
\input amstex


\magnification=1400


\loadmsbm

\input colordvi

\nopagenumbers
\parindent=0pt

\def\cgy{\GreenYellow}     % GreenYellow  Approximate PANTONE 388
\def\cyy{\Yellow}	  % Yellow  Approximate PANTONE YELLOW
\def\cgo{\Goldenrod}	  % Goldenrod  Approximate PANTONE 109
\def\cda{\Dandelion}	  % Dandelion  Approximate PANTONE 123
\def\capr{\Apricot}	  % Apricot  Approximate PANTONE 1565
\def\cpe{\Peach}		  % Peach  Approximate PANTONE 164
\def\cme{\Melon}		  % Melon  Approximate PANTONE 177
\def\cyo{\YellowOrange}	  % YellowOrange  Approximate PANTONE 130
\def\coo{\Orange}	  % Orange  Approximate PANTONE ORANGE-021
\def\cbo{\BurntOrange}	  % BurntOrange  Approximate PANTONE 388
\def\cbs{\Bittersweet}	  % Bittersweet  Approximate PANTONE 167
%\def\creo{\RedOrange}	  % RedOrange  Approximate PANTONE 179
\def\cma{\Mahogany}	  % Mahogany  Approximate PANTONE 484
\def\cmr{\Maroon}	  % Maroon  Approximate PANTONE 201
\def\cbr{\BrickRed}	  % BrickRed  Approximate PANTONE 1805
\def\crr{\Red}		  % Red  VERY-Approx PANTONE RED
\def\cor{\OrangeRed}	  % OrangeRed  No PANTONE match
\def\paru{\RubineRed}	  % RubineRed  Approximate PANTONE RUBINE-RED
\def\cwi{\WildStrawberry}  % WildStrawberry  Approximate PANTONE 206
\def\csa{\Salmon}	  % Salmon  Approximate PANTONE 183
\def\ccp{\CarnationPink}	  % CarnationPink  Approximate PANTONE 218
\def\cmag{\Magenta}	  % Magenta  Approximate PANTONE PROCESS-MAGENTA
\def\cvr{\VioletRed}	  % VioletRed  Approximate PANTONE 219
\def\parh{\Rhodamine}	  % Rhodamine  Approximate PANTONE RHODAMINE-RED
\def\cmu{\Mulberry}	  % Mulberry  Approximate PANTONE 241
\def\parv{\RedViolet}	  % RedViolet  Approximate PANTONE 234
\def\cfu{\Fuchsia}	  % Fuchsia  Approximate PANTONE 248
\def\cla{\Lavender}	  % Lavender  Approximate PANTONE 223
\def\cth{\Thistle}	  % Thistle  Approximate PANTONE 245
\def\corc{\Orchid}	  % Orchid  Approximate PANTONE 252
\def\cdo{\DarkOrchid}	  % DarkOrchid  No PANTONE match
\def\cpu{\Purple}	  % Purple  Approximate PANTONE PURPLE
\def\cpl{\Plum}		  % Plum  VERY-Approx PANTONE 518
\def\cvi{\Violet}	  % Violet  Approximate PANTONE VIOLET
\def\parp{\RoyalPurple}	  % RoyalPurple  Approximate PANTONE 267
\def\cbv{\BlueViolet}	  % BlueViolet  Approximate PANTONE 2755
\def\cpe{\Periwinkle}	  % Periwinkle  Approximate PANTONE 2715
\def\ccb{\CadetBlue}	  % CadetBlue  Approximate PANTONE (534+535)/2
\def\cco{\CornflowerBlue}  % CornflowerBlue  Approximate PANTONE 292
\def\cmb{\MidnightBlue}	  % MidnightBlue  Approximate PANTONE 302
\def\cnb{\NavyBlue}	  % NavyBlue  Approximate PANTONE 293
\def\crb{\RoyalBlue}	  % RoyalBlue  No PANTONE match
%\def\cbb{\Blue}		  % Blue  Approximate PANTONE BLUE-072
\def\cce{\Cerulean}	  % Cerulean  Approximate PANTONE 3005
\def\ccy{\Cyan}		  % Cyan  Approximate PANTONE PROCESS-CYAN
\def\cpb{\ProcessBlue}	  % ProcessBlue  Approximate PANTONE PROCESS-BLUE
\def\csb{\SkyBlue}	  % SkyBlue  Approximate PANTONE 2985
\def\ctu{\Turquoise}	  % Turquoise  Approximate PANTONE (312+313)/2
\def\ctb{\TealBlue}	  % TealBlue  Approximate PANTONE 3145
\def\caq{\Aquamarine}	  % Aquamarine  Approximate PANTONE 3135
\def\cbg{\BlueGreen}	  % BlueGreen  Approximate PANTONE 320
\def\cem{\Emerald}	  % Emerald  No PANTONE match
\def\cjg{\JungleGreen}	  % JungleGreen  Approximate PANTONE 328
\def\csg{\SeaGreen}	  % SeaGreen  Approximate PANTONE 3268
\def\cgg{\Green}	  % Green  VERY-Approx PANTONE GREEN
\def\cfg{\ForestGreen}	  % ForestGreen  Approximate PANTONE 349
\def\cpg{\PineGreen}	  % PineGreen  Approximate PANTONE 323
\def\clg{\LimeGreen}	  % LimeGreen  No PANTONE match
\def\cyg{\YellowGreen}	  % YellowGreen  Approximate PANTONE 375
\def\cspg{\SpringGreen}	  % SpringGreen  Approximate PANTONE 381
\def\cog{\OliveGreen}	  % OliveGreen  Approximate PANTONE 582
\def\pars{\RawSienna}	  % RawSienna  Approximate PANTONE 154
\def\cse{\Sepia}		  % Sepia  Approximate PANTONE 161
\def\cbr{\Brown}		  % Brown  Approximate PANTONE 1615
\def\cta{\Tan}		  % Tan  No PANTONE match
\def\cgr{\Gray}		  % Gray  Approximate PANTONE COOL-GRAY-8
\def\cbl{\Black}		  % Black  Approximate PANTONE PROCESS-BLACK
\def\cwh{\White}		  % White  No PANTONE match


