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\begin{center}
\vskip 1cm{\LARGE\bf On the Diophantine Equation \\
\vskip .1in
$x^4+y^4+z^4+t^4 = w^2$ 
}
\vskip 1cm
\large
Alejandra Alvarado  \\
Eastern Illinois University  \\
Department of Mathematics and Computer Science \\
600 Lincoln Avenue \\
Charleston, IL 61920 \\
USA\\
\href{mailto:aalvarado2@eiu.edu}{\tt aalvarado2@eiu.edu} \\
\ \\
Jean-Jo\"{e}l Delorme \\
6 rue des Emeraudes \\
69006 Lyon \\
France \\
\href{mailto:jean.joel.delorme@wanadoo.fr }{\tt jean.joel.delorme@wanadoo.fr} \\
\end{center}

\vskip .2 in

\begin{abstract}
To our knowledge, only three parametric solutions to the equation $x^4+y^4+z^4+t^4 = w^2$ were previously known.  In this paper, we study the equation $x^4+y^4+z^4+t^4 = (x^2+y^2+z^2-t^2)^2$.  We prove that it is possible to obtain infinitely many parametric solutions by finding points on an elliptic curve over a field $\mathbb{Q}(m)$ and we give several new parametric solutions.
\end{abstract}

\section{Introduction}
Jacobi and Madden \cite{JacMad} considered the equation 
\begin{equation}\label{JM}
x^4+y^4+z^4+t^4 = (x+y+z+t)^4.
\end{equation}  
They showed the existence of infinitely many integral solutions to \eqref{JM}.  This is a special case of the equation 
\begin{equation}\label{NE}
x^4+y^4+z^4+t^4 = w^4,
\end{equation}  
for which Elkies \cite{NE} found an infinite family of integral solutions when $t=0$.   In this paper, we consider a special case of a similar equation 
\begin{equation}\label{title}
x^4+y^4+z^4+t^4 = w^2.
\end{equation}


\section{Background}
Consider the equation  (\ref{title}).  We say that a solution is \emph{trivial} if at least three of the numbers $ x , y , z , t , w$ are zero, for instance $ ( x , y , z , t , w ) = ( x , 0 , 0 , 0 , x^2 ) $.   If two and only two of the numbers $ x , y , z , t $ are zero, the equation has no nontrivial solution  
since Fermat proved that the equation $ x^4 + y^4 = w^2 $ has no solution in nonzero integers. 

The first known parametric solution is nontrivial but very elementary: 
\begin{center}
$(x,  y,  z,  t,  w) = (a^2,  ab,  b^2,  ab,  a^4+b^4).$
\end{center}
In the next solution, found by Fauquembergue \cite{EF}, one of the numbers $ x , y , z , t, w $ is zero, for instance $ z = 0 $: 
\begin{center}
$(x,  y, z, t,  w) = (ac,  bc,  0,  ab,  a^4+a^2 b^2+b^4)$
\end{center}
where  $ a^2 + b^2 = c^2 $.
The following solution was also found by Fauquembergue \cite{EF}, again assuming  $a^2+b^2=c^2$:
\begin{align*}
(x,  y,  z,  t,  w) & = \left(2 a^2 b c^3,  2 a b^2 c^3,  ( a^2 - b^2 ) c^4,  2 a b ( a^4 + b^4 ), \right.\\ 
& \quad \left. ( a^6 + 2 a^5 b + 3 a^4 b^2 + 3 a^2 b^4 + 2 a b^5 + b^6 ) ( a^6 - 2 a^5 b + 3 a^4 b^2 + 3 a^2 b^4 - 2 a b^5 + b^6 ) \right).
\end{align*}
These three parametric solutions yield nontrivial numerical solutions, unless $ a b = 0 $. 


\section{The Equation $x^4+y^4+z^4+t^4 = (x^2 + y^2 + z^2 - t^2)^2$}
While investigating solutions to equation (\ref{title}), the second author noticed some interesting properties.  After some numerical results, he considered the following three cases.

\begin{itemize}
\item If $w=x^2+y^2+z^2+t^2$ then
$x^2y^2+x^2z^2+z^2t^2+y^2z^2+y^2t^2+z^2t^2=0$; hence, there are only trivial
solutions.

\item If $w=x^2+y^2-z^2-t^2$ then
$x^2y^2-x^2z^2-y^2z^2=t^2(x^2+y^2-z^2)$.  This is an interesting but
complicated case.  We leave this for future work.

\item If $w=x^2+y^2+z^2-t^2$ then
$x^2y^2+x^2z^2+y^2z^2=t^2(x^2+y^2+z^2)$.  This case also looked
interesting and will be discussed below, beginning with the following
proposition.  We believe our analysis of this case is new.
\end{itemize}


\begin{proposition}
If $ x^2 + y^2 + z^2 \not= 0 $, then $ ( x , y , z , t ) $ satisfies 
\begin{equation}\label{E1}
x^4 + y^4 + z^4 + t^4 = ( x^2 + y^2 + z^2 - t^2 )^2
\end{equation}
if and only if
\begin{equation}\label{E2} 
( x^2 + y^2 + z^2 ) ( y^2 z^2 + z^2 x^2 + x^2 y^2 )=\square 
\end{equation} and 
\begin{equation}\label{T2}
t^2 \ = \ \frac { y^2 z^2 + z^2 x^2 + x^2 y^2 } { x^2 + y^2 + z^2 } . 
\end{equation}
\label{prop1}
\end{proposition}

