% Add soln to question about G_{2-torsion} and G/2G for G given as
% product of various cyclic groups.
\input amssym.def 
\input amssym.tex
%\def\Bbb{\bf}
\def\notdiv{{\not\hskip-.5pt |\ }}
\def\ge{\geqslant}
\def\le{\leqslant}
\def\ctq{{\cal C}_{\lower 1pt\hbox{\eightsl tors}}({\Bbb Q})}
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%\hoffset=1truecm
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\font\eightsl=cmsl8
\font\frkkk=eufm10
\font\twelverm=cmr12
\font\tenrm=cmr10
\font\ninerm=cmr9
\font\ninebf=cmbx9
\font\eightrm=cmr8
\font\sixrm=cmr6
\font\scrpp=eusm10 
\font\frkk=eufm10
\font\deffont=cmssi10
\font\chaptitle=cmbx10 at 14 pt
\tolerance=10000
\def\sqr{\ifmmode\square\else{$\square$}\fi}
\def\square{\vcenter{
\hrule height.1mm
\hbox{\vrule width.1mm height2.2mm\kern2.18mm\vrule width.1mm}
\hrule height.1mm}}                  % This is a slimmer sqr.
%\def\sqr{$\vcenter{\hrule height .3mm
%\hbox {\vrule width .3mm height 2mm \kern 1.4mm
%\vrule width .3mm} \hrule height .3mm}$}
%
\null
%
%\vsize=19.5 true cm
%\hsize=11.5 true cm
%\vskip 5 true cm
%\def\leqslant{\le}
\def\c{{\cal C}}
\def\pk{\phi _\kappa}
\def\im{{\hbox{\sl im}}}
\def\hs{H_{\varsigma}}
\def\hpk{\hat \phi _\kappa}
\font\sc=cmssqi8 
\def\scc#1{\hbox{\sc #1}}
\def\sf{{\scc F}}
\def\pnbq{{\Bbb P}^n(\overline {\Bbb Q} )}
\def\hk{{\hat \kappa}}
\def\bq{{\overline {\Bbb Q}}}
\def\hq{{\hat q}}
\def\pv{\prod\limits_v }
\def\pnk{{\Bbb P}^n(K)}
\def\mnkvw{{\Bbb M}^n(K[{\bf v}^2,{\bf w}^2])}
\def\pnkv{{\Bbb P}^n(K[{\bf v}^2])}
\def\kj{\kappa (J)}
\def\qss{{({\Bbb Q}^*)^2}}
\def\rss{{({\Bbb R}^*)^2}}
\def\css{{({\Bbb C}^*)^2}}
\def \qmods {{\Bbb Q}^*/({\Bbb Q}^*)^2}
\def \rmods {{\Bbb R}^*/({\Bbb R}^*)^2}
\def \cmods {{\Bbb C}^*/({\Bbb C}^*)^2}
\def \qmodss { {\Bbb Q}^*/({\Bbb Q}^*)^2 \times 
{\Bbb Q}^*/({\Bbb Q}^*)^2 }
\def \qs{{\Bbb Q}^*}
\def\bbQ{\Bbb Q}
\def\bbZ{\Bbb Z}
\def\bbR{\Bbb R}
\def\bbC{\Bbb C}
%
\chaptitle
\noindent
\centerline{Elliptic Curves. Solutions to Sheet 0.}
\rm
\bigskip
\noindent
\noindent {\bf 1.}
\par\noindent {\bf (a).} There is an identity element
$\bigl( {1\atop 0} {0\atop 1} \bigr)$, but the matrix
$( {0\atop 0} {0\atop 0} \bigr)$, for example, does not have
an inverse, since there does not exist a matrix~$A$ such that
$A( {0\atop 0} {0\atop 0} \bigr) = ( {0\atop 0} {0\atop 0} \bigr)A
= \bigl( {1\atop 0} {0\atop 1} \bigr)$; so, not a group.
\par\noindent {\bf (b).} This is group, with identity
$\bigl( {0\atop 0} {0\atop 0} \bigr)$.
Closure and
associativity are easy to show.
The inverse of
$\bigl( {a\atop c} {b\atop d} \bigr)$
is
$\bigl( {-a\atop -c} {-b\atop -d} \bigr)$.
\medskip
\noindent {\bf 2.}
\par\noindent {\bf (a).} Not a homomorphism since, for example,
$\phi (1+3) = \phi (4) = 17$, but $\phi (1) \times \phi (3) = 2 \times 10 =
20$.
