin{proof}[Proof of (2.4)] ({\textbf{P3}}).
When \z {n=0} we obtain \q {
J_{2k}^2+J_{2k+2}^2=J_{0}^2+J_{2}^2+{\tau}_1\,J_{2k}\,
[{\eta}_0\,J_{2k}+8\,{\beta}_0]=
17\,J_{2k}^2+8\,J_{2k}+1,} since \z {J_0=0},
\z {J_1=1}, \z {{\beta}_0=1}, \z {{\eta}_0=17}
and \z {{\tau}_1=1}. The above equality is
equivalent to \z {J_{2k+2}^2=16\,J_{2k}^2+8\,
J_{2k}+1=(4\,J_{2k}+1)^2} which is true
because it follows from the relation \z
{J_{n+2}=4\,J_n+1}.

Assume that the relation (2.4) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)+1}\,J_{2k+2i}^2=
J_{2k+4r+4}^2+J_{2k+4r+6}^2+
\sum_{i=0}^{2r+1}\,J_{2k+2i}^2=
J_{2k+4r+4}^2+J_{2k+4r+6}^2+\\
\sum_{i=0}^{2r+1}\,J_{2i}^2+{\tau}_{r+1}\,J_{2\,k}
\left[{\eta}_r\,J_{2k}+ 8\,{\beta}_r\right]
=\sum_{i=0}^{2(r+1)+1}\,J_{2i}^2+{\tau}_{r+2}\,J_{2\,k}
\left[{\eta}_{r+1}\,J_{2k}+
8\,{\beta}_{r+1}\right],
\end{multline*}
where the last step uses Lemma 2 for \z
{n=r+1}.
\end{proof}

\section{Jacobsthal odd squares}
The initial step in an inductive proof of the
first part of our second theorem uses the
following lemma.
\begin{lem}\label{l3}
    For every \z {k\geqslant 0} the following
    identity holds
    \q {J_{2k+1}^2=
    16\,J_{2k}\,J_{2k-2}+8\,J_{2k}+1.}
\end{lem}
\begin{proof} ({\textbf{P2}}).
Let \z {M=2^k} and \z {P=(-1)^k}. The
difference of the left and the right hand side
is equal to \z {{\frac{\left( P-1 \right)
\left( P+1
\right)}{3}}\left(8\,M^2-5\,{P}^{2}+3
\right)=0.}
\end{proof}
The initial step in an inductive proof of the
second part of our second theorem uses the
following lemma.
\begin{lem}\label{l4}
    For every \z {k\geqslant 0} the following
    identity holds
    \q {J_{2k+1}^2+J_{2k+3}^2=10+
    8\,J_{2k}\,(34\,J_{2k-2}+15).}
\end{lem}
\begin{proof} ({\textbf{P2}}).
\z {10\left( P-1 \right) \left( P+1 \right)
(4\,{M}^{2}- 3\,{P}^{2}+1).}
\end{proof}

The following lemma is needed to accomplish
the inductive step in the proof of the first
part of our second theorem.

For any \z {n=0,\,1,\,2,\dots} let \z
{{\pi}_n={\frac{2^{4n+1}-1}{3}}=J_{4n+1}}.

\begin{lem}\label{l5}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline}
 \hspace{2cm}  8\,J_{2k}\,\left[2\,({\beta}_{n+1}\,
   {\gamma}_{n+1}-{\beta}_{n}\,{\gamma}_{n})\,
   J_{2k-2}+
{\beta}_{n+1}\,{\pi}_{n+1}-
{\beta}_{n}\,{\pi}_{n}\right]\\
    =J_{2k+4n+5}^2+J_{2k+4n+3}^2-
J_{4n+5}^2-J_{4n+3}^2.\hspace{2cm}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {5440}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {80}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k+1}+
\left( -1 \right) ^{4\,n} \right) (A+1)C\\+ {
\frac {1088}{9}}\,\left( 4\,\left( -1 \right)
^{4\,k}+5\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n}+
 \left( -1 \right) ^{8\,n+4\,k+1}\right).
\end{multline*}
\end{proof}
The following lemma is needed to accomplish
the inductive step in the proof of the second
part of our second theorem.

