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\headline={\ifnum\pageno<2 \hfil \else \smalltt INTEGERS: 
\smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4
(2004), \#A16 \hfil \folio \fi} 

\footline={\hfil}

{\smalltt INTEGERS: 
\smallrm
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4
(2004), \#A16}
\vskip .5in
 

\centerline{\uppercase{\bf Some fourth degree Diophantine 
equations in}}
\centerline{\uppercase{\bf Gaussian integers}}

\vskip 30pt


\centerline{{\bf S\'andor Szab\'o}}
\centerline{\smallit Institute of Mathematics and Informatics, University
of P\'ecs, Ifj\'us\'ag u. 6, 7624 P\'ecs, HUNGARY}
\vskip 40pt
\centerline{\smallit Received: 4/17/04, Revised: 9/10/04,
 Accepted: 10/20/04,
Published: 10/25/04}

\vskip 30pt

\baselineskip=5pt
\centerline{\bf Abstract}

For certain choices of the coefficients $a$, $b$, $c$
the solutions of the Diophantine equation $ax^4+by^4=cz^2$ 
in Gaussian integers satisfy $xy=0$.

\baselineskip=15pt
\parindent=15pt
\vskip 30pt



\line{\bf 1. Introduction \hfil}
 
The solution $(x_0,y_0,z_0)$ of the equation $ax^4+by^4=cz^2$ 
is called {\sl trivial} if $x_0=0$ or $y_0=0$.
P. Fermat showed that the equation $x^4+y^4=z^2$ has 
only trivial solutions in integers.
D. Hilbert [2] extended this result by showing that the 
equation $x^4+y^4=z^2$ has only trivial 
solutions in a larger domain, namely in the integers of $Q(\sqrt{-1})$.
In fact from his proof, it follows that the equation 
$x^4-y^4=z^2$ also has only trivial solutions.
J. T. Cross [1] gave a new proof for Hilbert's result.
We consider the following eight equations $x^4+my^4=z^2$, where
$m=\pm 2^n$, $0\leq n\leq 3$.
The equations $x^4-2y^4=z^2$, $y^4+8y^4=z^2$ have 
nontrivial solutions in integers as shown by the solutions 
$(3,2,7)$, $(1,1,3)$, respectively.
We will show that the remaining six equations have 
only trivial solutions in the integers of the quadratic field $Q(\sqrt{-1})$.
The $m=\pm 1$ case is covered by Hilbert's result, so we 
will deal only with four cases.
It is worthwhile to point out that the equation 
$x^4+2y^4=z^2$ has nontrivial solution in 
$Z[\sqrt{\pm 2}]$, as the solution 
$(1,\sqrt{\pm 2},3)$ shows.

It is proved in [3], among various similar results, that the
equation $x^4-py^4=z^2$ has only trivial solutions in integers,
where $p$ is a prime $p\equiv\pm 3,-5\pmod{16}$. 
We will show that the equations $x^4-py^4=z^2$,
$x^4-p^3y^4=z^2$ have only trivial solutions in the Gaussian integers, where $p$ is a prime $p\equiv 3\pmod{8}$.
We would like to point out that the equations
$x^4+py^4=z^2$, $x^4+p^2y^4=z^2$
have nontrivial integer solutions when $p=3$ as 
shown by the solutions $(1,1,2)$, $(2,1,5)$, respectively.

It is shown in [4] (Theorem 117, p. 230), that the equation $x^4-y^4=pz^2$ 
has only trivial solutions in integers, where $p$ is a prime 
$p\equiv 3\pmod{8}$.
It is shown in [3] (p. 23) that the equation $x^4-py^4=z^2$ 
has only trivial solutions in integers, where $p$ is a prime 
$p\equiv \pm 3,-5\pmod{16}$.
Motivated by these results, we will show that the equations
$x^4-y^4=pz^2$, $x^4-p^2y^4=z^2$ have only trivial solutions 
in Gaussian integers if $p$ is a rational prime 
$p\equiv 3\pmod{8}$. 

We list the properties of $Q(\sqrt{-1})$ which play part later.
Let $i=\sqrt{-1}$ and $\omega=1+i$.
The ring of integers of $Q(i)$ is
$Z[i]=\{u+vi: u,v\in Z\}$ which is a unique factorization
domain.
The units of $Z[i]$ are $1$, $i$, $-1$, $-i$.
The norm of $\omega$ is $2$ and consequently $\omega$ is a prime in
$Z[i]$.
The prime factorization of $2$ is $(-i)\omega^2$.
We will use the ideals formed by the Gaussian integer multiples of
$\omega^n$, $1\leq n\leq 6$.
Note that $\omega^2$, $\omega^4$, $\omega^6$ are associates of
$2$, $4$, $8$, respectively, and so they span the same ideals.
We will prefer to use the terminology connected with congruences
instead of with ideals.

We will use the next observation several times.
If $\alpha$ is an integer in $Q(\sqrt{-1})$ and $\alpha\equiv 1\pmod{\omega}$,
then $\alpha^2\equiv 1\pmod{\omega^2}$ and $\alpha^4\equiv 1\pmod{\omega^6}$.
In order to verify the first claim, write $\alpha$ in the form
$\alpha=k\omega+1$, where $k\in Z[i]$ and compute $\alpha^2$.
Since $\alpha^2=k^2\omega^2+2k\omega+1$, it follows that
$\alpha^2\equiv 1\pmod{\omega^2}$.
In order to verify the second claim write $\alpha$ in the form
$\alpha=k\omega^2+l$, $k,l\in Z[i]$ and compute $\alpha^4$.
$$\alpha^4=(k\omega^2)^4+4(k\omega^2)^3l+6(k\omega^2)^2l^2
           +4(k\omega^2)l^3+l^4$$
This shows that $\alpha^4\equiv l^4\pmod{\omega^6}$.
Since $0$, $1$, $i$, $1+i$ form a complete set of representatives
modulo $\omega^2$ and since $\alpha\equiv 1\pmod{\omega}$ we can choose
$l$ to be either $1$ or $i$.
Therefore $\alpha^4\equiv 1\pmod{\omega^6}$.

\vskip 20pt


\line{\bf 2. The equation $x^4-dy^4=z^2$ \hfil}


\noindent
{\bf Theorem 1.}
Let $p$ be a rational prime $p\equiv 3\pmod{8}$ and let $d=p$ or $d=p^3$.
The equation $x^4-dy^4=z^2$ has only trivial solution in $Z[i]$.

\vskip.1in

\noindent
{\it Proof.}
We divide the proof into 5 smaller steps.

(1) If $(x_0,y_0,z_0)$ is a nontrivial solution of the equation
$x^4-dy^4=z^2$, then we may assume that $x_0$, $y_0$, $z_0$ are
pairwise relatively primes.

Let $(x_0,y_0,z_0)$ be a solution of the 
equation $x^4-dy^4=z^2$. 
We will call the quantity $N(x_0)N(y_0)N(z_0)$ the {\sl height} of the 
solution. 
Here $N(u)$ is the norm of $u$.
Choose a nontrivial solution $(x_0,y_0,z_0)$ such that  
the height of the solution is minimal.

Suppose first that $x_0$ and $y_0$ are not relatively prime.
Let $g$ be the greatest common divisor of $x_0$ and $y_0$ in $Z[i]$.
As $x_0\not=0$, it follows that $g\not=0$.
Dividing $x_0^4-dy_0^4=z_0^2$ by $g^4$ we get
$(x_0/g)^4-d(y_0/g)^4=(z_0/g^2)^2$.
This equation holds in $Q(i)$.
The left hand side of the equation is an element of $Z[i]$.
Consequently the right hand side of the equation belongs to $Z[i]$.
Thus, $(x_0/g,y_0/g,z_0/g^2)$ is also a nontrivial solution of the
equation $x^4-dy^4=z^2$ in $Z[i]$.
Clearly the height of this solution is smaller than the 
height of $(x_0,y_0,z_0)$.
Hence, we may assume that $x_0$ and $y_0$ are relatively prime in $Z[i]$.
We claim that if there is a prime $q$ of $Z[i]$ such that 
$q|x_0$ and $q|z_0$, then $q|y_0$.
In order to verify the claim assume that $q$ is prime 
that divides $x_0$ and $z_0$. 
From the equation $x_0^4-dy_0^4=z_0^2$ it follows that 
$q^2|dy_0^4$.
If $d=p$, then $q|y_0$ because $p$ itself is a prime in $Z[i]$. 
If $d=p^3$, then it may be the case that $q=p$. 
But in this case $p^3|x_0^4$ and $p^3|dy_0^4$ so 
$p^3|z_0^2$, therefore $p^2|z_0$; hence, $p^4|z_0^2$ 
and since $p^4|x_0^4$, it follows that $p^4|dy_0^4$ 
and we can conclude that $p|y_0$.