\voffset=-.5in
\hoffset=-.5in
\hsize = 7.5 true in
\vsize=10.9 true in

%\voffset=1.2in
%\hoffset=-.5in
%\hsize = 10.2 true in
%\vsize=8 true in

\overfullrule=0pt


\def\ctln{\centerline}
\def\u{\underbar}
\def\ssk{\smallskip}
\def\msk{\medskip}
\def\bsk{\bigskip}
\def\hsk{\hskip.1in}
\def\hhsk{\hskip.2in}
\def\dsl{\displaystyle}
\def\hskp{\hskip1.5in}

\def\moda{\medspace {\underset a\to \equiv} \medspace}
\def\modb{\medspace {\underset b\to \equiv} \medspace}
\def\modc{\medspace {\underset c\to \equiv} \medspace}

\def\lra{$\Leftrightarrow$ }


\ctln{\bf Math 445 Number Theory}

\ssk

\ctln{November 17, 2004}

\msk


\crr{{\it Theorem:} If $abc$ is square-free, then $ax^2+by^2+cz^2=0$ has a 
(non-trfvial!) solution
$x,y,z\in{\Bbb Z}$ \lra\  $a,b,c$ do not all have the same sign, and each of 
the equations }

\crr{\hhsk $w^2\equiv -ab\pmod{c} ,w^2\equiv -ac\pmod{b}, w^2\equiv -bc\pmod{a}$
\hhsk have solutions.}

\msk

($\Leftarrow$ :) After possible renaming variables and taking negatives, we may assume
that $a>0$ and $b,c<0$ . Suppose $r^2\equiv -ab\pmod{c}$ and $aA\equiv 1\pmod{c}$ .
Then for any $x,y\in{\Bbb Z}$, mod $c$ we have  
$ax^2+by^2+cz^2\equiv ax^2+by^2\equiv aA(ax^2+by^2)\equiv A(a^2x^2+aby^2)=A(a^2x^2-r^2y^2)=A(ax-ry)(ax+ry) \equiv
(x-Ary+0z)(ax+ry+0z)$ . Similarly, mod $b$ (with $s^2\equiv -ac$) we have $ax^2+by^2+cz^2\equiv (x+0y-Asz)(ax+0y+sz)$ and, 
mod $a$ (with $t^2\equiv -bc$ and $bB\equiv 1$) we have $ax^2+by^2+cz^2\equiv (0x+y-Btz)(0x+by+tz)$ .