\par\noindent {\bf (b).} This is a homomorphism since, for all
$v,w \in \bbQ$, we have $\phi (v+w) = \sqrt{2}(v+w) = \sqrt{2}\, v
+ \sqrt{2}
\, w = \phi (v) + \phi (w)$. The map is injective since, for all
$v,w \in \bbQ$, if $\phi (v) = \phi (w)$, then $\sqrt{2}\, v = \sqrt{2}\, w$,
and so $v=w$. The map is not surjective since, for example,~2
is not in the image of~$\phi$ (since there is no $v\in \bbQ$
such that $\sqrt{2}\, v = 2$). Hence the map is not bijective.
Finally, $v\in \hbox{ker}\phi \iff \sqrt{2}\, v = 0 \iff v=0$,
and so $\hbox{ker}\phi = \{ 0 \}$.
\par\noindent {\bf (c).} This is a homomorphism since, for all
$v,w\in \bbZ$, we have $\phi (v+w) = 2(v+w) = 2v + 2w =
\phi (v) + \phi (w)$. The map is not injective since, for example,
the elements $0$ and $3$ (which are distinct in $\bbZ$)
both map to $0$ in $\bbZ / 3\bbZ$; therefore the map is
also not bijective. The map is surjective
since $0\mapsto 0$, $1\mapsto 2$ and $2\mapsto 1$ so that all
three elements of $\bbZ / 3\bbZ$ are in the image of~$\phi$.
Finally, $v\in \hbox{ker}\phi \iff 2v=0 \hbox{ in } \bbZ / 3\bbZ
\iff 3\vert 2v \iff 3\vert v \hbox{ in }\bbZ$.
Hence the kernel of~$\phi$ is $\{ \ldots  -3 ,0,3  \ldots \} = 3\bbZ$.
\medskip
\noindent {\bf 3.} 
\par\noindent {\bf (a).}
$3=(1/27)\times 81$ and $81 = 9^2\in \qss$, so $3=1/27$ in $\qmods$.
\ \ $-4=4\times (-1)$ and $-1 \not\in \qss$, so
$-4\not= 4$ in $\qmods$. \ \ Finally, $3=(5/6)\times (18/5)$,
and $18/5 \not\in \qss$, so $3\not= 5/6$ in $\qmods$.
\par\noindent {\bf (b).} In $\qmods$:
\ \ $-2/27 = (-2/27)\times 9^2 = -6$,\ \ 
$16 = 16 \times (1/4)^2 = 1$, \ \ $12 = 12 \times (1/2)^2 = 3$,\ \ 
and $1/3 = (1/3)\times 3^2 = 3$.
\par\noindent {\bf (c).} In $\qmods$:
$6\times 10 = 60 = 60\times (1/2)^2 = 15$,\ \ 
$10 / 21 = (10/21)\times 21^2 = 210$, \ \
$15^{101} = 15\times (15^{50})^2 = 15$,\ \ 
and $3^{-1} = 3^{-1}\times 3^2 = 3$.
\par\noindent {\bf (d).}
Whenever $p_1,p_2$ are distinct primes, we have that $p_1 = p_2 \times
(p_1/p_2)$, where $p_1/p_2 \not\in \qss$. Hence, $p_1 \not= p_2$
in $\qmods$. This means that the primes are all distinct in $\qmods$,
and so $\qmods$ is infinite (note that this does not describe all
elements in $\qmods$, but the above argument is sufficient
to show that the size of~$\qmods$ is infinite).
In $\rmods$, every $r > 0$ is equal to~$1$ (since such an $r$ is
a member of $\rss$), and similarly any~$r<0$ is equal to~$-1$.
But $1 \not= -1$ in $\rmods$, since $-1 \not\in \rss$. Hence,
$\rmods$ has exactly 2 distinct elements: $1$ and $-1$. 
Finally, everything in $\Bbb C^*$ is in $\css$ (since every
complex number is a square of some complex number), and so
$\cmods$ has exactly 1 element.
\medskip
\noindent {\bf 4.}
\par\noindent {\bf (a).} A singular point $(x,y)$ must satisfy
the three equations:
$f(x,y) = x^4 + y^3 - 3 x^2 y = 0$, $g(x,y) = 
(\partial f / \partial X )(x,y) = 4 x^3 - 6 x y = 0$
and $h(x,y) = (\partial f / \partial Y )(x,y) =  
3 y^2 - 3 x^2 = 0$. From $h(x,y) = 0$, we get $(y+x)(y-x)=0$
and so $y=x$ or $y=-x$, which we now consider as separate cases.