This violates that $x_0$ and $y_0$ are relatively prime.
Similarly, if $q|y_0$ and $q|z_0$, then $q|x_0$ violating again that
$x_0$ and $y_0$ are relatively prime.
Thus we may assume that $x_0$, $y_0$, $z_0$ are pairwise relatively
prime.

(2) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4-dy^4=z^2$ in $Z[i]$ such that $x_0$, $y_0$, $z_0$ are pairwise
relatively prime.
Note that at most one of $x_0$, $y_0$, $z_0$ can be congruent to $0$
modulo $\omega$.
We consider the following four cases.
None of $x_0$, $y_0$, $z_0$ is congruent to $0$ modulo $\omega$ and
three cases depending on $x_0$, $y_0$, $z_0$ congruent to $0$ modulo
$\omega$ respectively.
Table 1 summarizes the cases.

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                           \hfil#\hfil\tabskip=5pt&
                \hfil\vrule\hfil#\tabskip=5pt&
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                           \hfil#\hfil\tabskip=5pt&
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              \noalign{\hrule}
              &        && $x_0\equiv$ && $y_0\equiv$ && $z_0\equiv$
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              \noalign{\hrule}
              & case 3 && $1$ && $0$ && $1$ && $\pmod{\omega}$  &\cr
              \noalign{\hrule}
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\centerline{Table 1}
\endinsert

In case 1, the equation $x_0^4-dy_0^4=z_0^2$ leads to the contradiction
$1-1\equiv 1\pmod{\omega}$.
In other words case 1 is merely the fact that 
``a sum of two odd numbers cannot be odd", where here, 
an odd number is a number which is not a multiple of 
$\omega$.
We will use this fact several times later without 
mentioning it explicitely.

In case 2, write the equation $x_0^4-dy_0^4=z_0^2$ in the equivalent
form $dy_0^4=(x_0^2-z_0)(x_0^2+z_0)$ and compute the greatest
common divisor of $(x_0^2-z_0)$ and $(x_0^2+z_0)$.
Let $g$ be this greatest common divisor.
Then $g|(x_0^2-z_0)$, $g|(x_0^2+z_0)$ implies $g|(2x_0^2)$, $g|(2z_0)$.
As $g|(dy_0^4)$, it follows that $g\not=0$.
If $q$ is a prime divisor of $g$, then $q\not|\omega$ and so
$q|x_0$, $q|z_0$.
But this cannot happen as $x_0$ and $z_0$ are relatively prime.
Thus, $g=1$.
The unique factorization property in $Z[i]$ gives that there are elements
$a$, $b$, $A$, $B$ and a unit $\varepsilon$ of $Z[i]$ such that
$$x_0^2-z_0=\varepsilon a^4A,\qquad x_0^2+z_0=\varepsilon^{-1}b^4B,$$
where
$$AB=d,\qquad a^4b^4=y_0^4.$$
Further, $A$ and $B$ are relatively prime, and so we may assume that either
$A=1$, $B=d$, or $A=d$, $B=1$.
By addition, we get that
$$2x_0^2=\varepsilon a^4A+\varepsilon^{-1}b^4B.$$
Let $x_0=\omega^tx_1$, $t\geq 1$, $x_1\equiv 1\pmod{\omega}$.
Writing $x_1$ in the form $x_1=k\omega+1$ and computing $x_1^2$,
$$x_1^2=k^2\omega^2+2k\omega+1,$$
we get that $x_0^2=\omega^{2t}(m\omega^2+1)$ with a suitable $m\in Z[i]$.

As $a|y_0^4$, it follows that $a\equiv 1\pmod{\omega}$.
We then get that $a^4\equiv 1\pmod{\omega^6}$.
Similarly, $b^4\equiv 1\pmod{\omega^6}$.

We focus our attention on the equation
$$(-i)\omega^2\omega^{2t}(m\omega^2+1)=
  \varepsilon a^4A+\varepsilon^{-1}b^4B$$
modulo $\omega^6$.
(We remind the reader that $2=(-i)\omega^2$.)
Let us first deal with the case when $A=d$ and $B=1$.
If $t=1$, then the equation reduces to
$$-i\omega^4\equiv\varepsilon(3)+\varepsilon^{-1}\pmod{\omega^6}.$$
As $\varepsilon$ varies over the units of $Z[i]$, we get 
the following contradictions.
$$\eqalign{
(-i)\omega^4&\equiv(1)(3)+(1)  ~~~~~\pmod{\omega^6},\cr
(-i)\omega^4&\equiv(i)(3)+(-i) ~~~\pmod{\omega^6},\cr
(-i)\omega^4&\equiv(-1)(3)+(-1)\pmod{\omega^6},\cr
(-i)\omega^4&\equiv(-i)(3)+(i) ~~~\pmod{\omega^6}.}$$
If $t\geq 2$, then the equation reduces to
$$0\equiv\varepsilon(3)+\varepsilon^{-1}\pmod{\omega^6}.$$
As $\varepsilon$ varies over the units of $Z[i]$, we get 
the following contradictions.
$$\eqalign{
0&\equiv(1)(3)+(1)  ~~~~~\pmod{\omega^6},\cr
0&\equiv(i)(3)+(-i) ~~~\pmod{\omega^6},\cr
0&\equiv(-1)(3)+(-1)\pmod{\omega^6},\cr
0&\equiv(-i)(3)+(i) ~~~\pmod{\omega^6}.}$$
The case when $A=1$ and $B=d$ can be settled in a similar way.
This shows that case 2 is not possible.

Next, we verify that case 4 is not possible either.
We write $z_0$ in the form $z_0=\omega^tz_1$, $t\geq 1$,
$z_1\equiv 1\pmod{\omega}$.
We focus our attention on the equation
$$x_0^4-dy_0^4=\omega^{2t}z_1^2$$
modulo $\omega^4$.
It leads to the contradictions $(1)-(3)(1)\equiv\omega^2$,
$(1)-(3)(1)\equiv 0\pmod{\omega^4}$ corresponding to $t=1$ or
$t\geq 2$.

(3) In case 3, let $(x_1,\omega^ty_1,z_1)$ be a solution of the equation
$x^4-dy^4=z^2$, where $t\geq 1$, $x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that $z_1\equiv 1\pmod{\omega^2}$.

In order to prove this claim, write $z_1$ in the form $z_1=k\omega^2+l$,
$k,l\in Z[i]$, and compute $z_1^2$.
$$z_1^2=k^2\omega^4+2k\omega^2l+l^2.$$
From this it follows that $z_1^2\equiv l^2\pmod{\omega^4}$. 
Since the elements $0$, $1$, $i$, $1+i$ form a complete set of
representatives modulo $\omega^2$, and since 
$z_1\equiv 1\pmod{\omega}$, we may choose $l$ to be $1$, or $i$.
Consequently, $z_1^2$ is congruent to $1$ or $-1$ modulo
$\omega^4$. 
The equation $x_1^4-d\omega^{4t}y_1^4=z_1^2$ gives
that $1\equiv z_1^2\pmod{\omega^4}$, and so 
$z_1\equiv 1\pmod{\omega^2}$.

(4) We will show that there are pairwise relatively prime 
elements $x_2$, $y_2$, $z_2$ of $Z[i]$, such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$ and
$(x_2,\omega^{t-1}y_2,z_2)$ is a solution of
the equation $x^4-dy^4=z^2$.