For any \z {n=0,\,1,\,2,\dots} let \z
{{\sigma}_n=5\,J_{4n+3}}.

\begin{lem}\label{l6}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline}
 \hspace{2cm}  8\,J_{2k}\,\left[2\,({\tau}_{n+2}\,
   {\eta}_{n+1}-{\tau}_{n}\,{\eta}_{n})\,
   J_{2k-2}+
{\tau}_{n+2}\,{\sigma}_{n+1}-
{\tau}_{n+1}\,{\sigma}_{n}\right]\\
    =J_{2k+4n+7}^2+J_{2k+4n+5}^2-
J_{4n+7}^2-J_{4n+5}^2.\hspace{2cm}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {87040}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {320}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k+1}+
\left( -1 \right) ^{4\,n} \right) (A+1)C\\+ {
\frac {17408}{9}}\,\left( 4\,\left( -1 \right)
^{4\,k}+5\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n}+
 \left( -1 \right) ^{8\,n+4\,k+1}\right).
\end{multline*}
\end{proof}

\begin{thm}\label{t2}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq
{\sum_{i=0}^{m}\,J_{2k+2i+1}^2=
\sum_{i=0}^{m}\,J_{2i+1}^2+8\,{\beta}_n\,J_{2\,k}
\left[2\,{\gamma}_n\,J_{2k-2}+
{\pi}_n\right],} if \z {m=2n} and \z
{n=0,\,1,\,2,\dots}  and \qq {
\sum_{i=0}^{m}\,J_{2k+2i+1}^2=
\sum_{i=0}^{m}\,J_{2i+1}^2+8\,{\tau}_{n+1}\,J_{2\,k}
\left[2\,{\eta}_n\,J_{2k-2}+{\sigma}_n\right],}
if \z {m=2n+1} and \z {n=0,\,1,\,2,\dots}.
\end{thm}
\begin{proof}[Proof of (3.3)] ({\textbf{P3}}).
When \z {n=0} we obtain \q {
J_{2k+1}^2=J_{1}^2+8\,{\beta}_0\,J_{2k}\,
[2\,{\gamma}_0\,J_{2k-2}+{\pi}_0]=1+8\,J_{2k}\,
[2\,J_{2k-2}+1],} because \z {J_1=1}, \z
{{\beta}_0=1}, \z {{\gamma}_0=1} and \z
{{\pi}_0=1}. But, this equality is true by
Lemma 3.

Assume that the relation (3.3) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)}\,J_{2k+2i+1}^2=
J_{2k+4r+5}^2+J_{2k+4r+3}^2+
\sum_{i=0}^{2r}\,J_{2k+2i+1}^2=
J_{2k+4r+5}^2+J_{2k+4r+3}^2+
\sum_{i=0}^{2r}\,J_{2i+1}^2\\+8\,{\beta}_n\,J_{2\,k}
\left[2\,{\gamma}_n\,J_{2k-2}+ {\pi}_n\right]
=\sum_{i=0}^{2(r+1)}\,J_{2i+1}^2+8\,{\beta}_{n+1}\,
J_{2\,k}\left[2\,{\gamma}_{n+1}\,J_{2k-2}+
{\pi}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 5 for \z
{n=r+1}.
\end{proof}
\begin{proof}[Proof of (3.4)] ({\textbf{P3}}).
The proof is by induction on \z {n}. When \z
{n=0} we obtain \q {
    J_{2k+1}^2+J_{2k+3}^2=
    J_{1}^2+J_{3}^2+
8\,{\tau}_{1}\,J_{2\,k}
\left[2\,{\eta}_0\,J_{2k-2}+{\sigma}_0\right]
= 10+8\,J_{2\,k}
\left[34\,J_{2k-2}+15\right],} since \z
{J_1=1}, \z {J_3=3}, \z {{\tau}_1=1}, \z
{{\eta}_0=17} and \z {{\sigma}_0=15}. The
above equality is true by Lemma 4.

Assume that the relation (3.4) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)+1}\,J_{2k+2i+1}^2=
J_{2k+4r+7}^2+J_{2k+4r+5}^2+
\sum_{i=0}^{2r+1}\,J_{2k+2i+1}^2=\\
J_{2k+4r+7}^2+J_{2k+4r+5}^2+
\sum_{i=0}^{2r+1}\,J_{2i+1}^2+8\,{\tau}_{r+1}\,
J_{2\,k}\left[2\,{\eta}_r\,J_{2k-2}+ {\sigma}_r\right]\\
=\sum_{i=0}^{2(r+1)+1}\,J_{2i+1}^2+8\,{\tau}_{r+2}\,
J_{2\,k}\left[2\,{\eta}_{r+1}\,J_{2k-2}+
{\sigma}_{r+1}\right],
\end{multline*}
where the last step uses Lemma 6 for \z
{n=r+1}.
\end{proof}


\section{Jacobsthal products}
For the first two steps in a proof by
induction of our next theorem we require the
following lemma.