\begin{proof}
We have
$$ x^4 + y^4 + z^4 + t^4 - ( x^2 + y^2 + z^2 - t^2 )^2  = 2 \big( ( x^2 + y^2 + z^2 ) t^2 - ( y^2 z^2 + z^2 x^2 + x^2 y^2 ) \big). $$
\end{proof}
Since  (\ref{E1}) is homogeneous of degree four, from now on we will write solutions to this equation as $(x:y:z:t)$.  The equation represents a surface in $\mathbb{P}^3$,
$$\mathcal{S}' =\left\{ (x:y:z:t)\in \mathbb{P}^3 \ | \ x^4+y^4+z^4+t^4 = (x^2 + y^2 + z^2 - t^2)^2  \right\}. $$
Assuming $t\neq 0$, we can view the surface in affine form by corresponding $(x,y,z)\leftrightarrow (x:y:z:1)$.  This gives a rather interesting looking surface in three dimensions; see Figure~\ref{Delorme}.
\begin{figure}[h]
\begin{center}
\includegraphics{Delorme.eps}
\caption{Plot of $\mathcal{S}' $ in affine space}\label{Delorme}
\end{center}
\end{figure}

The main focus of this paper is to answer the following question.
\begin{question}
How many rational curves $m\mapsto \left(x(m),y(m),z(m) \right)$ are on the surface $\mathcal{S}'$?
\end{question}

If $ x y z \not= 0 $, then (\ref{E2}) can be expressed as
$$ ( x^2 + y^2 + z^2 ) \left( \ \frac {1} {x^2} + \frac {1} {y^2} + \frac {1} {z^2} \ \right) = \square .$$
This leads to the following lemma.

\begin{lemma} \label{LEM}
If $ ( x: y: z :t ) $ is a solution to (\ref{E1}) such that $ x y z t \not= 0 $ 
then $\displaystyle \left( \frac {1}{x} : \frac {1}{y}: \frac {1}{z} : \frac {1}{t} \right)$ is also a solution to ({\ref{E1}}). 
\end{lemma}

\begin{proof}
Let $\displaystyle (X:Y:Z:T)=\left(\frac {1}{x}: \frac {1}{y}: \frac {1}{z}: \frac {1}{t} \right)$.  Then
\begin{align*}
X^2 + Y^2 + Z^2 & = \frac {1} { x^2 y^2 z^2 } \ ( y^2 z^2 + z^2 x^2 + x^2 y^2 ) \text{ and}\\
Y^2 Z^2 + Z^2 X^2 + X^2 Y^2 & = \frac {1} { x^2 y^2 z^2 } ( x^2 + y^2 + z^2 )\\
\end{align*} 
Hence 
$$\dfrac{ Y^2 Z^2 + Z^2 X^2 + X^2 Y^2 } { X^2 + Y^2 + Z^2 } = \dfrac{ x^2 + y^2 +z^2 } { y^2 z^2 + z^2 x^2 + x^2 y^2 } =\dfrac{1}{t^2}=T^2$$
\end{proof}
The following examples can be shown to be solutions to (\ref{E1}).
\begin{itemize}
\item The elementary solution $\mathcal{F}_0= (a^2:  ab:  b^2:  ab)$.

\item The first solution of Fauquembergue  $\mathcal{F}_1= (ac:  bc:  0:  ab)$ where $a^2+b^2=c^2$. 

\item The second solution of Fauquembergue $\mathcal{F}_2=  (2 a^2 b c^3: 2 a b^2 c^3:  ( a^2 - b^2 ) c^4:  
2 a b ( a^4 + b^4 ))$ where $a^2+b^2=c^2$ .
\end{itemize}
An application of the previous lemma to the second solution of Fauquembergue, we deduce a new solution to (\ref{E1}).
\begin{proposition}
If $ a^2 + b^2 =c^2$ and
\begin{align*}
x & =  a c  ( a^2 - b^2 )  ( a^4 + b^4 ) \\ 
y & =  b c   ( a^2 - b^2 )  ( a^4 + b^4 ) \\ 
z & =  2 a^2 b^2  ( a^4 + b^4 ) \\ 
t & =  a b (a^2+b^2)^2  ( a^2 - b^2 ) 
\end{align*}
then $(x:y:z:t)$ is a solution to (\ref{E1}).
\end{proposition} 
We will label this solution $\mathcal{D}_1$.  
For example, if $(a:b:c) = (4:3:5)$ then $(x:y:z:t)=(47180:35385:97056:52500)$ is a solution to (\ref{E1}).

\section{An Elliptic Curve over $\mathbb{Q}(m)$}

We begin this section by providing some background on elliptic
surfaces, which can be defined as a one-parameter algebraic family of
elliptic curves.  See Silverman \cite[Chapter 3]{S2}.