In order to verify the claim, write the equation
$x_1^4-d\omega^{4t}y_1^4=z_1^2$ in the form
$d\omega^{4t}y_1^4=(x_1^2-z_1)(x_1^2+z_1)$, and compute 
the greatest common divisor of $(x_1^2-z_1)$ and $(x_1^2+z_1)$.
Let $g$ be this greatest common divisor.
As $g|d\omega^{4t}y_1^4$, it follows that $g\not=0$.
Now $g|(x_1^2-z_1)$, $g|(x_1^2+z_1)$ implies that 
$g|2x_1^2$, $g|2z_1$.
If $q$ is a prime divisor of $g$ with $q\not|\omega$, 
we then get $q|x_1$, $q|z_1$.
But we know that this is not the case as $x_1$ and $z_1$ 
are relatively prime.
Thus, $g=\omega^s$, and $0\leq s\leq 2$ since $g|2$.
By (3) $z_1\equiv 1\pmod{\omega^2}$.
This together with $x_1^2\equiv 1\pmod{\omega^2}$ gives that
$(x_1^2-z_1)\equiv 0\pmod{\omega^2}$,
$(x_1^2+z_1)\equiv 0\pmod{\omega^2}$.
Therefore, $g=\omega^2$.
The unique factorization property in $Z[i]$ gives that
there are relatively prime elements $a,b\in Z[i]$ such that
$$x_1^2-z_1=\omega^2a,\qquad x_1^2+z_1=\omega^2b.$$
Let $a=\omega^ua_1$, $b=\omega^vb_1$.
So $d\omega^{4t}y_1^4=\omega^{u+v+4}a_1b_1$.
By the unique factorization property in $Z[i]$, there are elements
$a_2$, $b_2$, $A$, $B$ and a unit $\varepsilon$ in $Z[i]$ for which
$$x_1^2-z_1=\omega^{u+2}\varepsilon a_2^4A,\qquad
  x_1^2+z_1=\omega^{v+2}\varepsilon^{-1}b_2^4B,$$
$$4t=u+v+4,\qquad a_2^4b_2^4=y_1^4,\qquad AB=d.$$
Here, $a_2$, $b_2$ are prime to $\omega$, and $A$ is prime to $B$.
It follows that $a_2\equiv 1\pmod{\omega}$, 
$b_2\equiv 1\pmod{\omega}$.
We may choose $A$, $B$ such that either $A=d$, $B=1$, 
or $A=1$, $B=d$.
By addition, we get
$$2x_1^2=\omega^{v+2}\varepsilon^{-1}b_2^4B+
         \omega^{u+2}\varepsilon a_2^4A.$$
After dividing by $\omega^2$, we get
$$-ix_1^2=\omega^{v}\varepsilon^{-1}b_2^4B+
           \omega^{u}\varepsilon a_2^4A.$$
In the remaining part of the proof, we distinguish two 
cases depending on whether $A=1$, $B=d$, or $A=d$, $B=1$.

Let us deal with the case $A=1$, $B=d$ first.
We distinguish two subcases depending on whether 
$u=0$, $v=4t-4$, or $v=0$, $u=4t-4$.
When $u=0$, $v=4t-4$, we get
$$-ix_1^2=\omega^{4t-4}\varepsilon^{-1}b_2^4d+\varepsilon a_2^4.$$
If $4t-4=0$, then this relation reduces to
$$-i\equiv\varepsilon^{-1}+\varepsilon\pmod{\omega^2}.$$
But this is not possible as
$\varepsilon^{-1}+\varepsilon\equiv 0\pmod{\omega^2}$.
Thus, $4t-4\not=0$.
Now
$$-i\equiv\varepsilon\pmod{\omega^2}.$$
From this, it follows that $\varepsilon=\pm i$. 
By multiplying it by $-\varepsilon$ we get
$$(i\varepsilon)x_1^2=
\omega^{4t-4}(-\varepsilon^{-1}\varepsilon)b_2^4d+
  (-\varepsilon^2)a_2^4.$$
Note that $i\varepsilon$ is a square of an element of $Z[i]$,
say $i\varepsilon=\sigma^2$. 
Thus $(a_2,\omega^{t-1}b_2,\sigma x_1)$, $t\geq 2$ is 
a nontrivial solution of the equation $x^4-dy^4=z^2$.

When $v=0$, $u=4t-4$, we get
$$-ix_1^2=\varepsilon^{-1}b_2^4d+\omega^{4t-4}\varepsilon a_2^4.$$
If $4t-4=0$, then this reduces to
$$-i\equiv\varepsilon^{-1}+\varepsilon\pmod{\omega^2}.$$
But this is not possible as
$\varepsilon^{-1}+\varepsilon\equiv 0\pmod{\omega^2}$.
Thus $4t-4\not=0$.
Now
$$-i\equiv\varepsilon\pmod{\omega^2}.$$
From this, it follows that $\varepsilon=\pm i$.
By multiplying it by $\varepsilon^{-1}$, we get
$$(-i\varepsilon^{-1})x_1^2=(\varepsilon^{-2})b_2^4d+
  \omega^{4t-4}(\varepsilon^{-1}\varepsilon)a_2^4.$$
Note that $-i\varepsilon^{-1}$ is a square of an element of $Z[i]$, say
$-i\varepsilon^{-1}=\sigma^2$.
Thus $(\omega^{t-1}a_2,b_2,\sigma x_1)$, $t\geq 2$ is a nontrivial
solution of the equation $x^4-dy^4=z^2$.
By (2), this is not possible.
The case $A=d$, $B=1$ can be settled in a similar way.

(5) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4-dy^4=z^2$ in $Z[i]$.
By (2), there is a solution $(x_1,\omega^ty_1,z_1)$ with
$x_1,y_1,z_1\equiv 1\pmod{\omega}$, $t\geq 1$.
Choose a solution for which $t$ is minimal.
According to (4), there is a solution $(x_2,\omega^{t-1}y_2,z_2)$, where
$x_2,y_2,z_1\equiv 1\pmod{\omega}$, $t\geq 2$.
This contradicts the choice of $t$, and so completes the proof.

\vskip 20pt

\line{\bf 3. The equations $x^4-y^4=pz^2$ and $x^4-p^2y^4=z^2$ \hfil}

\noindent
{\bf Theorem 2.}
Let $p$ be a rational prime $p\equiv 3\pmod{8}$.
The equations $x^4-y^4=pz^2$ and $x^4-p^2y^4=z^2$ have only trivial solutions 
in $Z[i]$.


\noindent
{\it Proof.}
We divide the proof into 11 steps.
The first 3 steps deal with the equation $x^4-y^4=pz^2$, 
and the next 7 steps deal with the equation $x^4-p^2y^4=z^2$.

(1) If $(x_0,y_0,z_0)$ is a nontrivial solution of the equation
$x^4-y^4=pz^2$, we may then assume that $x_0$, $y_0$, $z_0$ are
pairwise relatively prime.

Choose a nontrivial solution $(x_0,y_0,z_0)$ of the equation
$x^4-y^4=pz^2$ with minimal height.
Suppose first that there is a prime $q$ in $Z[i]$ such that 
$q|x_0$, $q|y_0$. 
From $x_0^4-y_0^4=pz_0^2$ it follows that $q^4|pz_0^2$.
If $q\not|p$, then $q^4|z_0^2$.
Now $(x_0/q)^4-(y_0/q)^4=p(z_0/q^2)^2$ shows that 
$(x_0/q,y_0/q,z_0/q^2)$ is a nontrivial solution of the 
equation $x^4-y^4=pz^2$. 
The height of this solution is smaller than the height 
of $(x_0,y_0,z_0)$. 
This is a contradiction.
If $q|p$, then $q^3|z_0^2$. 
Again we conclude that that $q^4|z_0^2$ and then 
$(x_0/q,y_0/q,z_0/q^2)$ is a nontrivial solution of the 
equation $x^4-y^4=pz^2$.
The height again decreased. 
Thus we may assume that if $(x_0,y_0,z_0)$ is a nontrivial 
solution of the equation $x^4-y^4=pz^2$, then $x_0$ and 
$y_0$ are relatively prime.

Next suppose that there is a prime $q$ in $Z[i]$ such that 
$q|x_0$, $q|z_0$. 
It follows that $q|y_0$. 
This violates that $x_0$ and $y_0$ are relatively prime.