\par\noindent {\bf Case 1.} $y=x$. In this case,
$f(x,y)=0$ becomes $x^3(x-2) = 0$ (with solutions $x=0,2$)
and $g(x,y)=0$ becomes $2x^2(2x-3) = 0$ (with solutions $x=0,3/2$),
and so the only simultaneous solution for $x$ is $x=0$.
Substituting $x=0$ into $f(x,y)=0$ gives $y^3=0$ and so $y=0$.
Hence, $(0,0)$ is the only possibility within case 1.
\par\noindent {\bf Case 2.} $y=-x$. In this case, 
$f(x,y)=0$ becomes $x^3(x+2)=0$ (with solutions $x=0,-2$)
and $g(x,y)=0$ becomes $2x^2(2x+3) = 0$ (with solutions $x=0,-3/2$), 
and so the only simultaneous solution for $x$ is $x=0$.
As before, substituting $x=0$ into $f(x,y)=0$ gives $y^3=0$ and so $y=0$.
\par\noindent {\bf In either case}, $(0,0)$ is the only possible point 
which could be singular.
\par\noindent [{\it Alternative Method (harder than the above, but guaranteed
to work for any curve): Let
$\alpha (x) = \hbox{\it Res}(f,g)$ with respect to~$y$, and
$\beta (x) = \hbox{\it Res}(g,h)$ with respect to~$y$, and
finally use resultants with respect to~$x$ to show that~$x=0$
is the only common root of $\alpha(x)$ and $\beta(x)$}].
\par
Finally, note that~$(0,0)$ does indeed
satisfy $f(x,y)=g(x,y)=h(x,y)=0$ and so is a singular point.
Conclusion: $(0,0)$ is the only singular point.
\par $f(X+0,Y+0) = R_3(X,Y) + R_4(X,Y)$, where $R_3(X,Y) =
Y^3 - 3 X^2 Y = Y(Y - \sqrt{3}\, X)(Y + \sqrt{3}\, X)$.
So the three tangents at~$(0,0)$ are $Y=0$, $Y=\sqrt{3}\, X$
and $Y=-\sqrt{3}\, X$. 
\par\noindent {\bf (b).} A singular point $(x,y)$ must satisfy
the three equations:    
$f(x,y) = y^2 - x(x^2-1)^2 = 0$, $g(x,y) =
(\partial f / \partial X )(x,y) = -(x^2-1)(5x^2-1) = 0$
and $h(x,y) = (\partial f / \partial Y )(x,y) =
2y = 0$. From $h(x,y)=0$ we have $y=0$ and, on substituting
this into $f(x,y)=0$ we have $- x(x^2-1)^2 = 0$, and
so the only possible values for~$x$ are $x=0,1-1$. 
But from $g(x,y)=0$ we see that $x$ must also satisfy
$x=1-1,\sqrt{1/5}, -\sqrt{1/5}$. The only common solutions
are: $x=1,-1$. Substituting $x=1$ or $x=-1$ into $f(x,y)=0$
gives that $y=0$, and so the only possible points which could
be singular are: $(1,0)$ and $(-1,0)$. Finally, note that
$x=1,y=0$ does indeed satisfy $f(x,y)=g(x,y)=h(x,y)=0$
as does $x=-1,y=0$. Conclusion: $(1,0)$ and
$(-1,0)$ are the only singular points.
\par At $(0,0)$, we have $f(X+0,Y+0) = -X + \hbox{ terms of degree at
least 2}$, and so there is a unique tangent at the point~$(0,0)$,
namely the line~$X=0$. At $(1,0)$, we note that
$f(X+1,Y+0) = (Y^2-4X^2) + \hbox{ terms of degree at
least 3}$. Here, $(Y^2-4X^2) = L_1(X,Y)L_2(X,Y)$, where
$L_1(X,Y) = Y-2X$ amd $L_2(X,Y) = Y+2X$. Finally, the
two tangents to~$\c$ at $(1,0)$ are therefore
$L_1(X-1,Y-0) = Y-2(X-1)=0$ and $L_2(X-1,Y-0) = Y+2(X-1)=0$.