Let $C$ be a curve defined over a field $k$.  Consider a rational map $C\rightarrow \mathbb{P}^1$.  The collection of all such maps is denoted by $K=k(C)$. For example, if $C: a^2+b^2=c^2$ is defined over $\mathbb{Q}$, we have an isomorphism $C \rightarrow \mathbb{P}^1$ given by 
$$(a:b:c) \mapsto m=\dfrac{a}{c-b} \iff (a:b:c) = (2m:m^2-1:m^2+1).$$
Hence, $K=\mathbb{Q}(C)=\mathbb{Q}(m)$.  This is the field we will consider.

Consider a family of curves $$E_m : x_2^2 = x_1^3+A(m)x_1+B(m)$$ with rational functions $A(m), B(m) \in K$.  
If we rewrite our equation in homogeneous form, we form the elliptic surface
$$\mathcal{E} = \left\{ ((x_1: x_2: x_3),m) \in\mathbb{P}^2\times C | x_2^2x_3 = x_1^3+A(m)x_1x_3^2+B(m)x_3^3   \right\}. $$
We have a map $\pi: \mathcal{E}\rightarrow C$ defined by $((x_1:x_2:x_3),m)\mapsto m$.  
For $m\in\mathbb{Q}$ such that $4A(m)^3+27B(m)^2\neq 0$, the \emph{fiber} $$\mathcal{E}_m = \pi^{-1}(m) = \left\{ (x_1:x_2: 1)\in\mathbb{P}^2 |  x_2^2x_3 = x_1^3+A(m)x_1x_3^2+B(m)x_3^3   \right\}$$ is the elliptic curve $E_m$ over $\mathbb{Q}$.

We say that the elliptic surface is \emph{non-split} if the $j$-invariant
$$ j: C\rightarrow\mathbb{P}^1 \text{ defined by } j(m)=1728\dfrac{4A(m)^3}{4A(m)^3+27B(m)^2} $$
is a non-constant function. A \emph{parametrization of $\mathcal{E}$ by $C$}, or a \emph{section to $\pi$}, is a map $\sigma : C\rightarrow \mathcal{E}$ such that the composition $\pi\circ\sigma : m\mapsto m$ is the identity map on $C$.  There is always a trivial section on $\mathcal{E}$, namely the map $\sigma_0:m\mapsto \mathcal{O}_m=((0:1:0),m)$.   In general, the collection $\mathcal{E}(C)$ of all sections is an abelian group, where we define 
\begin{align*} 
\sigma_1(m) & = (P(m),m) & & &\\
 & & \Longrightarrow & & (\sigma_1\oplus \sigma_2)(m)  = (P(m)\oplus Q(m), m) .\\
\sigma_2(m) & = (Q(m),m) & &  &\\
\end{align*}
We often abuse notation and write $E: x_2^2=x_1^3+Ax_1+B$ as an elliptic curve over the function field $K=k(C)$.  In fact, we have an isomorphism between the group of points of $E$ over $K$ and the group of sections of $\mathcal{E}$ over $\mathbb{Q}$.

\begin{align*}
&E(K) && \tilde{\longrightarrow} && \ \ \mathcal{E}(C) \\
P(m) &= (x_1(m): x_2(m):x_3(m)) && \mapsto && \left[ \sigma: m \mapsto ((x_1(m): x_2(m): x_3(m)), m)  \right] \\
\end{align*}
It will be helpful to consider $\mathcal{E}$ as a two-dimensional
surface, where each section maps to a one-dimensional curve.  
The result \cite[Theorem 6.1, Chapter 3]{S2} asserts that $E(K)\simeq
\mathcal{E}(C)$ is a finitely generated abelian group whenever
$\mathcal{E}$ is a non-split surface.

Let $m_0\in\mathbb{Q}$ such that $E_0=\pi^{-1}(m_0)$ is an elliptic curve over $k=\mathbb{Q}$.  
Silverman's ``specialization theorem'' 
\cite[Theorem 11.4, Chapter 3]{S2} 
asserts that the map $\mathcal{E}(C)\rightarrow E_0(k)$ which sends a section $\sigma : m \mapsto (P(m),m)$ to the point $P_0=P(m_0)$ is injective for all but finitely many $m_0\in k$.  In particular, if $P_0$ is a point of finite (infinite) order in $E_0(k)$ for some $m_0\in k$, then $P(m)$ must be a point of finite (infinite) order in $E(K)$ as a function of $m$.  


Let us return to the condition
$$ ( x^2 + y^2 + z^2 ) ( y^2 z^2 + z^2 x^2 + x^2 y^2 ) = \square.$$  
If $a^2+b^2=c^2$ nonzero then we can express $(a:b:c)=(2m: m^2-1: m^2+1)$.  For the sake of space, we will leave our work below in terms of $a,b,c$.  If we impose the condition $(x,y)=(a,b)$, we obtain  
$$( c^2 + z^2 ) ( c^2 z^2 + a^2 b^2 ) = \square .$$
Dividing by $ c^2 $, we can consider the following equation,
\begin{equation}\label{E3}
h^2 = z^4 + \dfrac { a^4 + 3  a^2 b^2 + b^4 } { c^2 } \ z^2 + a^2 b^2   
\end{equation}
From the preceding examples $\mathcal{F}_0, \mathcal{F}_1, \mathcal{F}_2, \mathcal{D}_1$, we know four solutions to (\ref{E3}), 
\begin{align*}
(z,h)_{\mathcal{F}_0} & =  \left(\dfrac {b^2} {a}, \dfrac { b ( a^4 + a^2 b^2 + b^4 ) } { a^2 c }  \right); \\
(z,h)_{\mathcal{F}_1} & = (0, ab); \\
(z,h)_{\mathcal{F}_2} & =  \left(\dfrac { ( a^2 - b^2 ) c } { 2 \ a b }, \dfrac { (a^4 + b^4 ) c^2 } { 4 \ a^2 b^2 }\right); \\
(z,h)_{\mathcal{D}_1} & = \left( \dfrac{ 2 \ a^2 b^2 }{ ( a^2 - b^2 ) c}, \dfrac{ (a^4 + b^4 ) a b }{ ( a^2 - b^2 )^2 } \right).
\end{align*}