Finally suppose that there is a prime $q$ in $Z[i]$ such that 
$q|y_0$, $q|z_0$. 
We get that $q|x_0$. 
This is a contradiction as $x_0$ and $y_0$ are relatively 
prime.

(2) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4-y^4=pz^2$ in $Z[i]$ such that $x_0$, $y_0$, $z_0$ are pairwise
relatively primes.
Note that at most one of $x_0$, $y_0$, $z_0$ can be congruent to $0$
modulo $\omega$.
We consider the four cases summarized in Table 1.

We first show that case 1 is not possible.
To do this, write $z_0$ in the form $z_0=k\omega^2+l$, $k,l\in Z[i]$.
Computing $z_0^2$, $z_0^2=k^2\omega^4+2k\omega^2l+l^2$ shows that
$z_0^2\equiv l^2\pmod{\omega^4}$.
As $0$, $1$, $i$, $1+i$ is a complete set of representatives modulo
$\omega^2$, it follows that $l$ can be chosen to be $1$ or $i$.
From the equation $x_0^4-y_0^4=pz_0^2$, we get that
$0\equiv 3l^2\pmod{\omega^4}$.
But this is a contradiction as $l^2=\pm 1$.

We next show that case 3 is not possible either.
Let $y_0=\omega^ty_1$, where $t\geq 1$ and $y_1$ is prime to $\omega$.
Writing $z_0$ in the form $z_0=k\omega^2+l$, $k,l\in Z[i]$, from the 
equation $x_0^4-\omega^{4t}y_1^4=pz_0^2$, we get that
$1\equiv 3l^2\pmod{\omega^4}$.
In the case $l=1$, this leads to the contradiction
$1\equiv 3\pmod{\omega^4}$, and so we left with the $l=i$ choice.
Now writing $z_0$ in the form $z_0=r\omega^4+s$, $r,s\in Z[i]$ and
computing $z_0^2$, $z_0^2=r^2\omega^8+2r\omega^4s+s^2$ gives that
$z_0^2\equiv s^2\pmod{\omega^6}$.
From $z_0\equiv i\pmod{\omega^2}$, it follows that we can choose $s$
and $s^2$ in the way summarized by Table 2.

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                           \hfil#\hfil\tabskip=5pt&
                                #\hfil\vrule\tabskip=0pt\cr
              \noalign{\hrule}
              & $s$   && $i$  && $2+i$  && $3i$  && $2+3i$   &\cr 
              \noalign{\hrule}
              & $s^2$ && $-1$ && $3+4i$ && $-9$ && $-5+12i$ &\cr
              \noalign{\hrule}
              }
       }
   }$$
\centerline{Table 2}
\endinsert

From the equation $x_0^4-\omega^{4t}y_1^4=z_0^2$, we get
$1+\omega^{4t}\equiv s^2\pmod{\omega^6}$.
In the case $t=1$, this leads to the contradictions
$$1-4\equiv -1\pmod{\omega^6},\qquad
  1-4\equiv 3+4i\pmod{\omega^6}.$$
In the case $t\geq 2$, we arrive at the contradictions
$$1\equiv -1\pmod{\omega^6},\qquad
  1\equiv 3+4i\pmod{\omega^6}.$$
(Note that Table 2 shows four possibilities but modulo 
$\omega^6=8$ there are only two posibilities.)

Finally, notice that multiplying the equation
$x_0^4-y_0^4=pz_0^2$ by $(-1)$ gives $y_0^4-x_0^4=p(iz_0)^2$,
and so case 2 reduces to case 3. 

(3) In case 4, let $(x_1,y_1,\omega^tz_1)$ be a solution of the equation
$x^4-y^4=pz^2$, where $t\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$,
and $x_1$, $y_1$, $z_1$ are pairwise relatively primes.
We will show that there are pairwise relatively prime elements 
$x_2$, $y_2$, $z_2$ of $Z[i]$ such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$, and either
$(\omega^{t-2}x_2,y_2,z_2)$ or 
$(x_2,\omega^{t-2}y_2,z_2)$ is a solution of
the equation $x^4-p^2y^4=z^2$.

In order to verify the claim, write the equation
$x_1^4-y_1^4=p\omega^{2t}z_1^2$ in the form
$p\omega^{2t}z_1^2=(x_1^2-y_1^2)(x_1^2+y_1^2)$, and 
compute the greatest
common divisor of $(x_1^2-y_1^2)$ and $(x_1^2+y_1^2)$.
Let $g$ be this greatest common divisor.
As $g|p\omega^{2t}z_1^2$ it follows that $d\not=0$.
$g|(x_1^2-y_1^2)$, $g|(x_1^2+y_1^2)$ implies that $g|2x_1^2$, $g|2y_1^2$.
If $q$ is a prime divisor of $g$ with $q\not|\omega$,  
we then get $q|x_1$, $q|y_1$.
But we know that this is not the case as $x_1$ and $y_1$ 
are relatively prime.
Thus, $g=\omega^s$ and $0\leq s\leq 2$ since $g|2$.
As
$(x_1^2-y_1^2)\equiv (x_1^2+y_1^2)\equiv 0\pmod{\omega^2}$,
it follows that $g=\omega^2$.
The unique factorization property in $Z[i]$ gives that
there are relatively prime elements $a,b\in Z[i]$ such that
$$x_1^2-y_1^2=\omega^2a,\qquad x_1^2+y_1^2=\omega^2b.$$
Let $a=\omega^ua_1$, $b=\omega^vb_1$.
So, $p\omega^{2t}z_1^2=\omega^{u+v+4}a_1b_1$.
By the unique factorization property in $Z[i]$, there are elements
$a_2$, $b_2$, $a_3$, $b_3$ and a unit $\varepsilon$ in $Z[i]$ for which
$$x_1^2-y_1^2=\omega^{u+2}\varepsilon a_2^2a_3,\qquad
  x_1^2+y_1^2=\omega^{v+2}\varepsilon^{-1}b_2^2b_3,$$
$$2t=u+v+4,\qquad a_2^2b_2^2=z_1^2,\qquad a_3b_3=p.$$
Here $a_2$, $b_2$ are prime to $\omega$.
It follows that $a_2\equiv b_2\equiv 1\pmod{\omega}$.
By addition and subtraction, we get
$$\eqalign{
2x_1^2&=\omega^{v+2}\varepsilon^{-1}b_2^2b_3+
        \omega^{u+2}\varepsilon a_2^2a_3,\cr
2y_1^2&=\omega^{v+2}\varepsilon^{-1}b_2^2b_3-
        \omega^{u+2}\varepsilon a_2^2a_3.}$$
After dividing by $\omega^2$, they give
$$\eqalign{
(-i)x_1^2&=\omega^{v}\varepsilon^{-1}b_2^2b_3+
           \omega^{u}\varepsilon a_2^2a_3,\cr
(-i)y_1^2&=\omega^{v}\varepsilon^{-1}b_2^2b_3-
           \omega^{u}\varepsilon a_2^2a_3.}$$
By multiplying the two equations together and multiplying 
the result by $\varepsilon^2$, we get
$$-\varepsilon^2 x_1^2y_1^2=\omega^{2v}b_2^4b_3^2-\omega^{2u}a_2^4a_3^2.$$
We distinguish two cases depending on whether $a_3=1$, $b_3=p$, 
or $a_3=p$, $b_3=1$.
In the case $a_3=1$, $b_3=p$ 
$-\varepsilon^2 x_1^2y_1^2=\omega^{2v}b_2^4p^2-\omega^{2u}a_2^4$
we distinguish two subcases depending on whether 
$u=0$, $v=2t-4$, or $u=2t-4$, $v=0$.
When $u=0$, $v=2t-4$, we get
$-\varepsilon^2 x_1^2y_1^2=\omega^{4t-8}b_2^4p^2-a_2^4$.
Thus, $(a_2,\omega^{t-2}b_2,\varepsilon x_1y_1)$, is a 
nontrivial solution of the equation $x^4-p^2y^4=z^2$.