\medskip \noindent {\bf 5.} If $(x,y)$ is singular
on $Y^2 = X^3 + AX + B$, then $(x,y)$ satisfies the three equations:     
$f(x,y) = y^2 - (x^3 + Ax + B) = 0$, $g(x,y) = 
(\partial f / \partial X )(x,y) = -(3x^2+A) = 0$ 
and $h(x,y) = (\partial f / \partial Y )(x,y) =  
2y = 0$. From $h(x,y)=0$, we have $y=0$; substituting $y=0$
into $f(x,y)=0$, we deduce $x^3 + Ax + B = 0$. From $g(x,y)=0$,
we see that $x$ also satisfies $3x^2+A=0$; that is, $x$
is a common root of $p_1(x) = x^3 + Ax + B$ and its derivative
$p_2(x) = 3x^2+A$; that is, $x$ satisfies:
$p_1(x) = x^3 + Ax + B = 0$
and $p_2(x) = 3x^2+A = 0$. Then $x$ would satisfy: 
$p_3(x) = 3 p_1(x) - x p_2(x) = 2Ax + 3B = 0$,
and so: $p_4(x) = 2Ap_2(x) - 3xp_3(x) = -9Bx + 2A^2$,
and so: $p_5(x) = 9Bp_3(x) + 2Ap_4(x) = 4A^3 + 27B^2 = 0$.
But we are told that $4A^3 + 27B^2\not= 0$ and so no
such~$x$ can exist; therefore there are no singular points.
[The above is from first principles.
An easier alternative
method is to compute: Disc($x^3 + Ax + B$) is $4A^3 + 27B^2$;
if this is nonzero, no such~$x$ can exist, and so
no singular point can exist.] 
\medskip\noindent {\bf 6.} 
\par\noindent {\bf (a).} Let $f(X,Y) = Y^2 - X^5$. Imagine
that $f(X,Y) = g(X,Y)h(X,Y)$, where $g,h$ are non-constant
polynomials defined over~$\Bbb C$. Since $f(X,Y)$ has degree~2
in~$Y$, there are three possibilities: $g$ has degree $0$ in $Y$
and $h$ has degree $2$ in $Y$; $g$ has degree $1$ in $Y$
and $h$ has degree $1$ in $Y$; $g$ has degree $2$ in $Y$
and $h$ has degree $0$ in $Y$.
\par\noindent {\bf Case 1.} $g$ has degree $0$ in $Y$
and $h$ has degree $2$ in $Y$. Then we can write:
$g(X,Y) = \phi_0(X)$, where $\phi_0(X)$ is a polynomial
in~$X$, and $h(X,Y) = \theta_2(X) Y^2 + \theta_1(X) Y + \theta_0(X)$,
where each $\theta_i(X)$ is a polynomial in~$X$. Then, equating the
coefficients of $Y^2$ in the
equation $f(X,Y) = g(X,Y)h(X,Y)$ gives that $1 = \phi_0(X) \theta_2(X)$,
and so both $\phi_0(X)$ and $\theta_2(X)$ must be constants.
But then $g(X,Y)$ would be constant, contradicting out initial
assumption that $g,h$ are non-constant.
\par\noindent {\bf Case 2.} $g$ has degree $1$ in $Y$ 
and $h$ has degree $1$ in $Y$. Then we can write:
$g(X,Y) = \phi_1(X) Y + \phi_0(X)$, where each $\phi_i(X)$ is a polynomial  
in~$X$, and $h(X,Y) = \theta_1(X) Y + \theta_0(X)$, where
each $\theta_i(X)$ is a polynomial
in~$X$. Then, equating the 
coefficients of $Y^2$ in the
equation $f(X,Y) = g(X,Y)h(X,Y)$ gives that $1 = \phi_1(X) \theta_1(X)$,
and so both $\phi_1(X)$ and $\theta_1(X)$ must be constants (and
multiplicative inverses);
say $\phi_1(X) = r$ and $\theta_1(X) = 1/r$, where $r\in {\Bbb C}$
is a nonzero constant. Then the equation $f(X,Y) = g(X,Y)h(X,Y)$
becomes: $Y^2 - X^5 = Y^2 + \bigl( r\theta_0(X) + (1/r)\phi_0(X)\bigr) Y
+ \theta_0(X)\phi_0(X)$. Equating coefficients of~$Y$
gives that $r\theta_0(X) + (1/r)\phi_0(X) = 0$ and
so $\phi_0(X) = -r^2 \theta_0(X)$. Hence the equation
becomes: $Y^2 - X^5 = Y^2 - r^2 \bigl( \theta_0(X) \bigr)^2$.