\begin{comment}
If we impose the condition $( x , y ) = (a,b)= ( 2m , m^2-1 )$, we obtain
$$\left( (m^2+1)^2 + z^2 \right) \left( (m^2+1)^2 z^2 + ((2m)(m^2-1))^2 \right) = \square .$$
Dividing by $ (m^2+1)^2 $, we can consider the following equation,
\begin{equation}\label{E3}
h^2 = z^4 + \dfrac {m^8+8m^6-2m^4+8m^2+1  } { (m^2+1)^2 }  z^2 + \left((2m)(m^2-1)\right)^2
\end{equation}
From the preceding examples, $\mathcal{F}_0, \mathcal{F}_1, \mathcal{F}_2, \mathcal{D}_1,$ we know four solutions $(z,h)$ to (\ref{E3}), imposing the condition $a^2+b^2=c^2$ and  $ ( x , y ) = ( a , b ) $.  In terms of the
parameter $m$ we obtain 
\begingroup
\addtolength{\jot}{1em}
\begin{align*}
(z,h)_{\mathcal{F}_0} &  =\left(\dfrac {(m^2-1)^2} {2m}, \dfrac { (m^2-1)(m^4-2m^3+2m^2+2m+1)(m^4+2m^3+2m^2-2m+1) } { 4m^2 (m^2+1) }  \right) ; \\  
(z,h)_{\mathcal{F}_1} &  = (0, 4m(m^2-1)) ; \\ 
(z,h)_{\mathcal{F}_2} & = \left(\dfrac { -(m^2+1)(m^2+2m-1)(m^2-2m-1) } { 4m(m^2-1) }, \dfrac { (m^2+1)^2(m^8-4m^6+22m^4-4m^2+1) } { 16m^2(m^2-1)^2 }\right) ; \\ 
(z,h)_{\mathcal{D}_1} &  = \left( \dfrac{ 8m^2(m^2-1)^2 }{ -(m^2+1)(m^2+2m-1)(m^2-2m-1) }, \dfrac{ 2m(m^2-1)(m^8-4m^6+22m^4-4m^2+1)}{ ((m^2+2m-1)(m^2-2m-1))^2 } \right) . \\
\end{align*}
\endgroup
\end{comment}

We will show that, in fact, there are infinitely many parametric solutions to equation (\ref{E1}) by showing there are infinitely many parametric solutions to (\ref{E3}).
\begin{theorem}\label{ParaSol}
Parametric solutions of the equation $$x^4+y^4+z^4+t^4=(x^2+y^2+z^2-t^2)^2$$ may be obtained by finding points on an elliptic curve over the field $\mathbb{Q}(m)$.
\end{theorem}
\begin{proof}
By Proposition~\ref{prop1} and assuming $a^2+b^2=c^2$ nonzero, consider equation (\ref{E3}). Let $ A  =  \dfrac { a^4 + 3 a^2 b^2 + b^4 } { 4(a^2+b^2) }  $ and $  B  =  \dfrac {a^2 b^2} {4} $, so that (\ref{E3}) can be expressed as 
\begin{equation}\label{ZH}
h^2 = z^4 + 4  A z^2 + 4  B 
\end{equation}
If we have a rational solution to (\ref{ZH}) then we get a rational solution to 
\begin{equation}\label{UV}
v^2 = u^3 + \alpha u^2 + \beta u
\end{equation}
where $ \alpha = - 2 A $ and $ \beta = A^2 - B $, by 
$$ u  = \dfrac {1} {2}  ( z^2 + 2 A - h ) \ \text{ and } \ v =  \dfrac {1} {2} z ( z^2 + 2 A - h ).  $$
Conversely, assuming $(u,v)\neq (0,0)$,  a  rational solution to (\ref{UV})  leads to a rational solution to (\ref{ZH}) by 
$$ z  =  \dfrac {v} {u}   \ \text{ and } \  h  =  \dfrac { v^2 } { u^2 } + 2  A - 2 u .$$
The discriminant  of (\ref{UV}) is  
$$( \alpha^2 - 4 \beta) \beta^2  =  4 B ( A^2 - B )^2 
= \dfrac { a^2 b^2 ( a^4 + a^2 b^2 + b^4 )^4  } { 256 (a^2+b^2)^4 } $$
which is nonzero since at least one of $a, b$ are nonzero.
Hence (\ref{UV}) defines an elliptic curve over the field $ \mathbb{Q} (m) $ and
every point $ ( u , v ) $ on this elliptic curve yields a point $ ( z , h ) $ satisfying (\ref{E3}), which implies a solution $ (x,y,z,t) $ to (\ref{E1}).
\end{proof}