When $u=2t-4$, $v=0$, we get
$-\varepsilon^2 x_1^2y_1^2=b_2^4p^2-\omega^{4t-8}a_2^4$.
Thus, $(\omega^{t-2}a_2,b_2,\varepsilon x_1y_1)$, is a 
nontrivial solution of the equation $x^4-p^2y^4=z^2$.
The case $a_3=p$, $b_3=1$ can be settled in a similar way.

We now turn our attention to the equation $x^4-p^2y^4=z^2$.

(4) If $(x_0,y_0,z_0)$ is a nontrivial solution of the equation
$x^4-p^2y^4=z^2$, we may then assume that $x_0$, $y_0$, $z_0$ are
pairwise relatively prime.

Choose a nontrivial solution $(x_0,y_0,z_0)$ of the equation 
$x^4-p^2y^4=z^2$ with minimal height.
First suppose that $x_0$ and $y_0$ are not relatively 
prime.
Let $g$ be a greatest common divisor of $x_0$ and $y_0$ in 
$Z[i]$. 
The left hand side of the equation 
$(x_0/g)^4-p^2(y_0/g)^4=(z_0/g^2)^2$ is an element of $Z[i]$ 
so the right hand side of the equation is an element of 
$Z[i]$. 
Therefore $(x_0/g,y_0/g,z_0/g^2)$ is a nontrivial solution 
of $x^4-p^2y^4=z^2$. 
By the minimality of the height we may assume that $x_0$ 
and $y_0$ are relatively prime.

Next suppose that there is a prime $q$ in $Z[i]$ such that 
$q|y_0$, $q|z_0$. 
In this case we get $q|x_0$. 
This is a contradiction since $x_0$ and $y_0$ are relatively 
prime.

Finally suppose that there is a prime $q$ in $Z[i]$ such 
that $q|x_0$, $q|z_0$. 
It follows that $q^2|p^2y_0^4$.
If $q|y_0$, then $x_0$ and $y_0$ are not relatively prime. 
This is not the case so $q\not|y_0$. 
It follows that $q^2|p^2$. 
Hence $q$ and $p$ are associates.
Set $x_0=px_1$, $z_0=pz_1$. 
From $p^4x_1-p^2y_0^4=p^2z_1$ we get 
$$\eqalign{
p^2x_1^2-y_0^4&=z_1^2,\cr
y_0^4-p^2x_1^4&=-z_1^2.}$$
Therefore $(y_0,x_1,iz_1)$ is a nontrivial solution of 
the equation $x^4-p^2y^4=z^2$. 
By the minimality of the height we may assume that 
$x_0$ and $z_0$ are relatively prime.

(5) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4-p^2y^4=z^2$ in $Z[i]$ such that $x_0$, $y_0$, $z_0$ are 
pairwise relatively prime.
We consider the four cases listed in Table 1.
In case 1, the equation $x_0^4-p^2y_0^4=z_0^2$ gives 
the contradiction $1-1\equiv 1\pmod{\omega}$.

(6) In case 2, multiply the equation $x^4-p^2y^4=z^2$ by 
$(-1)$ to get $-x^4+p^2y^4=(iz)^2$.
Let $(\omega^tx_1,y_1,z_1)$ be a solution of the equation
$-x^4+p^2y^4=z^2$, where $t\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that $z_1\equiv 1\pmod{\omega^2}$.

In order to prove this claim, write $z_1$ in the form $z_1=k\omega^2+l$,
$k,l\in Z[i]$.
It follows that $z_1^2\equiv l^2\pmod{\omega^4}$, and we 
may choose $l$ to be $1$ or $i$.
Consequently, $z_1^2$ is congruent to $1$ or $-1$ modulo $\omega^4$.
The equation $-\omega^{4r}x_1^4+p^2y_1^4=z_1^2$ gives that 
$1\equiv z_1^2\pmod{\omega^4}$, and so $z_1\equiv 1\pmod{\omega^2}$. 

(7) In case 2 let $(\omega^rx_1,y_1,z_1)$ be a solution of the equation
$-x^4+p^2y^4=z^2$, where $r\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that there are pairwise relatively prime elements 
$x_2$, $y_2$, $z_2$ of $Z[i]$ such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$ and 
$(x_2,\omega^{r-1}y_2,z_2)$ is a solution of
the equation $x^4-y^4=pz^2$.

In short, case 2 leads to a nontrivial solution of the equation
$x^4-y^4=pz^2$ corresponding to cases 1--3.
By step (2), no such solution exists, and so case 2 of equation
$x^4-p^2y^4=z^2$ is not possible.

From the equation $\omega^{4t}x_1^4=(py_1^2-z_1)(py_1^2+z_1)$, 
we can deduce that
$$(-i)py_1^2=
\omega^v\varepsilon^{-1}b_2^4+\omega^u\varepsilon a_2^4,$$
$$4t=u+v+4,\qquad a_2b_2=x_1^4.$$
We distinguish two cases depending on whether $u=0$, $v=4t-4$, or
$u=4t-4$, $v=0$.
In the case $u=0$, $v=4t-4$, 
$(-i)py_1^2=\omega^{4t-4}\varepsilon^{-1}b_2^4+\varepsilon a_2^4$.
If $4t-4=0$, then it reduces to 
$-i\equiv\varepsilon^{-1}+\varepsilon\pmod{\omega^2}$.
But this a contradiction as 
$\varepsilon^{-1}+\varepsilon\equiv 0\pmod{\omega^2}$.
The details are given in Table 3.

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   }$$
\centerline{Table 3}
\endinsert
 
Thus $4t-4\geq 2$, and $-i\equiv\varepsilon\pmod{\omega^2}$.
From this, it follows that $\varepsilon=\pm i$, that is, either
$\varepsilon=i$, $\varepsilon^{-1}=-i$, or
$\varepsilon=-i$, $\varepsilon^{-1}=i$.
In the first case
$$\eqalign{
(-i)py_1^2&=\omega^{4t-4}(-i)b_2^4+(i)a_2^4,\cr
    py_1^2&=\omega^{4t-4}b_2^4-a_2^4,}$$
and so $(\omega^{t-1}b_2,a_2,y_1)$ is a solution of the equation
$x^4-y^4=pz^2$.
In the second case
$$\eqalign{
(-i)py_1^2&=\omega^{4t-4}(i)b_2^4+(-i)a_2^4,\cr
    py_1^2&=-\omega^{4t-4}b_2^4+a_2^4;}$$
hence $(a_2,\omega^{t-1}b_2,y_1)$ is a solution of the equation
$x^4-y^4=pz^2$.
The case $u=4t-4$, $v=0$ can be settled in a similar way.

(8) In case 4, let $(x_1,y_1,\omega^tz_1)$ be a solution of the equation
$x^4-p^2y^4=z^2$, where $t\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
It follows that $z_1\equiv 1\pmod{\omega^2}$.

(9) In case 3, let $(x_1,\omega^ty_1,z_1)$ be a solution of the equation
$x^4-p^2y^4=z^2$, where $r\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$,
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that there are pairwise relatively prime elements 
$x_2$, $y_2$, $z_2$ of $Z[i]$ such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$, and either
$(\omega^{t-1}x_2,y_2,z_2)$, or $(x_2,\omega^{t-1}y_2,z_2)$ is a solution of
the equation $x^4-p^2y^4=z^2$.