But, $r^2 \bigl( \theta_0(X) \bigr)^2$ is a polynomial of
even degree in~$X$, whereas $X^5$ is of odd degree, a contradiction.
\par\noindent {\bf Case 3.} $g$ has degree $2$ in $Y$ 
and $h$ has degree $0$ in $Y$. This gives the same contradiction
as in Case 1, but with $g,h$ interchanged.
\par In summary, our initial assumption, that 
$f(X,Y) = g(X,Y)h(X,Y)$, where $g,h$ are non-constant
polynomials defined over~$\Bbb C$, lead to a contradiction
in all cases. Hence, it is not possible to write $f(X,Y)$
as such a product, and so $f(X,Y)$ is irreducible over~$\Bbb C$.
Of course, this means that $f(X,Y)$ is irreducible over~$\Bbb Q$, also.
\par\noindent {\bf (b).} Let $f(X,Y) = Y^3 - X^3 = (X-Y)(X^2 + XY + Y^2)
= (X-Y)(X-\omega Y)(X-\omega^2 Y)$, where $\omega = e^{2\pi / 3}
= (1+\sqrt{-3})/2$. So, the irreducible components over~$\Bbb C$
are: $X-Y$, $X-\omega Y$ and $X+\omega Y$, whereas the irreducible
components over~$\Bbb Q$ are $X-Y$ and $X^2 + XY + Y^2$ (since 
$t^2+t+1$ cannot be factorised over~$\Bbb Q$).
\par\noindent {\bf (c).} This is irreducible over~$\Bbb C$
(and so irreducible over~$\Bbb Q$) by the same argument as in (a).
Note that this argument applies to any curve of the
form $Y^2 - Q(X) = 0$, where $Q(X)$ is not the square of
a polynomial.
\medskip
\noindent
{\bf 7.}
\par\noindent {\bf (a).} Trying
simple substitutions
of the form where $X,Y$ are replaced by $aX+bY$ and $cX+dY$ in
$2X^2 - Y^2 = 1$
gives: $(2 a^2 - c^2 )X^2 + (2 b^2 - d^2)Y^2 + (4ab - 2cd) XY = 1$;
we can now notice that the coefficients on the LHS can be
made to be $1,1,-6$ by taking (for example) $a=1, b=-1, c=1, d=1$.
Be careful to note that $(X,Y) \mapsto (X-Y, X+Y)$ is then
a birational transformation from ${\cal D}: X^2 + Y^2 - 6XY = 1$
to ${\cal C}: 2X^2 - Y^2 = 1$ [since $X^2 + Y^2 - 6XY = 1$
can be rewritten as $2(X-Y)^2 - (X+Y)^2 = 1$, and so
a point $(X,Y)$ on $\cal D$ maps to a point $(X-Y, X+Y)$
in $\c$].
The inverse map from $\c$ to $\cal D$ is $(X,Y) \mapsto
\bigl( (X+Y)/2, (Y-X)/2 \bigr)$. 
\par\noindent {\bf (b).} Let
$\c : Y^2=(X+2)^6(X^3+1)$ and ${\cal D}: Y^2 = X^3 + 1$. Then
$\c$ can be rewritten as $ \bigl( Y/(X+2)^3 \bigr)^2 = X^3 + 1$
and so there is a birational map $(X,Y) \mapsto \bigl( X, Y/(X+2)^3
\bigr)$
from $\c$ to $\cal D$. This is a birational transformation,
since there is the inverse map $(X,Y) \mapsto \bigl( X, Y (X+2)^3\bigr)$.
\par\noindent {\bf (c).} The map $(X,Y) \mapsto (\sqrt{2}\, X , Y)$
give a birational transformation over $\Bbb C$ from $Y^2=2X^2$
to~$Y^2=X^2$, with inverse map $(X,Y) \mapsto
(X/\sqrt{2}, Y)$. There is no birational transformation over~$\Bbb Q$,
since $Y^2 = X^2$ has infinitely many $\bbQ$-rational points of the form
$(a/b, \pm a/b)$, where $a,b \in \bbZ$, whereas $Y^2 = 2X^2$
has $(0,0)$ as its only $\bbQ$-rational point
(since $\sqrt{2}$ is irrational).