We will show how to obtain new solutions, by adding points on the elliptic curve.  We write $E$ for the elliptic curve (\ref{UV}) over $\mathbb{Q}(m)$,  $+$ for the addition of points on the curve (\ref{UV}), and $P_S$ will denote a point on $E$ yielding a solution $S$ to (\ref{E1}).  The addition of points on an elliptic curve is described in Silverman \cite{S1}.  Note that instead of writing $P+P$ we will write $2P$.

\begin{example}  The solutions $ \mathcal{F}_0 ,  \ \mathcal{F}_1 ,\  \mathcal{F}_2 , \text{ and } \mathcal{D}_1 $ to (\ref{E1}) are provided
by the following points on (\ref{UV}):

\begingroup
\addtolength{\jot}{1em}
\begin{align*}
P_{ \mathcal{F}_0} & = \left( \ \frac { ( c - b ) \ ( a^2 - a b  + b^2 ) \ ( a^2 + a b + b^2 ) } { 4 \ ( c + b ) \ ( a^2 + b^2 ) } \ , 
 \frac { b^2 \ ( c - b ) \ ( a^2 - a b  + b^2 ) \ ( a^2 + a b + b^2 ) } { 4 \ a \ ( c + b ) \ ( a^2 + b^2 ) } \ \right) \\
P_{ \mathcal{F}_1} & = \left( \ \frac { ( a^2 - a b  + b^2 )^2 } { 4 \ ( a^2 + b^2 ) } \ , \ 0 \ \right) \\
P_{ \mathcal{F}_2}  & = \left( \ \frac { a^2 b^2 } { 4 \ ( a^2 + b^2 ) } \ , \ \frac { a b \ ( a^2 -  b^2 ) } { 8 c } \ \right) \\
P_{ \mathcal{D}_1} & = \left( \ \frac { ( a - b )^2 \ ( a^2 - a b  + b^2 )^2 } { 4 \ ( a + b )^2 \ ( a^2 + b^2 ) } \ , 
\ \frac { a^2 b^2 \ ( a - b ) \ ( a^2 + a b  + b^2 )^2 } { 2 c \ ( a + b )^3 \ ( a^2 + b^2 ) } \ \right) 
\end{align*}
\endgroup
\end{example}

Let us remind the reader of the Lutz-Nagell theorem, which will be used in the proof of Theorem~\ref{thm8} along with the ``specialization theorem''.
\begin{theorem}
Let $E$ be given by $y^2=x^3+Ax+B$ with $A,B\in\mathbb{Z}$.  Let $P=(x,y)\in E(\mathbb{Q})$.  Suppose $P$ has finite order.  Then $x,y\in \mathbb{Z}$.  If $y\neq 0$ then $y^2$ divides $4A^3+27B^2$.
\label{thm8}
\end{theorem}

\begin{theorem}
There exists infinitely many points on (\ref{UV}).
\end{theorem}
\begin{proof}
Let $N=\dfrac{a^2-ab+b^2}{2c}$ and $L=\dfrac{a^2+ab+b^2}{2c}$.
Then  (\ref{UV}) can be expressed as 
\begin{equation}\label{UV2} E: \quad v^2 = u\left(u-N^2\right)\left(u-L^2 \right) \end{equation}
It can be shown that the rank of this elliptic curve over $\mathbb{Q}(m)$ is at least one.   
To show the rank is at least one, specialize at say, $(a,b)=(3,4)$.  
Then the point $P= (36/25, 21/10)$ is on the curve $$E_1:\quad v^2 = u^3 - \dfrac{769}{50}u^2 + \dfrac{231361}{10000}u.$$  
In order to use the Lutz-Nagell theorem, we need to express $E_1$ in Weierstrass form with integral coefficients:  $$E_1':\quad y^2 = x^3 - 1538x^2 + 231361x .$$
The point $P$ on $E_1$ corresponds to $P' = (144,2100)$ on $E_1'$, and thus $$2P'=(70980625/28224 , 389867877575/4741632).$$  Since $2P'$ is not an integral point, $P'$ is not a point of finite order on $E_1'$, so by the ``specialization theorem'' the rank of (\ref{UV2}) is of positive rank, these are the same ideas as in Ulas \cite{MU}.  Also note, calculations found at least 17 distinct points on $E$.  The maximum number of torsion points is 16, so the rank must be at least one.
\end{proof}

\begin{remark}
On $E$, the torsion points  are $(0,0), \ (N^2,0), \ (L^2, 0), $ and $\mathcal{O}$, the point at infinity.
\end{remark}


\begin{figure}[h]
\begin{center}
\includegraphics[scale=1.5]{GroupLaw2.eps}
\caption{$E:  v^2=u^3+\alpha u^2+\beta u$}\label{grouplaw1}
\end{center}\end{figure}