\begin{lem}\label{l7}
For every \z {k\geqslant 0} the following
equalities hold \qq {J_{2k+1}=8\,J_{2k-2}+3.}
\qq {J_{2k}\,J_{2k+1}+J_{2k+2}\,J_{2k+3}=
3+J_{2k}\,(136\,J_{2k-2}+55).}
\end{lem}
\begin{proof}[Proof of (4.1)]
    By the formula \z {J_k={\frac{2^k-(-1)^k}{3}}} we have
    \begin{multline*}
    J_{2k+1}={\frac{2^{2k+1}-(-1)^{2k+1}}{3}}=
{\frac{2^{2k+1}+1}{3}}
=8\cdot{\frac{2^{2k-2}-(-1)^{2k-2}}{3}}
+3=8\,J_{2k-2}+3.
\end{multline*}
\end{proof}
\begin{proof}[Proof of (4.2)] ({\textbf{P2}}).
\z {{\frac{\left( P-1 \right) \left( P+1
\right)}{3}}\left(55\,M^2-46\,{P}^{2}+9
\right).}
\end{proof}

With the following lemma we shall make the
inductive step in the proof of the first part
of our third theorem.
\begin{lem}\label{l8}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline} \hspace{2cm}
J_{2k}\,\left[8\,({\beta}_{n+1}\,
   {\gamma}_{n+1}-{\beta}_{n}\,{\gamma}_{n})\,
   J_{2k-2}+
{\beta}_{n+1}\,{\mu}_{n+1}-
{\beta}_{n}\,{\mu}_{n}\right]=\\
J_{2k+4n+5}\,J_{2k+4n+4}+J_{2k+4n+3}\,J_{2k+4n+2}-
J_{4n+5}\,J_{4n+4}-J_{4n+3}\,J_{4n+2}.
\hspace{1cm}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
 {\frac
{2720}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+{\frac {20}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k}+ \left(
-1 \right) ^{4\,n+1} \right) (A+1)C+\\ { \frac
{544}{9}}\,\left( 4\,\left( -1 \right)
^{4\,k}+5\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n+1}+
 \left( -1 \right) ^{8\,n+4\,k}\right).
\end{multline*}
\end{proof}
With the following lemma we shall make the
inductive step in the proof of the second part
of our third theorem.
\begin{lem}\label{l9}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equality holds
\begin{multline}
\hspace{1.3cm}
J_{2k}\,\left[8\,({\tau}_{n+2}\,
   {\eta}_{n+1}-{\tau}_{n+1}\,{\eta}_{n})\,
   J_{2k-2}+
{\tau}_{n+2}\,{\nu}_{n+1}-
{\tau}_{n+1}\,{\nu}_{n}\right]=\\
J_{2k+4n+5}\,J_{2k+4n+4}+J_{2k+4n+7}\,J_{2k+4n+6}-
J_{4n+5}\,J_{4n+4}-J_{4n+7}\,J_{4n+6}.
\hspace{1cm}
\end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {43520}{9}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {80}{9}}\,
\left( \left( -1 \right) ^{4\,n+2\,k}+ \left(
-1 \right) ^{4\,n+1} \right)
\left(A+1\right)C\\+ { \frac
{8704}{9}}\,\left( 4\,\left( -1 \right)
^{4\,k}+5\, \left( -1 \right)
^{1+2\,k}+1\right) B+{\frac{2}{9}}\,\left(
 \left( -1 \right) ^{8\,n+1}+
 \left( -1 \right) ^{8\,n+4\,k}\right).
\end{multline*}
\end{proof}
For any \z {n=0,\,1,\,2,\dots} let \z
{{\mu}_n={\frac{2^{4n+3}+1}{3}}=J_{4n+3}} and
\z
{{\nu}_n={\frac{5\,\left(2^{4n+5}+1\right)}{3}}}
\z {=5\,J_{4n+5}}.
\begin{thm}\label{t3}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq
{\sum_{i=0}^{m}\,J_{2k+2i}\,J_{2k+2i+1}=
\sum_{i=0}^{m}\,J_{2i}\,J_{2i+1}+{\beta}_n\,J_{2\,k}
\left[8\,{\gamma}_n\,J_{2k-2}+
{\mu}_n\right],} if \z {m=2n} and \z
{n=0,\,1,\,2,\dots}  and \qq {
\sum_{i=0}^{m}\,J_{2k+2i}\,J_{2k+2i+1}=
\sum_{i=0}^{m}\,J_{2i}\,J_{2i+1}+{\tau}_{n+1}\,J_{2\,k}
\left[8\,{\eta}_n\,J_{2k-2}+{\nu}_n\right],}
if \z {m=2n+1} and \z {n=0,\,1,\,2,\dots}.
\end{thm}

\begin{proof}[Proof of (4.5)] ({\textbf{P3}}).
For \z {n=0} the relation (4.5) is \q
{J_{2k}\,J_{2k+1}=J_0\,J_1+{\beta}_0\,J_{2k}(
8\,{\gamma}_0\,J_{2k-2}+{\mu}_0)=J_{2k}
(8\,J_{2k-2}+3)} which is true since \z
{J_{2k+1}=8\,J_{2k-2}+3} by the relation (4.1)
in Lemma~7.