Form the equation
$p^2\omega^{4t}y_1^4=(x_1^2-z_1)(x_1^2+z_1)$,
we deduce again that
$$(-i)x_1^2=\omega^{v}\varepsilon^{-1}b_2^4b_3+
            \omega^{u}\varepsilon a_2^4a_3,$$
$$4t=u+v+4,\qquad a_2^4b_2^4=y_1^4,\qquad a_3b_3=p^2.$$
As $a_3$ and $b_3$ are relatively prime, we may assume 
that there are two cases depending on whether $a_3=1$, $b_3=p^2$, 
or $a_3=p^2$, $b_3=1$.
In the case $a_3=1$, $b_3=p^2$,
$$(-i)x_1^2=\omega^{v}\varepsilon^{-1}b_2^4p^2+
            \omega^{u}\varepsilon a_2^4.$$
We distinguish two subcases depending on whether $u=0$, $v=4t-4$, 
or $u=4t-4$, $v=0$.
When $u=0$, $v=4t-4$, we get
$(-i)x_1^2=\omega^{4t-4}\varepsilon^{-1}b_2^4p^2+\varepsilon a_2^4$.
If $4t-4=0$, then this reduces to
$-i\equiv\varepsilon^{-1}+\varepsilon\pmod{\omega^2}$, which is not
possible.
Thus $4t-4\geq 2$ and
$(-i)\equiv-\varepsilon\pmod{\omega^2}$.
From this, it follows that $\varepsilon=\pm i$, that is, either
$\varepsilon=i$, $\varepsilon^{-1}=-i$ or 
$\varepsilon=-i$, $\varepsilon^{-1}=i$.
In the first case, we get
$$\eqalign{
(-i)x_1^2&=\omega^{4t-4}(-i)b_2^4p^2+(i)a_2^4,\cr
   -x_1^2&=-\omega^{4t-4}b_2^4p^2+a_2^4.}$$
Thus $(a_2,\omega^{t-1}b_2,ix_1)$, is a nontrivial
solution of the equation $x^4-p^2y^4=z^2$.
In the second case, we get
$$\eqalign{
(-i)x_1^2&=\omega^{4t-2}(i)b_2^4p^2+(-i)a_2^4,\cr
    x_1^2&=-\omega^{4t-4}b_2^4p^2+a_2^4.}$$
Therefore $(a_2,\omega^{t-1}b_2,x_1)$ is a nontrivial
solution of the equation $x^4-p^2y^4=z^2$.

When $u=4t-4$, $v=0$, we get
$(-i)x_1^2=\varepsilon^{-1}b_2^4p^2+\omega^{4t-4}\varepsilon a_2^4$.
If $4t-4=0$, then this reduces to
$-i\equiv\varepsilon^{-1}+\varepsilon\pmod{\omega^2}$, which is not
possible.
Thus $4t-4\geq 2$ and
$(-i)\equiv\varepsilon^{-1}\pmod{\omega^2}$.
From this, it follows that $\varepsilon=\pm i$, that is, either
$\varepsilon=i$, $\varepsilon^{-1}=-i$, or 
$\varepsilon=-i$, $\varepsilon^{-1}=-i$.
In the first case, we get
$$\eqalign{
(-i)x_1^2&=(-i)b_2^4p^2+\omega^{4t-4}(i)a_2^4,\cr
   -x_1^2&=-b_2^4p^2+\omega^{4t-4}a_2^4.}$$
Thus $(\omega^{t-1}a_2,b_2,ix_1)$ is a nontrivial
solution of the equation $x^4-p^2y^4=z^2$.
In the second case, we get
$$\eqalign{
(-i)x_1^2&=(i)b_2^4p^2+\omega^{4t-4}(-i)a_2^4,\cr
    x_1^2&=-b_2^4p^2+\omega^{4t-4}a_2^4,}$$
and so $(\omega^{t-1}a_2,b_2,x_1)$ is a nontrivial
solution of the equation $x^4-p^2y^4=z^2$.
 
The case when $a_3=p^2$, $b_3=1$ can be completed in a similar way.

(10) In case 4, let $(x_1,y_1,\omega^tz_1)$ be a solution of the equation
$x^4-p^2y^4=z^2$, where $t\geq 1$, 
$x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$,
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that there are pairwise relatively prime elements 
$x_2$, $y_2$, $z_2$ of $Z[i]$ such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$ and either
$(\omega^{t-2}x_2,y_2,z_2)$ or 
$(x_2,\omega^{t-2}y_2,z_2)$ is a solution of
the equation $x^4-y^4=pz^2$.

The conclusion of these steps is that in case 4 we end up with a nontrivial
solution of the equation $x^4-y^4=pz^2$ corresponding to 
one of cases 1--3.
Since by step (2) this is not possible, it follows that case 4 of
equation $x^4-p^2y^4=z^2$ is not possible either.

From the equation
$\omega^{2t}z_1^2=(x_1^2-py_1^2)(x_1^2+py_1^2)$,
we can deduce that
$$\eqalign{
 (-i)x_1^2&=\omega^{v}\varepsilon^{-1}b_2^2+
            \omega^{u}\varepsilon a_2^2,\cr
(-i)py_1^2&=\omega^{v}\varepsilon^{-1}b_2^2-
            \omega^{u}\varepsilon a_2^2,}$$
$$2t=u+v+4,\qquad a_2^2b_2^2=z_1^2.$$
By multiplying the first two equations above together and 
multiplying the result by $\varepsilon^2$, we get
$$-\varepsilon^2 px_1^2y_1^2=\omega^{2v}b_2^4-\omega^{2u}a_2^4.$$
We distinguish two cases depending on whether $u=0$, $v=2t-4$, 
or $u=2t-4$, $v=0$.
When $u=0$, $v=2t-4$, we get
$-\varepsilon^2 px_1^2y_1^2=\omega^{4t-8}b_2^4-a_2^4$.
Thus $(a_2,\omega^{t-2}b_2,\varepsilon x_1y_1)$, is a 
nontrivial solution of the equation $x^4-y^4=pz^2$.

When $u=2t-4$, $v=0$, we get
$-\varepsilon^2 px_1^2y_1^2=b_2^4-\omega^{4t-8}a_2^4$.
Thus, $(\omega^{t-2}a_2,b_2,\varepsilon x_1y_1)$ is a 
nontrivial solution of the equation $x^4-y^4=pz^2$.

(11) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4-p^2y^4=z^2$ in $Z[i]$.
By steps (5), (7) and (10), cases 1, 2 and 4 are not
possible, and so there is a solution $(x_1,\omega^ty_1,z_1)$ with
$x_1,y_1,z_1\equiv 1\pmod{\omega}$, $t\geq 1$.
Choose a solution for which $t$ is minimal.
According to step (9), there is a solution either
of the form $(\omega^{t-1}x_2,y_2,z_2)$, or 
of the form $(x_2,\omega^{t-1}y_2,z_2)$, where
$x_2,y_2,z_2\equiv 1\pmod{\omega}$, $t\geq 2$.
The first case is not possible.
The second case contradicts the choice of $t$, and so we conclude that
the equation $x^4-p^2y^4=z^2$ has no nontrivial solutions in $Z[i]$.

By step (3), a nontrivial solution of the equation $x^4-y^4=pz^2$ leads
to a nontrivial solution of the equation $x^4-p^2y^4=z^2$.
Thus the equation $x^4-y^4=pz^2$ does not have nontrivial solutions
in $Z[i]$.
This completes the proof.

\vskip.1in

We may describe the combinatorial content of our argument 
in the following way.
We consider a directed graph ${\Gamma}$ whose 
vertices and edges are labeled.
To a nontrivial solution $(x_0,y_0,z_0)$ of the 
equation $E:ax^4+by^4=cz^2$, we assign a type $T$ 
depending on the residues of $x_0$, $y_0$, $z_0$ 
modulo $\omega$. 
(There are only a limited number of possibilities 
for $T$.)
The nodes of ${\Gamma}$ are the pairs $(E,T)$, where 
$E$ is an equation and $T$ is a possible solution type. 
We label the node $(E,T)$ impossible if the equation 
$E$ has no nontrivial solution of type $T$. 
We draw an arrow from the node $(E_1,T_1)$ to the 
node $(E_2,T_2)$ if a nontrivial solution of $E_1$ 
with type $T_1$ gives rise to a nontrivial solution 
of $E_2$ with type $T_2$.
We label an arrow with a $-$ sign if a quantity 
associated with a solution decreases.
If each path starting with a node $(E,T_i)$ terminates 
at a node labeled impossible or eventually reaches 
$(E,T_i)$ again but the edges are labeled with $-$ 
signs, then the equation cannot have nontrivial solutions.