\begin{theorem}
Every  solution to the equation 
$$x^4+y^4+z^4+t^4=(x^2+y^2+z^2-t^2)^2\quad\quad (\ref{E1})$$ such that $(x,y)=(a,b)$, $a$ and $b$ nonzero, proceeds from exactly two points on $E$ different from $(0,0)$.  If one of them is $P=(u,v)$, with $u\neq 0$, then the other one is $P'=(u',v')=\lambda (u,v)$, with $\lambda=\dfrac{\beta}{u^2}$.
\end{theorem}
\begin{proof}
Although $E$ is defined over a function field, Figure~\ref{grouplaw1} provides some intuition for our proof.
Let $P=(u, v)$ be a point on $E$, with $ u\not= 0$, which yields a solution to (\ref{E1}).  
If there exists a  point $P'=(u' , v')$ on $E$ different from $(0,0)$ yielding the same solution to (\ref{E1}) as $ ( u , v ) $, then $ z' = z $ so $u'\neq 0$ and $  \dfrac { v' } { u' } = \dfrac {v} {u} $. 
Thus there exists a  rational $ \lambda $ such that $ u' = \lambda u $ and $ v' = \lambda v $ and 
$$ 0 
= - v'^2 + u'^3 + \alpha u'^2 + \beta u'
= - \lambda^2 v^2 + \lambda^3 u^3 + \alpha \lambda^2 u^2 + \beta \lambda u . $$
By substituting for $v^2$, we have 
$$ 0 
= - \lambda^2 ( u^3 + \alpha u^2 + \beta u ) + \lambda^3 u^3 + \alpha^2 \lambda^2 u^2 + \beta \lambda u = \lambda \ ( \lambda - 1 ) \ u \ ( \lambda u^2 - \beta ) .$$
Since $ u' \not= 0 $, then $ \lambda \not= 0 $.  Assuming $ ( u' , v' ) \not= ( u , v ) $ implies $ \lambda \not= 1 $. 
Thus  $  \lambda = \dfrac { \beta } { u^2 }  $.  

To show $(u',v')$ is on $E$, notice
$$u'^{3} + \alpha u'^2 + \beta u' 
= \frac { \beta^3 } { u^3 } + \alpha \ \frac { \beta^2 } { u^2 } + \frac { \beta^2 } { u } 
= \frac { \beta^2 } { u^4 } \left( \beta u + \alpha u^2 + u^3 \right)
= \frac { \beta^2 } { u^4 } v^2
= \left( \frac { \beta } { u^2 } v \right)^2
= v'^2.$$
Both of these points yield the same solution  to (\ref{E1}) since $x=a, y=b$ and $z=z'$.
This defines a  solution $ (x:y:z:t) $, except possibly for the sign of $ t $. 
\end{proof}
\begin{remark}
If $ P_{ \mathcal{F}_2 } =  ( u , v ) $, we find that $ 2 P_{ \mathcal{F}_0 } = ( u' , v' ) $, both yielding the same solution $ \mathcal{F}_2 $.

\end{remark}
From the previous two theorems we deduce the following corollary.
\begin{corollary}
There exists infinitely many parametric solutions to \ $ x^4 + y^4 + z^4 + t^4 = ( x^2 + y^2 + z^2 - t^2 )^2 $. 
\end{corollary}

\section{Obtaining new parametric solutions}
Before obtaining new parametric solutions, we can interpret Lemma \ref{LEM} in terms of points in $E(K)$ where $K=\mathbb{Q}(m)$ and $E$ as described earlier.
\begin{proposition}
Let $ ( u , v ) \in E(K) $ such that  $ ( u , v ) \notin \big\{ \mathcal{O}, ( 0 , 0 ) , ( N^2 , 0 ) , ( L^2 , 0 ) \big\} $, and 
let $ ( u' ,v' ) , ( u'' , v'' ) \in E(K) $ such that: \\
$$ ( u , v ) + ( u' , v' ) = ( N^2 , 0 ) \text{ and } ( u , v ) + ( u'' , v'' ) = ( L^2 , 0 ) .$$ 
 If $ ( u , v ) $ yields a solution $ ( x :y:z: t ) $ to (\ref{E1}) such that $xyzt\neq 0$, 
then $ ( u' , v' ) $ and $ ( u'' , v'' ) $ both yield $  \left( \dfrac {1} {y} : \dfrac {1} {x} : \dfrac {1} {z} : \dfrac {1} {t}  \right) $, except perhaps the signs of $ \dfrac {1} {z} $ and $ \dfrac {1} {t} $.  
\end{proposition}

\begin{figure}[h]
\begin{center}
\includegraphics[scale=1.5]{GroupLaw.eps}
\caption{$E:  v^2=u^3+\alpha u^2+\beta u$}\label{grouplaw2}
\end{center}\end{figure}