Assume that the relation (4.5) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)}\,J_{2k+2i}\,J_{2k+2i+1}=
J_{2k+4r+4}\,J_{2k+4r+5}+J_{2k+4r+2}\,J_{2k+4r+3}
 +\sum_{i=0}^{2r}\,J_{2k+2i}\,J_{2k+2i+1}=\\
J_{2k+4r+4}\,J_{2k+4r+5}+J_{2k+4r+2}\,J_{2k+4r+3}
+\sum_{i=0}^{2r}\,J_{2i}\,J_{2i+1}+{\beta}_n\,J_{2\,k}
\left[8\,{\gamma}_n\,J_{2k-2}+ {\mu}_n\right]\\
=\sum_{i=0}^{2(r+1)}\,J_{2i}\,J_{2i+1}+{\beta}_{n+1}\,
J_{2\,k}\left[8\,{\gamma}_{n+1}\,J_{2k-2}+
{\mu}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 8 for \z
{n=r+1}.
\end{proof}
\begin{proof}[Proof of (4.6)] ({\textbf{P3}}).
For \z {n=0} the relation (4.6) is
\begin{multline*}
    J_{2k}\,J_{2k+1}+J_{2k+2}\,J_{2k+3}=J_0\,J_1+
J_2\,J_3+{\tau}_1\,J_{2k}(
8\,{\eta}_0\,J_{2k-2}+{\nu}_0)
    =3+J_{2k}
(136\,J_{2k-2}+55)
\end{multline*}
which is true by (4.2) in Lemma 7.

Assume that the relation (4.6) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{ 2(r+1)+1}\,J_{2k+2i}\,J_{2k+2i+1}=
J_{2k+4r+4}\,J_{2k+4r+5}+J_{2k+4r+6}\,J_{2k+4r+7}
 +\sum_{i=0}^{2r+1}\,J_{2k+2i}\,J_{2k+2i+1}=\\
J_{2k+4r+4}\,J_{2k+4r+5}+J_{2k+4r+6}\,J_{2k+4r+7}
+\sum_{i=0}^{2r+1}\,J_{2i}\,J_{2i+1}+{\tau}_{n+1}\,J_{2\,k}
\left[8\,{\eta}_n\,J_{2k-2}+ {\nu}_n\right]\\
=\sum_{i=0}^{2(r+1)+1}\,J_{2i}\,J_{2i+1}+
{\tau}_{n+2}\,
J_{2\,k}\left[8\,{\eta}_{n+1}\,J_{2k-2}+
{\nu}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 9 for \z
{n=r+1}.
\end{proof}

\section{Alternating Jacobsthal even squares}
In this section we look for formulas that give
closed forms for alternating sums of squares
of Jacobsthal numbers with even indices.
\begin{lem}\label{l10}
For every \z {k\geqslant 0} we have \qq
{J_{2k}=4\,J_{2k-2}+1.}
\end{lem}
\begin{proof}
By the formula \z {J_k={\frac{2^k-(-1)^k}{3}}}
we get \q {
4\,J_{2k-2}+1=4\cdot{\frac{2^{2k-2}-(-1)^{2k-2}}{3}}+
1={\frac{2^{2k}-(-1)^{2k}}{3}}=J_{2k}.}
\end{proof}