\medskip
\noindent {\bf 8.}
\par\noindent {\bf (a).} The discriminant of~$X^4-2$ is the same
as $\hbox{Res}(X^4-2, 4X^3)$, which is the determinant of
the matrix:
$$ \pmatrix{
0&0&1&0&0&0&-2\cr
0&1&0&0&0&-2&0\cr
1&0&0&0&-2&0&0\cr
0&0&0&4&0&0&0\cr
0&0&4&0&0&0&0\cr
0&4&0&0&0&0&0\cr
4&0&0&0&0&0&0\cr
}.
$$
\noindent The determinant is $\hbox{Disc}(X^4-2) = 2^{11}$.
\par\noindent {\bf (b).} The resultant of $X^3 - a$ and $X^2 - b$
is the determinant of the matrix:
$$ \pmatrix{
0&1&0&0&-a\cr
1&0&0&-a&0\cr
0&0&1&0&-b\cr
0&1&0&-b&0\cr
1&0&-b&0&0\cr
}.
$$
\noindent The determinant is $\hbox{Res}(X^3-a,X^2-b) = a^2 - b^3$.
\medskip\noindent {\bf 9.} First note that, if there is
an intersection point of $X^3 + Y^3 = Z^3$ and $X^2 + Y^2 = Z^2$
with $Z=0$, then $X^3 = -Y^3$ and $X^2 = -Y^2$; squaring
both sides of the first equation, and cubing both sides of
the second equation one deduces: $X^6 = Y^6$ and $X^6 = -Y^6$;
adding these gives $2X^6 = 0$, and so $X=0$, and so $Y=0$.
But the point $(0,0,0)$ is not allowed in projective
coordinates, so we conclude that there are no points of
intersection with $Z=0$. We can therefore consider only
points with $Z\not= 0$.
\par
Given that $Z\not= 0$, we can therefore use the affine
models $x^3 + y^3 = 1$ and $x^2 + y^2 = 1$.
\par\noindent [Each point $(x,y)$ 
of intersection on $x^3 + y^3 = 1$ and $x^2 + y^2 = 1$ will give
the point $(x,y,1)$ of intersection on
$X^3 + Y^3 = Z^3$ and $X^2 + Y^2 = Z^2$; conversely, each
point $(X,Y,Z)$ of intersection on
$X^3 + Y^3 = Z^3$ and $X^2 + Y^2 = Z^2$ with $Z\not= 0$ (which
we have established must be the case)
will give the point $(X/Z, Y/Z)$ of intersection on
$x^3 + y^3 = 1$ and $x^2 + y^2 = 1$].
\par
Let $(x,y)$ be a point of intersection on $x^3 + y^3 = 1$
and $x^2 + y^2 = 1$. Then $x^6 = (1-y^3)^2$ and
$x^6 = (1-y^2)^3$, and so $y$ must satisfy:
$(1-y^3)^2 = (1-y^2)^3$, which can be rewritten as:
\medskip
\hskip 5truecm $y^2(y-1)^2(2y^2 + 4y + 3) = 0$ \hfill $(*)$
\medskip
\noindent [Note that $(*)$ can also be derived as
$\hbox{Res}( x^2 + y^2 - 1, x^3 + y^3 - 1)$ with respect
to the variable~$x$].
\par
The solutions in~$y$
are: $0,1,-1+\sqrt{-1/2},-1-\sqrt{-1/2}$. When $y=0$, then $x$
simultaneously satisfies
$x^3 = 1 - 0^3$ and $x^2  = 1- 0^2$ so $x=1$, giving rise to the
point of intersection: $P_1 = (1,0)$. When $y=1$, then $x$
simultaneously satisfies
$x^3  = 1- 1^3$ and $x^2 = 1 - 1^2$ so $x=0$, giving rise to the 
point of intersection: $P_2 = (0,1)$. 
When $y = -1+\sqrt{-1/2}$, then $x$
simultaneously satisfies
$x^3 = 1 - (-1+\sqrt{-1/2})^3$ and
$x^2 = 1 - (-1+\sqrt{-1/2})^2$, and so
$x = x^3/x^2 = \bigl( 1 - (-1+\sqrt{-1/2})^3 \bigr) /
\bigl( 1 - (-1+\sqrt{-1/2})^2 \bigr) = -1-\sqrt{-1/2}$;
this gives rise to the point of intersection:
$P_3 = (-1-\sqrt{-1/2},-1+\sqrt{-1/2})$.