\begin{proof}
From the group law on equation (\ref{UV}), we deduce \\
$$ ( u' , v' )   =  \left( \dfrac { N^2 ( u - L^2 ) } { u - N^2 }  , \dfrac { N^2 ( N^2 - L^2 )  v } { ( u - N^2 )^2 }  \right), \quad 
( u'' , v'' )  = \left( \ \frac { L^2 ( u - N^2 ) } { u - L^2 } , \dfrac { L^2  ( L^2 - N^2 )  v } { ( u - L^2 )^2 } \right) .$$\\
If $(u',v')$ yields $(x':y':z': t')$ then $ z'=\dfrac{v'}{u'} = \dfrac { ( N^2 - L^2 ) v } { ( u - N^2 )  ( u - L^2 ) } .$
From the relations $ N^2 - L^2 = - a b $  and  $ v^2 = u  ( u - N^2 )  ( u - L^2 ) $, we conclude 
$$z' =  -  a b  \dfrac { u } { v } =  -  \dfrac { ab } { z } .$$
Thus if $(u,v)$ yields $(x:y:z:t)=(a:b:z:t)$, then $(u' ,v' )$ yields 
$$(x':y' :z' :t')=\left(a: b : \dfrac{ab}{z}: \dfrac{ab}{t} \right) = \left( \dfrac {1} {b}: \dfrac {1} {a} :  \dfrac {1} {z} : \dfrac{1}{t} \right) = \left( \dfrac {1} {y} : \dfrac {1} {x} :  \dfrac {1} {z} : \dfrac{1}{t}\right), $$ 
where the signs of $z'$ and $t'$ may be positive or negative. The proof for $ z''$ is similar.  
\end{proof}

Next let $ P_{\mathcal{D}_2} = P_{\mathcal{F}_0} + P_{\mathcal{D}_1} $.  If $a^2+b^2=c^2$, we find $ P_{\mathcal{D}_2} = ( u , v ) $ with 

\begin{align*}
u & = \dfrac { ( c + a ) (a^4 + a^2 b^2 + b^4) ( 2 a^3 - a^2 c + b^2 c )^2 } 
{ 4  ( c - a )  ( a^2 + b^2 )  ( 2 a^3 + a^2 c - b^2 c )^2 } \\
v & = \dfrac { a^2 b  ( a^4 + a^2 b^2 + b^4 )  ( 2 a^3 - a^2 c + b^2 c )  ( 2 b^3 - a^2 c + b^2 c )  ( 2 b^3 + a^2 c - b^2 c ) } 
{ 4  ( c - a )^2  ( a^2 + b^2 )  ( 2 a^3 + a^2 c - b^2 c )^3 } \\
\end{align*}
By the same methods used in the proof of Theorem~\ref{ParaSol}, we deduce the following solution $ \mathcal{D}_2 $: 

\begin{proposition}
If  $ a^2 + b^2 = c^2 $  and if
\begin{align*}
x & = ab  ( 2 \ a^3 - a^2 c + b^2 c ) ( 2 \ a^3 + a^2 c - b^2 c ) ( 2 \ a b^2 + a^2 c  + b^2 c )  ( - 2 \ a b^2 + a^2 c  + b^2 c ) \\
y & = b^2 ( 2 \ a^3 - a^2 c + b^2 c )  ( 2 \ a^3 + a^2 c - b^2 c )  ( 2 \ a b^2 + a^2 c  + b^2 c )  ( - 2 \ a b^2 + a^2 c  + b^2 c )  \\
z & = a^2 ( 2 \ b^3 - a^2 c + b^2 c )  ( 2 \ b^3 + a^2 c - b^2 c )  ( 2 \ a b^2 + a^2 c  + b^2 c )  ( - 2 \ a b^2 + a^2 c  + b^2 c )  \\ 
t & = ab   ( 2 \ a^3 - a^2 c + b^2 c )  ( 2 \ a^3 + a^2 c - b^2 c )  ( 2 \ a^2 b + a^2 c  + b^2 c )  ( - 2 \ a^2 b + a^2 c  + b^2 c )  
\end{align*}
then $ \mathcal{D}_2 = ( x : y : z : t ) $ is a solution to (\ref{E1}). 
\end{proposition}
\begin{example}
Since $(a:b:c)=(2m: m^2-1: m^2+1)$, if $m=2$, then 
\begin{align*}
x & = 1 \ 899 \ 301 \ 428 \\
y & = 1 \ 424 \ 476 \ 071 \\
z & =  282 \ 491 \ 696 \\ 
t & = 1 \ 165 \ 848 \ 372 
\end{align*}
\end{example}
Next let $ P_{\mathcal{D}_3} = 2 P_{\mathcal{D}_1} $. We find that $ P_{\mathcal{D}_3} = ( u , v ) $ with: 
$$u  =  \frac { ( a^2 + b^2 )  ( a^4 + b^4 )^2 } 
{ 16  a^2 b^2  ( a - b )^2  ( a + b )^2 },  \quad
 v =  \frac 
{ P  Q  R  S  ( a^4 + b^4 ) } 
{ 64  a^3 b^3 c  ( a - b )^3 ( a + b )^3 } $$
where
\begin{align*}
P & = - a^3 + a^2 b + a b^2 + b^3,  &  Q & = a^3 - a^2 b + a b^2 + b^3  \\
R & =  a^3 + a^2 b - a b^2 + b^3, &  S & = a^3 + a^2 b + a b^2 - b^3  .
\end{align*}
From this we deduce  $ \mathcal{D}_3 $:
\begin{proposition}\label{D3}
If  $ a^2 + b^2 = c^2 $  and if
\begin{align*}
x & = a^2 b c ( a^2 - b^2 ) ( a^2 + b^2 ) ( a^4 + b^4 )  G \\
y & = a b^2 c ( a^2 - b^2 ) ( a^2 + b^2 ) ( a^4 + b^4 ) G \\
z & = P  Q  R S G \\ 
t & = 4 a b  ( a^2 - b^2 ) ( a^2 + b^2 )^2 ( a^4 + b^4 ) \ H 
\end{align*}
with  $ P , Q, R,  S $ as above, and
\begin{align*}
G & =  a^{12} + 6 a^{10} b^2  - a^8 b^4 + 4 a^6 b^6  - a^4 b^8 + 6  a^2 b^{10} + b^{12} \\
H & = a^{12} - 2 a^{10} b^2 + 7 a^8 b^4 + 4  a^6 b^6 + 7 a^4 b^8 - 2  a^2 b^{10} + b^{12}, 
\end{align*}
then $\mathcal{D}_2 =(x:y:z:t) $ is a solution to (\ref{E1}). 
\end{proposition}