\begin{lem}\label{l11}
    For every \z {k\geqslant 0} we have
    \qq {J_{2k+2}^2-J_{2k}^2=
    1+J_{2k}\,(60\,J_{2k-2}+23).}
\end{lem}
\begin{proof} ({\textbf{P2}}).
\z {{\frac{\left( P-1 \right) \left( P+1
\right)}{3}}\left(23\,M^2-20\,{P}^{2}+3
\right).}
\end{proof}
Let \z {{\tau}_0^*=1} and \z
{{\tau}_{n+1}^*-{\tau}_{n}^*=
2^{4n+3}\,(50\cdot 2^{4n}-1),} for \z
{n=0,\,1,\,2,\dots}. For any \z
{n=0,\,1,\,2,\dots} let \z
{{\gamma}_n^*={\frac{2^{8n+4}+1}{17}}=
{\frac{j_{8n+4}}{17}}}.
\begin{lem}\label{l12}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
    \hspace{1cm}J_{2k}\,(4\,({\gamma}_{n+1}^*-
    {\gamma}_{n}^*)\,J_{2k-2}+{\tau}_{n+1}^*-
    {\tau}_{n}^*)=J_{2k+4n+4}^2-J_{2k+4n+2}^2-
    J_{4n+4}^2+J_{4n+2}^2.\hspace{0.6cm}
        \end{multline}
    \end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {400}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {8}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k} -1
\right) AC+\\{ \frac {80}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {8}{3}
}\,\left(\left( -1 \right) ^{2\,k}+\left( -1
 \right) ^{4\,n+1} \right) C.
\end{multline*}
\end{proof}
Let \z {{\beta}_0^*=23} and \z
{{\beta}_{n+1}^*-{\beta}_{n}^*=2^{4n+5}\left(200\cdot
2^{4n}-1\right)} for any \z
{n=0,\,1,\,2,\dots}. For the same values of \z
{n}, let \z
{{\eta}_n^*={\frac{2^{8n+8}-1}{17}}={\frac{3}{17}}\,
J_{8n+8}}.
\begin{lem}\label{l13}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
    \hspace{1cm}J_{2k}\,(4\,({\eta}_{n+1}^*-
    {\eta}_{n}^*)\,J_{2k-2}+{\beta}_{n+1}^*-
    {\beta}_{n}^*)=J_{2k+4n+6}^2-J_{2k+4n+4}^2+
   J_{4n+4}^2-J_{4n+6}^2.\hspace{0.5cm}
        \end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {6400}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {32}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k} -1
\right) AC+\\{ \frac {1280}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {32}{3}
}\,\left(\left( -1 \right) ^{2\,k}+\left( -1
 \right) ^{4\,n+1} \right) C.
\end{multline*}
\end{proof}
\begin{thm}\label{t4}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq
{\sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i}^2=
\sum_{i=0}^{m}\,(-1)^i\,J_{2i}^2+J_{2\,k}
\left[4\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^*_n\right],} if \z {m=2n} and \z
{n=0,\,1,\,2,\dots}  and \qq {
\sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i}^2=
\sum_{i=0}^{m}\,(-1)^i\,J_{2i}^2-J_{2\,k}
\left[4\,{\eta}^*_n\,J_{2k-2}+
{\beta}^*_{n}\right],} if \z {m=2n+1} and \z
{n=0,\,1,\,2,\dots}.
\end{thm}

\begin{proof}[Proof of (5.5)] ({\textbf{P3}}).
For \z {n=0} the relation (5.5) is \q
{J_{2k}^2=J_{0}^2+J_{2k}\,(4\,{\gamma}_0^*\,
J_{2k-2}+{\tau}_0^*)=J_{2k}\,(4\, J_{2k-2}+1)}
(i. e., the relation (5.1) multiplied by \z
{J_{2k}}) which is true by Lemma 10.

Assume that the relation (5.5) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{2(r+1)}\,(-1)^i\cdot
J_{2k+2i}^2=Q_{2k+4r+4}^2-Q_{2k+4r+2}^2+
\sum_{i=0}^{2r}\,(-1)^i\cdot
J_{2k+2i}^2\\
=Q_{2k+4r+4}^2-Q_{2k+4r+2}^2+
\sum_{i=0}^{2r}\,(-1)^i\cdot J_{2i}^2+
J_{2\,k} \left[4\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^*_n\right]=\\
\sum_{i=0}^{2(r+1)}\,(-1)^i\cdot J_{2i}^2+
J_{2\,k} \left[4\,{\gamma}^*_{n+1}\,J_{2k-2}+
{\tau}^*_{n+1}\right],
\end{multline*}
where the last step uses Lemma 12.
\end{proof}

\begin{proof}[Proof of (5.6)] ({\textbf{P3}}).
For \z {n=0} the relation (5.6) is \q {
    J_{2k}^2-J_{2k+2}^2=(J_0^2-J_2^2)-
J_{2\,k} \left[4\,{\eta}^*_0\,J_{2k-2}+
{\beta}^*_{0}\right]
    =-1-J_{2\,k}
\left[60\,J_{2k-2}+ 23\right],} which is true
by Lemma 11.

Assume that the relation (5.6) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{2(r+1)+1}\,(-1)^i\cdot
J_{2k+2i}^2=J_{2k+4r+4}^2-J_{2k+4r+6}^2+
\sum_{i=0}^{2r+1}\,(-1)^i\cdot
J_{2k+2i}^2\\
=J_{2k+4r+4}^2-J_{2k+4r+6}^2+
\sum_{i=0}^{2r+1}\,(-1)^i\cdot J_{2i}^2-
J_{2\,k} \left[4\,{\eta}^*_n\,J_{2k-2}+
{\beta}^*_{n}\right]\\
=\sum_{i=0}^{2(r+1)+1}\,(-1)^i\cdot J_{2i}^2-
J_{2\,k} \left[4\,{\eta}^*_{n+1}\,J_{2k-2}+
{\beta}^*_{n+1}\right],
\end{multline*}
where the last step uses Lemma 13.
\end{proof}