Similarly, the final possibility for~$y$,
namely $y=-1-\sqrt{-1/2}$ gives rise
to the point $P_4 = (-1+\sqrt{-1/2},-1-\sqrt{-1/2})$.
We have shown that $P_1,P_2,P_3,P_4$ are the only possible
points of intersection, and substitution into
$x^3 + y^3 = 1$
and $x^2 + y^2 = 1$ shows that they are all indeed points
of intersection, so we have a complete list. B\'ezout's
Theorem tells us that there should be 6 points of intersection,
so some of the multiplicities must be greater than 1.
At $P_1 = (1,0)$, we see that $\hbox{d}x/\hbox{d}y = 0$
on both curves, but that $\hbox{d}^2x/\hbox{d}y^2$ differ
(since $\hbox{d}^2x/\hbox{d}y^2 = 0$ on $x^3 + y^3 = 1$
but is nonzero on $x^2 + y^2 = 1$). We conclude that $P_1 = (1,0)$
is a point of intersection of multiplicity exactly~2.
Similarly (but using $\hbox{d}y/\hbox{d}x$ and
$\hbox{d}^2y/\hbox{d}x^2$), we see (by symmetry) that
$P_2 = (0,1)$
is a point of intersection of multiplicity exactly~2.
We can now apply B\'ezout's Theorem to see that $P_3,P_4$
must each be points of intersection of multiplicity only~1
(since the total of all multiplicities must be~6);
or one can check directly that
the value of $\hbox{d}y/\hbox{d}x$ 
on $x^3 + y^3 = 1$ at $P_3$ is distinct from
the value of $\hbox{d}y/\hbox{d}x$ 
on $x^2 + y^2 = 1$ at $P_3$ (and similarly for $P_4$).
\par
On the original projective curves,
$X^3 + Y^3 = Z^3$ and $X^2 + Y^2 = Z^2$, the six points of
intersection are therefore:
\par\noindent \hskip 4cm $(1,0,1)$, with multiplicity 2.
\par\noindent \hskip 4cm $(0,1,1)$, with multiplicity 2.
\par\noindent \hskip 4cm $(-1-\sqrt{-1/2},-1+\sqrt{-1/2},1)$,
with multiplicity 1.
\par\noindent \hskip 4cm $(-1+\sqrt{-1/2},-1-\sqrt{-1/2},1)$,
with multiplicity 1.
\medskip
\noindent {\bf 10.}
{\bf (a).} From the rule that $2$ is a quadratic residue
residue mod~$p$ ($p\not=2$) iff $p\equiv \pm 1$ modulo~$8$, we have
that $2$ is a quadratic residue modulo~$1009$
(since 1009 is prime and $1009 \equiv 1$
modulo~$8$).
\par By quadratic reciprocity, since $3$ and $1009$ are both
odd primes, and since at least one of them is congruent
to~$1$ modulo~$4$, we have $({3\over 1009}) = ({1009\over 3})
= ({1\over 3}) = 1$, since $1\equiv 1^2$ mod~$5$.
Hence, $3$ is a quadratic residue modulo~$1009$.
\par Again applying quadratic reciprocity,
$({5\over 1009}) = ({1009\over 5})
= ({4\over 5}) = 1$, since $4\equiv 2^2$ mod~$5$.
Hence, $5$ is a quadratic residue modulo~$1009$.
\par 
$({10\over 1009}) = ({2\over 1009}) ({5\over 1009}) = 1\cdot 1 = 1$.
Hence, $10$ is a quadratic residue modulo~$1009$.
\par
$({15\over 1009}) = ({3\over 1009}) ({5\over 1009}) = 1\cdot 1 = 1$. 
Hence, $15$ is a quadratic residue modulo~$1009$.
\par\noindent {\bf (b).} When $p\equiv 1$ mod~$4$, we have
by quadratic reciprocity that
$({3\over p}) = ({p\over 3})$, which is~1~ when $p\equiv 1$ mod~$3$,
and is $-1$ when $p\equiv 2$ mod~$3$. When $p\equiv 3$ mod~$4$,
we have
$({3\over p}) = -({p\over 3})$, which is~-1~ when $p\equiv 1$ mod~$3$, 
and is $1$ when $p\equiv 2$ mod~$3$. In summary: 
\par\noindent $({3\over p}) = 1$
when $p$ satisfies both $p\equiv 1$ mod~$4$ and $p\equiv 1$ mod~$3$,
that is, when $p\equiv 1$ mod~$12$.