Using the Chinese Remainder Theorem, we can solve $\alpha\moda 1,\alpha\modb 1,\alpha\modc 0$ , 
$\beta\moda -A,\beta\modb 0,\beta\modc 1$ , $\gamma\moda 0,\gamma\modb -As,\gamma\modc -Bt$ ,
$\delta\moda a,\delta\modb a,\delta\modc 0$ , $\epsilon\moda r,\epsilon\modb 0,\epsilon\modc b$ , 
and $\eta\moda 0,\eta\modb s,\eta\modc t$ . Then,
 \crr{mod $abc$ , $ax^2+by^2+cz^2\equiv (\alpha x+\beta y + \gamma z)(\delta x + \epsilon y+ \eta z)$ .}
\hhsk
Then we need a

\ssk

{\it Lemma :} If $\lambda,\mu,\nu\in{\Bbb R}$ and positive, with $\lambda\mu\nu = M\in{\Bbb Z}$ , 
then for any $\alpha,\beta,\gamma\in{\Bbb Z}$, $\alpha x+\beta y+\gamma z\equiv 0\pmod{M}$ 
has a solution with $x,y,z\in{\Bbb Z}$ , $(x,y,z)\neq (0,0,0)$, 
and $|x|\leq \lfloor\lambda\rfloor, |y|\leq\lfloor\mu\rfloor, |z|\leq\lfloor\nu\rfloor$ . 

\ssk

The proof is simply that,
among $0\leq x\leq \lfloor\lambda\rfloor, 0\leq y\leq\lfloor\mu\rfloor,0\leq z\leq\lfloor\nu\rfloor$, 
we have $(1+\lfloor\lambda\rfloor)(1+\lfloor\mu\rfloor)(1+\lfloor\nu\rfloor) > \lambda\mu\nu = M$
triples $(x,y,z)$ to choose from, so $\alpha x+\beta y+\gamma z\equiv \alpha x_1+\beta y_1+\gamma z_1$ for
some pair of triples, and so $\alpha(x-x_1)+\beta(y-y_1)+\gamma(z-z_1)\equiv 0$.

\msk

Setting $\lambda=\sqrt{bc}, \mu=\sqrt{-ac},\nu=\sqrt{-ab}$, we then can solve 
$\alpha x+\beta y + \gamma z\equiv 0\pmod{abc}$ 
(so $ax^2+by^2+cz^2\equiv 0\pmod{abc}$) with $|x|\leq \sqrt{bc},|y|\leq\sqrt{-ac},|z|\leq\sqrt{-ab}$ .
But since $abc$ is square-free, none of these square roots are integers (unless they are 1). So
$x^2\leq bc, y^2\leq -ac, z^2\leq -bc$, and equality occurs for any only if the corresponding right-hand
side is 1.

\ssk

Then, unless $b=c=-1$, we have $x^2<bc$ and $abc|ax^2+by^2+cz^2$ with $ax^2+by^2+cz^2\leq ax^2<abc$ and 
$ax^2+by^2+cz^2 \geq by^2+cz^2 > b(-ac) + c(-ab) = -2abc$ . [The last inequality is
reversed, since $b,c<0$. It is strict, unless $a=1$ as well.] So $ax^2+by^2+cz^2 = 0$
 or $=-abc$ . In the first case we are done; in the second, setting $X=-by+xz,Y=ax+yz,Z=z^2+ab$
we have 
$aX^2+bY^2+cZ^2=a(-by+xz)^2+b(ax+yz)^2+c(z^2+ab)^2 = 
(ab^2y^2-2abxyz+ax^2z^2)+(a^2bx^2+2abxyz+by^2z^2)+(cz^4+2abcz^2+a^2b^2c) = 
(ax^2+by^2+cz^2)z^2+ab^2y^2+a^2bx^2+2abcz^2+a^2b^2c = -abcz^2+ab^2y^2+2abcz^2+a^2bx^2+a^2b^2c
=ab(ax^2+by^2+cz^2)+a^2b^2c = (ab)(-abc)+(ab)(abc) = 0$. This gives a non-trivial solution, unless
$0=-by+xz,0=ax+yz,0=z^2+ab$, so $z^2=-ab$, so $a=1,b=-1$ since $ab$ is square-free; and then
$(x,y,z)=(1,1,0)$ is a solution.

\msk

Finally, in the special case $b=c=-1$, we have $w^2\equiv -bc=-1\pmod{a}$, has a solution, so 
every prime factor $p$ of $a$ also has $w^2\equiv -1\pmod{p}$, so $p\equiv -1\pmod{4}$ 
for every prime factor, so $y^2+z^2=a$ has a solution, so $(1,y,z)$ is a solution to 
$ax^2+by^2+cy^2=ax^2-y^2-z^2=0$, as desired.



\vfill
\end