\section{Alternating Jacobsthal odd squares}
In this section we look for formulas that give
closed forms for alternating sums of squares
of Jacobsthal numbers with odd indices.
\begin{lem}\label{l14}
For every \z {k\geqslant 0} we have \qq
{J_{2k+1}^2=1+8\,J_{2k}\left[
2\,J_{2k-2}+1\right].}
\end{lem}
\begin{proof} ({\textbf{P2}}).
\z {{\frac{\left( P-1 \right) \left( P+1
\right)}{3}}\left(8\,M^2-5\,{P}^{2}+3
\right).}
\end{proof}

\begin{lem}\label{l15}
    For every \z {k\geqslant 0} we have
    \qq {J_{2k+3}^2-J_{2k+1}^2=
    8+8\,J_{2k}\,(30\,J_{2k-2}+13).}
\end{lem}
\begin{proof} ({\textbf{P2}}).
\z {{\frac{\left( P-1 \right) \left( P+1
\right)}{3}}\left(13\,M^2-10\,{P}^{2}+3
\right).}
\end{proof}
Let \z {{\tau}_0^{**}=1} and \z
{{\tau}_{n+1}^{**}-{\tau}_{n}^{**}= 101\cdot
2^{4n+1}+25\cdot 2^{4n+3}\,(2^{4n}-1)} for \z
{n=0,\,1,\,2,\dots}.
\begin{lem}\label{l16}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
8\,J_{2k}\,(2\,({\gamma}_{n+1}^*-
    {\gamma}_{n}^*)\,J_{2k-2}+{\tau}_{n+1}^{**}-
    {\tau}_{n}^{**})=
J_{2k+4n+5}^2-J_{2k+4n+3}^2-
    J_{4n+5}^2+J_{4n+3}^2.\hspace{0.5cm}
        \end{multline}
    \end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {1600}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {16}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k+1} +1
\right) AC+\\{ \frac {320}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {16}{3}
}\,\left(\left( -1 \right) ^{2\,k+1}+\left( -1
 \right) ^{4\,n} \right) C.
\end{multline*}
\end{proof}
%For any \z {n=0,\,1,\,2,\dots} let \z
%{{\eta}_n^*={\frac{2^{8n+8}-1}{17}}={\frac{3}{17}}\,
%J_{8n+8}}.

Let \z {{\beta}_0^{**}=13} and \z
{{\beta}_{n+1}^{**}-{\beta}_{n}^{**}= 401\cdot
2^{4n+3}+100\cdot
2^{4n+5}\left(2^{4n}-1\right)} for any \z
{n=0,\,1,\,2,\dots}.
\begin{lem}\label{l17}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
8\,J_{2k}\,(2\,({\eta}_{n+1}^*-
    {\eta}_{n}^*)\,J_{2k-2}+{\beta}_{n+1}^{**}-
    {\beta}_{n}^{**})=
J_{2k+4n+7}^2-J_{2k+4r+5}^2+
   J_{4n+5}^2-J_{4r+7}^2.\hspace{0.8cm}
        \end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {25600}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {64}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k+1} +1
\right) AC+\\{ \frac {5120}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {64}{3}
}\,\left(\left( -1 \right) ^{2\,k+1}+\left( -1
 \right) ^{4\,n} \right) C.
\end{multline*}
\end{proof}
\begin{thm}\label{t5}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq
{\sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i+1}^2=
\sum_{i=0}^{m}\,(-1)^i\,J_{2i+1}^2+8\,J_{2\,k}
\left[2\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^{**}_n\right],} if \z {m=2n} and \z
{n=0,\,1,\,2,\dots}  and \qq {
\sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i+1}^2=
\sum_{i=0}^{m}\,(-1)^i\,J_{2i+1}^2-8\,J_{2\,k}
\left[2\,{\eta}^*_n\,J_{2k-2}+
{\beta}^{**}_n\right],} if \z {m=2n+1} and \z
{n=0,\,1,\,2,\dots}.
\end{thm}

\begin{proof}[Proof of (6.5)] ({\textbf{P3}}).
For \z {n=0} the relation (6.5) is \q
{J_{2k+1}^2=J_1^2+8\,J_{2k}(2\,{\gamma}_0^*\,
J_{2k-2}+{\tau}_0^{**})} (i. e., the relation
(6.1)) which is true by Lemma 14.

Assume that the relation (6.5) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{2(r+1)}\,(-1)^i\,
J_{2k+2i+1}^2=J_{2k+4r+5}^2-J_{2k+4r+3}^2+
\sum_{i=0}^{2r}\,(-1)^i\,
J_{2k+2i+1}^2\\
=J_{2k+4r+5}^2-J_{2k+4r+3}^2+
\sum_{i=0}^{2r}\,(-1)^i\,
J_{2i+1}^2+8\,J_{2\,k}
\left[2\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^{**}_n\right]\\
=\sum_{i=0}^{2(r+1)}\,(-1)^i\,
J_{2i+1}^2+8\,J_{2\,k}
\left[2\,{\gamma}^*_{n+1}\,J_{2k-2}+
{\tau}^{**}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 16.
\end{proof}

\begin{proof}[Proof of (6.6)] ({\textbf{P3}}).
The proof is again by induction on \z {n}. For
\z {n=0} the relation (6.6) is \z
{J_{2k+1}^2-J_{2k+3}^2=J_1^2-J_3^2-
8\,J_{2\,k} \left[2\,{\eta}^*_0\,J_{2k-2}+
{\beta}^{**}_0\right]} which is true by Lemma
15 since \z {{\eta}^*_0=15} and \z
{{\beta}^{**}_0=13}.