\vskip 20pt

\line{\bf 4. The equation $x^4+my^4=z^2$ \hfil}

If $(x_0,y_0,z_0)$ is a nontrivial solution either one of 
the equations
$$x^4+4y^4=z^2,\qquad x^4-4y^4=z^2,\qquad x^4-8y^4=z^2,$$
then $(x_0,\omega y_0,z_0)$ is a nontrivial solution one of the
equations
$$x^4-y^4=z^2,\qquad x^4+y^4=z^2,\qquad x^4+2y^4=z^2,$$
respectively, as the first three equations can be written 
in the forms
$$x^4-\omega^4y^4=z^2,\qquad x^4+\omega^4y^4=z^2,\qquad
  x^4+2\omega^4y^4=z^2,$$
respectively.
Thus, it will be enough to prove that the equation $x^4+2y^4=z^2$ has
only trivial solutions in $Z[i]$.


\noindent
{\bf Theorem 3.}
The equation $x^4+2y^4=z^2$ has only trivial solutions in $Z[i]$.

\noindent
{\it Proof.}
We divide the proof into $5$ steps.

(1) If $(x_0,y_0,z_0)$ is a nontrivial solution of the equation
$x^4+2y^4=z^2$, then we may assume that $x_0$, $y_0$, $z_0$ are
pairwise relatively prime.

(1.a) Choose a nontrivial solution $(x_0,y_0,z_0)$ of the equation
$x^4+2y^4=z^2$ with minimal height. 
Suppose first that $x_0$ and $y_0$ are not relatively prime 
and let $g$ be a greatest common divisor of $x_0$ and $y_0$. 
The left hand side of the equation 
$(x_0/q)^4+2(y_0/g)^4=(z_0/g^2)^2$ is an element of $Z[i]$ and so 
the right hand side of the equation is an element of $Z[i]$. 
Hence $(x_0/g,y_0/g,z_0/g^2)$ is a nontrivial solution of 
$x^4+2y^4=z^2$. 
By the minimality of the height we may assume that $x_0$ and 
$y_0$ are relatively prime.

Assume next that there is a prime $q$ in $Z[i]$ such that 
$q|x_0$, $q|z_0$. 
It follows that $q|x_0$. 
This is a contradiction since $x_0$, $y_0$ are relatively 
prime.

Finally suppose there is a prime $q$ in $Z[i]$ such that 
$q|x_0$, $q|z_0$. 
We get that $q^2|2y_0$. 
If $q|y_0$ we get the contradiction that $x_0$ and $y_0$ are 
not relatively prime. 
Thus $q\not|y_0$ and consequently $q^2|2$. 
We get that $q$ is an associate of $\omega$.
Set $x_0=\omega x_1$, $z_0=\omega z_1$. 
From $\omega^4x_1^4+2y_0^4=\omega^2 z_1^2$ we get 
$$\eqalign{
\omega^4 x_1^4-i\omega^2y_0^4&=\omega^2z_1^2,\cr
        \omega^2 x_1^4-iy_0^4&=z_1^2,\cr
     (-i)\omega^2 x_1^4+y_0^4&=(-i)z_1^2,\cr
                 y_0^4+2x_1^4&=-iz_1^2.}$$
Therefore $(y_0,x_1,iz_1)$ is a nontrivial solution of the 
equation $x^4+2y^4=iz^2$.

(1.b) Pick a nontrivial solution of $(x_0,y_0,z_0)$ of the 
equation $x^4+2y^4=iz^2$ with minimal height. 
Using the argument we have seen in step (1.a) we may assume 
that $x_0$ and $y_0$ are relatively prime.
The assumption that there is a prime $q$ with $q|y_0$, 
$q|z_0$ gives the contradiction that $x_0$ and $y_0$ are 
not relatively prime.
 
Finally suppose that there is a prime $q$ such that 
$q|x_0$, $q|z_0$. 
We get that $q^2|2y_0^4$. 
Here $q|y_0$ leads to the contradiction that $x_0$ and 
$y_0$ are not relatively prime. 
Thus $q^2|2$ and so $q$ and $\omega$ are associates. 
Set $x_0=\omega x_1$, $z_0=\omega z_1$. 
From $\omega^4 x_1^4+2y_0^4=i\omega^2z_1^2$ we get 
$$\eqalign{
\omega^4 x_1^4-i\omega^2y_0^4&=i\omega^2z_1^2,\cr
        \omega^2 x_1^4-iy_0^4&=iz_1^2,\cr
     (-i)\omega^2 x_1^4-y_0^4&=iz_1^2,\cr
                 2x_1^4-y_0^4&=iz^2\cr
                 y_0^4-2x_1^4&=-iz_1^2.}$$
Hence $(y_0,x_1,iz_1)$ is a nontrivial solution of the 
equation $x^4-2y^4=iz^2$.

(1.c) Choose a nontrivial solution $(x_0,y_0,z_0)$ of the 
equation $x^4-2y^4=iz^2$ with minimal height.
As before we may assume that $x_0$ and $y_0$ are relatively 
prime.
If there is a prime $q$ with $q|y_0$, $q|z_0$ we get the 
contradiction that $x_0$ and $y_0$ are not relatively prime. 

Finally consider the case when there is a prime $q$ with 
$q|x_0$, $q|z_0$. 
It follows that $q^2|2y_0^4$. 
If $q|y_0$, then we get that $x_0$ and $y_0$ are not relatively 
prime. 
This is not the case. 
So $q^2|2$ and we get that $q$ and $\omega$ are associates. 
Setting $x_0=\omega x_1$, $z_0=\omega z_1$ from 
$\omega^4-2y_0^4=i\omega^2z_1^2$ we get 
$$\eqalign{
\omega^4 x_1^4+i\omega^2y_0^4&=i\omega^2z_1^2,\cr
        \omega^2 x_1^4+iy_0^4&=iz_1^2,\cr
     (-i)\omega^2 x_1^4+y_0^4&=z_1^2,\cr
                 y_0^4+2x_1^4&=z_1^2.}$$
Hence $(y_0,x_1,z_1)$ is a nontrivial solution of 
$x^4+2y^4=z^2$. 
The minimality of the height in (1.a) gives that we may 
assume that $x_0$, $y_0$, $z_0$ are relatively prime 
in (1.a).

(2) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4+2y^4=z^2$ in $Z[i]$ such that $x_0$, $y_0$, $z_0$ are pairwise
relatively prime.
Note that at most one of $x_0$, $y_0$, $z_0$ can be congruent to $0$
modulo $\omega$.
We consider the four cases listed in Table 1.

In cases 2 and 4, the equation $x_0^4+2y_0^4=z_0^2$ leads to the 
contradictions $0+0\equiv 1\pmod{\omega}$ and $1+0\equiv 0\pmod{\omega}$,
respectively.

We next show that case 1 is not possible either.
To do this, write $z_0$ in the form $z_0=k\omega^2+l$, $k,l\in Z[i]$.
Computing $z_0^2$, $z_0^2=k^2\omega^4+2k\omega^2l+l^2$ shows that
$z_0^2\equiv l^2\pmod{\omega^4}$.
As $0$, $1$, $i$, $1+i$ is a complete set of representatives modulo
$\omega^2$, it follows that $l$ can be chosen to be $1$ or $i$.
From the equation $x_0^4+2y_0^4=z_0^2$ we get that
$1+2\equiv l^2\pmod{\omega^4}$.
In the case $l=1$ this leads to the contradiction
$1+2\equiv 1\pmod{\omega^4}$ and so we left with the choice $l=i$.
Now writing $z_0$ in the form $z_0=r\omega^4+s$, $r,s\in Z[i]$ and
computing $z_0^2$, $z_0^2=r^2\omega^8+2r\omega^4s+s^2$ gives that
$z_0^2\equiv s^2\pmod{\omega^6}$.
From $z_0\equiv i\pmod{\omega^2}$, it follows that we can choose $s$
and $s^2$ in the way summarized by Table 2.
From the equation $x_0^4+2y_0^4=z_0^2$, we get the contradictions
$$3\equiv -1\pmod{\omega^6},\qquad
   3\equiv 3+4i\pmod{\omega^6}.$$

(3) In case 3, let $(x_1,\omega^ry_1,z_1)$ be a solution of the equation
$x^4+2y^4=z^2$, where $r\geq 1$, $x_1\equiv y_1\equiv z_1\equiv 1\pmod{\omega}$,
and $x_1$, $y_1$, $z_1$ are pairwise relatively prime.
We will show that $z_1\equiv 1\pmod{\omega^2}$.