Now put $ P_{\mathcal{D}_4} = P_{\mathcal{D}_3} + P_{\mathcal{F}_1} $. We find  $ P_{\mathcal{D}_4} = ( u , v ) $, with
$$u =  \frac{ ( a^2 - a b + b^2 )^2 P^2  S^2 } { 4  ( a^2 + b^2 )^2  Q^2  R^2 } , \quad
v  =  \frac{ a^2 b^2 c ( a^2 -  b^2 )  ( a^2 - a b + b^2 )^2    ( a^4 + b^4 ) P S} { Q^3  R^3 } $$
From this we deduce the following solution $ \mathcal{D}_4 $:
\begin{proposition}
If  $ a^2 + b^2 = c^2 $  and if
\begin{align*}
x & = a c P Q R S H \\
y & = b cP QR S H \\
z & = 4 a^2b^2 ( a^2-b^2) ( a^2 + b^2)^2 ( a^4 + b^4 )  H \\
t & = a b  P Q R S G 
\end{align*}
with the same $ P, Q,  R, S, G, H $ as in Proposition~\ref{D3}, then $\mathcal{D}_4= ( x : y : z : t ) $ is a solution to (\ref{E1}).
\end{proposition}
The degree of this solution is $ 26 $ in $ ( a:b: c ) $, and hence $ 52 $ in the homogeneous coordinates $(m:n)$ if we express $(a:b:c)=(2 m n: m^2 - n^2 : m^2 + n^2 )$ for nonzero integers $m, n$. 
\begin{remark}
The values $ z $ for  $ P_{ \mathcal{D}_3 } $ and $ z' $ for $ P_{ \mathcal{D}_4 } $ satisfy $ z z' = a b $, so 
except perhaps exchanging $ ( x' , y' ) $ and $ ( y' , x' ) $, $ \mathcal{D}_4 $ is deduced from $ \mathcal{D}_3 $
by replacing $ ( x , y , z ) $ by $\left( \dfrac {1} {x} , \dfrac {1} {y} , \dfrac {1} {z} \right)$.
\end{remark}
In summary, parametric solutions $(x:y:z:t)$ to (\ref{E1}) with their degree are shown in Table~\ref{eqtable}.
\begin{table}[H]
\[
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \text{solution} &  \mathcal{F}_0  &  \mathcal{F}_1  & \mathcal{F}_2  &  \mathcal{D}_1  &  \mathcal{D}_2  &  \mathcal{D}_3  &  \mathcal{D}_4  \\
\hline \text{degree } & 4 & 4 & 12 & 16 & 28 & 48 & 52 \\
\hline
\end{array} 
\]
\caption{Degree of parametric solutions}\label{eqtable}
\end{table}

\section{Acknowledgment}
The first author is thankful to Professor Edray Goins who helped visual
this interesting problem.  Both authors are thankful to Professor
Andrew Bremner for introducing us.  We are also grateful to the unknown
referee for their careful reading.

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\bibitem{NE} Noam D.~Elkies, On $A^4+B^4+C^4=D^4$, \textit{Math. Comp.}
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\bibitem{EF} Elie Fauquembergue, Question 266,
\textit{L'Interm\'{e}diaire des Math\'{e}maticiens} {\bf 1} (1894)
148.

\bibitem{JacMad} Lee W.~Jacobi and Daniel J.~Madden, , On
$a^4+b^4+c^4+d^4=(a+b+c+d)^4$, \textit{Amer. Math. Monthly} {\bf 115}
(2008), 220--236.

\bibitem{TP} Titus Piezas III, \textit{A Collection of Algebraic
Identities}, \newline
\url{https://sites.google.com/site/tpiezas/Home}.

\bibitem{S2} Joseph H.~Silverman, \textit{Advanced Topics in the
Arithmetic of Elliptic Curves}, Springer, 1994.

\bibitem{S1} Joseph H.~Silverman, \textit{The Arithmetic of Elliptic
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\bibitem{MU}  Maciej Ulas,   Rational points in arithmetic progressions
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193--207.
 \end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11D25; Secondary 11G05.

\noindent \emph{Keywords: } 
Diophantine equation, elliptic curve, elliptic surface.

\bigskip
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\vspace*{+.1in}
\noindent
Received August 11 2014;
revised version received October 5 2014.
Published in {\it Journal of Integer Sequences}, November 7 2014.

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


                                                                                