Assume that the relation (6.6) is true for \z
{n=r}. Then
\begin{multline*}
 \sum_{i=0}^{2(r+1)+1}\,(-1)^i\,
J_{2k+2i+1}^2=J_{2k+4r+5}^2-J_{2k+4r+7}^2+
\sum_{i=0}^{2r+1}\,(-1)^i\, J_{2k+2i+1}^2\\
=J_{2k+4r+5}^2-J_{2k+4r+7}^2+
\sum_{i=0}^{2r+1}\,(-1)^i\,
J_{2i+1}^2-8\,J_{2\,k}
\left[2\,{\eta}^*_n\,J_{2k-2}+
{\beta}^{**}_n\right]\\
=\sum_{i=0}^{2(r+1)+1}\,(-1)^i\,
J_{2i+1}^2-8\,J_{2\,k}
\left[2\,{\eta}^*_{n+1}\,J_{2k-2}+
{\beta}^{**}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 17.
\end{proof}


\section{Alternating Jacobsthal products}

\begin{lem}\label{l18}
    For every \z {k\geqslant 0} we have
    \qq {J_{2k+3}\,J_{2k+2}-
    J_{2k+1}\,J_{2k}=3+J_{2k}\,\left[
    120\,J_{2k-2}+49\right].}
\end{lem}
\begin{proof} ({\textbf{P2}}).
\z {{\frac{\left( P-1 \right) \left( P+1
\right)}{3}}\left(49\,M^2-40\,{P}^{2}+9
\right).}
\end{proof}
\begin{lem}\label{l19}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
\hspace{1cm}
J_{2k}\,\left(8\,({\gamma}_{n+1}^*-
    {\gamma}_{n}^*)\,J_{2k-2}+{\tau}_{n+1}^{***}-
    {\tau}_{n}^{***}\right)=\\
   J_{2k+4n+5}\,J_{2k+4n+4}-
   J_{2k+4n+3}\,J_{2k+4n+2}-J_{4n+5}\,J_{4n+4}
   +J_{4n+3}\,J_{4n+2}.
   \hspace{1cm}
        \end{multline}
    \end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {800}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {4}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k} -1
\right) AC+\\{ \frac {160}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {4}{3}
}\,\left(\left( -1 \right) ^{2\,k}+\left( -1
 \right) ^{4\,n+1} \right) C.
\end{multline*}
\end{proof}
Let \z {{\beta}_0^{***}=49} and \z
{{\beta}_{n+1}^{***}-{\beta}_{n}^{***}=
799\cdot 2^{4n+4}+25\cdot
2^{4n+9}\left(2^{4n}-1\right)} for any \z
{n=0,\,1,\,2,\dots}.
\begin{lem}\label{l20}
    For every \z {k\geqslant 0} and every
    \z {n\geqslant 0} we have
\begin{multline}
\hspace{1cm}J_{2k}\,(8\,({\eta}_{n+1}^*-
    {\eta}_{n}^*)\,J_{2k-2}+{\beta}_{n+1}^{***}-
    {\beta}_{n}^{***})=\\
   J_{2k+4n+6}\,J_{2k+4n+7}-
   J_{2k+4n+4}\,J_{2k+4n+5}+J_{4n+5}\,J_{4n+4}
   -J_{4n+6}\,J_{4n+7}.\hspace{1cm}
        \end{multline}
\end{lem}
\begin{proof} ({\textbf{P1}}).
\begin{multline*}
{\frac {12800}{3}}\,\left(  \left( -1 \right)
^{1+2\,k}+1 \right) AB+ {\frac {16}{3}}\,
\left( \left( -1 \right) ^{4\,n+2\,k} -1
\right) AC+\\{ \frac {2560}{3}}\,\left(
4\,\left( -1 \right) ^{4\,k}+5\, \left( -1
\right) ^{1+2\,k}+1\right) B +{\frac {16}{3}
}\,\left(\left( -1 \right) ^{2\,k}+\left( -1
 \right) ^{4\,n+1} \right) C.
\end{multline*}
\end{proof}