In order to prove this claim, write $z_1$ in the form $z_1=k\omega^2+l$,
$k,l\in Z[i]$, and compute $z_1^2$.
$$z_1^2=k^2\omega^4+2k\omega^2l+l^2.$$
From this, it follows that $z_1^2\equiv l^2\pmod{\omega^4}$
Since the elements $0$, $1$, $i$, $1+i$ form a complete set of representatives
modulo $\omega^2$, and since $z_1\equiv 1\pmod{\omega}$, we 
may choose $l$ to be $1$ or $i$.
Consequently, $z_1^2$ is congruent to $1$ or $-1$ modulo $\omega^4$.
The equation $x_1^4+2\omega^{4r}y_1^4=z_1^2$ gives that 
$1\equiv z_1^2\pmod{\omega^4}$, and so $z_1\equiv 1\pmod{\omega^2}$. 

(4) We will show that there are pairwise relatively prime elements 
$x_2$, $y_2$, $z_2$ of $Z[i]$, such that
$x_2\equiv y_2\equiv z_2\equiv 1\pmod{\omega}$, and 
$(x_2,\omega^{r-1}y_2,z_2)$ is a solution of
the equation $x^4+2y^4=z^2$.

In order to verify the claim, write the equation
$x_1^4+2\omega^{4r}y_1^4=z_1^2$ in the form
$2\omega^{4r}y_1^4=(z_1-x_1^2)(z_1+x_1^2)$, and compute the greatest
common divisor of $(z_1-x_1^2)$ and $(z_1+x_1^2)$.
Let $g$ be this greatest common divisor.
As $g|\omega^{4r}y_1^4$, it follows that $g\not=0$.
Since $g|(z_1-x_1^2)$, $g|(z_1+x_1^2)$ we get that $g|2x_1^2$, $g|2z_1$.
If $q$ is a prime divisor of $g$ with $q\not|\omega$, we then get
$q|x_1$, $q|z_1$.
But we know that this is not the case as $x_1$ and $z_1$ 
are relatively prime.
Thus, $g=\omega^s$ and $0\leq s\leq 2$ since $g|2$.
By step (3) $z_1\equiv 1\pmod{\omega^2}$.
This together with $x_1^2\equiv 1\pmod{\omega^2}$, gives that
$(z_1-x_1^2)\equiv (z_1+x_1^2)\equiv 0\pmod{\omega^2}$.
Therefore $g=\omega^2$.
The unique factorization property in $Z[i]$ implies that
there are relatively prime elements $a,b\in Z[i]$ such that
$$z_1-x_1^2=\omega^2a,\qquad z_1+x_1^2=\omega^2b.$$
Let $a=\omega^ua_1$, $b=\omega^vb_1$.
So $(-i)\omega^{4r+2}y_1^4=\omega^{u+v+4}a_1b_1$.
By the unique factorization property in $Z[i]$, there are elements
$a_2$, $b_2$ and units $\varepsilon$, $\mu$ in $Z[i]$ for which
$$z_1-x_1^2=\omega^{u+2}\varepsilon a_2^4,\qquad
  z_1+x_1^2=\omega^{v+2}\mu b_2^4,$$
$$4r+2=u+v+4,\qquad a_2^4b_2^4=y_1^4,\qquad\varepsilon\mu=-i.$$
Here, $a_2$, $b_2$ are prime to $\omega$.
It follows that $a_2\equiv b_2\equiv 1\pmod{\omega}$.
By addition, we get
$$2x_1^2=\omega^{v+2}\mu b_2^4-
         \omega^{u+2}\varepsilon a_2^4.$$
After dividing it by $\omega^2$, we get 
$$(-i)x_1^2=\omega^{v}\mu b_2^4-
            \omega^{u}\varepsilon a_2^4.$$
We distinguish two cases depending on whether $u=0$, $v=4r-2$, or
$v=0$, $u=4r-2$.
When $u=0$, $v=4r-2$, we get
$$(-i)x_1^2=\omega^{4r-2}\mu b_2^4-\varepsilon a_2^4.$$
This reduces to 
$(-i)\equiv-\varepsilon\pmod{\omega^2}$.
From this, it follows that $\varepsilon=\pm i$, that is, either
$\varepsilon=i$, $\mu=-1$, or $\varepsilon=-i$, $\mu=1$.
In the first case, we get
$$\eqalign{
(-i)x_1^2&=\omega^{4r-2}(-1)b_2^4-(i)a_2^4,\cr
    x_1^2&=(-i)\omega^2\omega^{4r-4}b_2^4+a_2^4,\cr
    x_1^2&=2\omega^{4r-4}b_2^4+a_2^4.}$$
Thus, $(a_2,\omega^{r-1}b_2,x_1)$ is a nontrivial
solution of the equation $x^4+2y^4=z^2$.
In the second case, we get
$$\eqalign{
(-i)x_1^2&=\omega^{4r-2}(-1)b_2^4-(i)a_2^4,\cr
   -x_1^2&=(-i)\omega^2\omega^{4r-4}b_2^4+a_2^4,\cr
   -x_1^2&=2\omega^{4r-4}b_2^4+a_2^4.}$$
Therefore, $(a_2,\omega^{r-1}b_2,ix_1)$ is a nontrivial
solution of the equation $x^4+2y^4=z^2$.

When $v=0$, $u=4r-2$, we get
$$(-i)x_1^2=\mu b_2^4-\omega^{4r-2}\varepsilon a_2^4.$$
This reduces to 
$(-i)\equiv\mu\pmod{\omega^2}$.
From this, it follows that $\mu=\pm i$, that is, either
$\varepsilon=-1$, $\mu=i$, or $\varepsilon=1$, $\mu=-i$.
In the first case, we get
$$\eqalign{
(-i)x_1^2&=(i)b_2^4-\omega^{4r-2}(-1)a_2^4,\cr
   -x_1^2&=b_2^4+(-i)\omega^2\omega^{4r-4}a_2^4,\cr
   -x_1^2&=b_2^4+2\omega^{4r-4}a_2^4.}$$
Thus, $(b_2,\omega^{r-1}a_2,ix_1)$ is a nontrivial
solution of the equation $x^4+2y^4=z^2$.
In the second case, we get
$$\eqalign{
(-i)x_1^2&=(-i)b_2^4-\omega^{4r-2}a_2^4,\cr
    x_1^2&=b_2^4+(-i)\omega^2\omega^{4r-4}a_2^4,\cr
    x_1^2&=b_2^4+2\omega^{4r-4}a_2^4,}$$
and so $(b_2,\omega^{r-1}a_2,x_1)$ is a nontrivial
solution of the equation $x^4+2y^4=z^2$.

(5) Let $(x_0,y_0,z_0)$ be a nontrivial solution of the equation
$x^4+2y^4=z^2$ in $Z[i]$.
By step (2), there is a solution $(x_1,\omega^ry_1,z_1)$ with
$x_1,y_1,z_1\equiv 1\pmod{\omega}$, $r\geq 1$.
Choose a solution for which $r$ is minimal.
According to step (4), there is a solution 
$(x_2,\omega^{r-1}y_2,z_2)$, where
$x_2,y_2,z_2\equiv 1\pmod{\omega}$, $r\geq 2$.
This contradicts the choice of $r$ and so completes the proof.

\vskip 20pt


\baselineskip=10pt


\line{\bf References \hfil}

\noindent
1. J. T. Cross,
In the Gaussian integers $\alpha^4+\beta^4\not=\gamma^4$,
{\it Math. Magazine}, {\bf 66} (1993), 105--108.



\noindent
2. D. Hilbert,
Jahresbericht d. Deutschen Math.-Vereinigung, 4, 1894-1895, 
517--525.



\noindent
3. L. J. Mordell,
{\it Diophantine Equations}, Academic Press, New York, 1969, 116--117.


\noindent
4. T. Nagell,
{\it Introduction to Number Theory}, Chelsea Publ. Comp., New York,
1981. 


%\noindent
%Mathematics Subject Classification (1991)

%\noindent
%Primary 11F06; Secondary 11E08

%\noindent
%Key words and phrases: Diophantine equations, Gaussian integers

%\vskip.2in


%\vfill
\end