\begin{thm}\label{t6}
For every \z {m\geqslant 0} and \z {k\geqslant
0} the following equalities hold \qq {
    \sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i}\,J_{2k+2i+1}=
    \sum_{i=0}^{m}\,(-1)^i\,J_{2i}\,J_{2i+1}+
    J_{2\,k}\left[8\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^{***}_n\right],} if \z {m=2n} and \z
{n=0,\,1,\,2,\dots}  and \qq {
\sum_{i=0}^{m}\,(-1)^i\,J_{2k+2i}\,J_{2k+2i+1}=
\sum_{i=0}^{m}\,(-1)^i\,J_{2i}\,J_{2i+1}-J_{2\,k}
\left[8\,{\eta}^*_n\,J_{2k-2}+
{\beta}^{***}_n\right],} if \z {m=2n+1} and \z
{n=0,\,1,\,2,\dots}.
\end{thm}

\begin{proof}[Proof of (7.4)] ({\textbf{P3}}).
For \z {n=0} the relation (7.4) is \q
{J_{2k}\,J_{2k+1}=J_0\,J_1+
    J_{2\,k}\left[8\,{\gamma}^*_0\,J_{2k-2}+
{\tau}^{***}_0\right]=
J_{2\,k}\left[8\,J_{2k-2}+3\right]} which is
true by the relation (4.1) in Lemma 7 since \z
{J_0=0}, \z {J_1=1}, \z {{\gamma}^*_0=1} and
\z {{\tau}^{***}_0=3}.

Assume that the relation (7.4) is true for \z
{n=r}. Then
\begin{multline*}
\hspace{-0.2cm}\sum_{i=0}^{2(r+1)}(-1)^i
J_{2k+2i}J_{2k+2i+1}=\sum_{i=0}^{2r}(-1)^i
J_{2k+2i}J_{2k+2i+1}+ J_{2k+4r+4}J_{2k+4r+5}-
J_{2k+4r+2}J_{2k+4r+3}\\
=J_{2k+4r+4}J_{2k+4r+5}-
J_{2k+4r+2}J_{2k+4r+3}+\sum_{i=0}^{2r}(-1)^i
J_{2i}J_{2i+1} +J_{2\,k}
\left[8\,{\gamma}^*_n\,J_{2k-2}+
{\tau}^{***}_n\right]\\
=\sum_{i=0}^{2(r+1)}(-1)^i J_{2i}J_{2i+1}
+J_{2\,k} \left[8\,{\gamma}^*_{n+1}\,J_{2k-2}+
{\tau}^{***}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 19.
\end{proof}

\begin{proof}[Proof of (7.5)] ({\textbf{P3}}).
For \z {n=0} the relation (7.5) is
\begin{multline*}
    J_{2k}\,J_{2k+1}-J_{2k+2}\,J_{2k+3}=
    J_0\,J_1-
J_2\,J_3-\\
    J_{2k}\left[8\,{\eta}^*_0\,J_{2k-2}
+{\beta}^{***}_0\right] =-3-J_{2k}
\left[120\,J_{2k-2} +49\right],
\end{multline*}
(i. e., the relation (7.1)) which is true by
Lemma 18.

Assume that the relation (7.5) is true for \z
{n=r}. Then
\begin{multline*}
    \sum_{i=0}^{2(r+1)+1}\,(-1)^i\,
J_{2k+2i}\,J_{2k+2i+1}=\sum_{i=0}^{2r+1}\,(-1)^i\,
J_{2k+2i}\,J_{2k+2i+1}+
J_{2k+4r+4}\,J_{2k+4r+5}-\\
J_{2k+4r+6}\,J_{2k+4r+7}
=J_{2k+4r+4}\,J_{2k+4r+5}-
J_{2k+4r+6}\,J_{2k+4r+7}+\sum_{i=0}^{2r+1}\,(-1)^i\,
J_{2i}\,J_{2i+1} -\\ J_{2\,k}
\left[8\,{\eta}^*_n\,J_{2k-2}+
{\beta}^{***}_n\right]
=\sum_{i=0}^{2(r+1)+1}\,(-1)^i\,
J_{2i}\,J_{2i+1} -J_{2\,k}
\left[8\,{\eta}^*_{n+1}\,J_{2k-2}+
{\beta}^{***}_{n+1}\right],
\end{multline*}
where the last step uses Lemma 20.
\end{proof}




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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject
Classification}: Primary 11B39; Secondary 11Y55, 05A19.

\noindent \emph{Keywords:} Jacobsthal numbers,
Jacobsthal-Lucas numbers, integer sequences,
sum of squares, combinatorial identity, Maple.



\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001045} and \seqnum{A014551}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent Received May 17 2006; revised version received January 19 2007.
Published in {\it Journal of Integer Sequences}, January 19 2007.

\bigskip
\hrule
\bigskip

\noindent Return to \htmladdnormallink{Journal
of Integer Sequences home
page}{http://www.math.uwaterloo.ca/JIS/}.

\end{document}
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