\par\noindent $({3\over p}) = -1$
when $p$ satisfies both $p\equiv 1$ mod~$4$ and $p\equiv 2$ mod~$3$,
that is, when $p\equiv 5$ mod~$12$.
\par\noindent $({3\over p}) = -1$
when $p$ satisfies both $p\equiv 3$ mod~$4$ and $p\equiv 1$ mod~$3$,
that is, when $p\equiv 7$ mod~$12$.
\par\noindent $({3\over p}) = 1$
when $p$ satisfies both $p\equiv 3$ mod~$4$ and $p\equiv 2$ mod~$3$,
that is, when $p\equiv 11$ mod~$12$.
\par
Hence, $3$ is a quadratic residue mod~$p$ when $p\equiv 1$ or $11$
mod~$12$, and $3$ is not a quadratic residue mod~$p$ when $p\equiv 5$ or $7$
mod~$12$. This describes what happens for all primes apart from
$p = 2,3$ (since any prime $p \not= 2,3$ must be congruent to one
of $1,5,7,11$ mod~$12$). Finally, when $p=2$ or~$3$ we have that $3$
is a quadratic residue mod~$p$.
\medskip
It is always true by quadratic reciprocity that, for any prime $p \not= 2$,
$({5\over p}) = ({p\over 5})$, since $5\equiv 1$ mod~$4$; that
is $5$ is a quadratic residue mod~$p$ iff $p$ is a quadratic residue mod~$5$.
But then $({p\over 5}) = 1$ when $p\equiv 1,4$ mod~$5$ (since $1\equiv 1^2$
and $4=2^2$ mod~$5$) and $({p\over 5}) = -1$ when $p\equiv 2,3$
mod~$5$ (since none of $0^2, 1^2, 2^2, 3^2, 4^2$ are congruent to
either of $2$ or $3$ mod~$5$). In summary, $5$ is a quadratic
residue mod~$p$ when $p\equiv 1,4$ mod~$5$, and $5$ is not a quadratic
residue mod~$p$ when $p\equiv 2,3$ mod~$5$. This
describes what happens for all primes apart from 
$p = 2,5$ (since any prime $p \not= 5$ must be congruent to one  
of $1,2,3,4$ mod~$5$). Finally, when $p=2$ or~$5$ we have that $5$
is a quadratic residue mod~$p$ (since it is congruent
mod~$p$ to $1^2$ and $0^2$,
respectively).
\medskip
We know $({10 \over p}) = ({2 \over p})({5 \over p})$. For $p\not= 2,5$
we can use the standard
rules that $({2 \over p}) = 1$ when $p \equiv 1,7$ mod~$8$,
that $({2 \over p}) = -1$ when $p \equiv 3,5$ mod~$8$;
also, $({5 \over p}) = 1$ when $p \equiv 1,4$ mod~$5$, and
$({5 \over p}) = -1$ when $p \equiv 2,3$ mod~$5$ (done above).
Note that $({10 \over p}) = 1$ when $({2 \over p})=({5 \over p})=1$
or $({2 \over p})=({5 \over p})=-1$ and $({10 \over p}) = -1$
when $({2 \over p})$ and $({5 \over p})$ have opposite sign.
\par
In a similar manner to the first example, we therefore have
that (for any $p\not= 2,5$), $({10 \over p}) = 1$ when $p \equiv 
1,3,9,13,27,31,37,39$ mod~$40$, and $({10 \over p}) = -1$ when $p \equiv  
7,11,17,19,21,23,29,33$ mod~$40$. This  
describes what happens for all primes apart from  
$p = 2,5$ (since any prime $p \not= 5$ must be congruent to one   
of $1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39$ mod~$40$).
Finally, when $p=2$ or~$5$ we have that $10$ 
is a quadratic residue mod~$p$ (since it is congruent
mod~$p$ to $0^2$).
\medskip
\noindent {\bf 11.} Suppose there were integers $a,b,c$, not all~$0$
such that $2a^2 + 5b^2 = c^2$. Without loss of generality,
we can divide through by the square of any common factor of $a,b,c$, and
assume that $\hbox{gcd}(a,b,c) = 1$. Reducing the equation
modulo~$5$ gives: $2a^2 \equiv c^2$ (mod~$5$).
\par Imagine that
$5$ does not divide~$a$; then there would exist~$\lambda \in \Bbb Z$
such that $\lambda a \equiv 1$ (mod~$5$); multiplying both
sides by~$\lambda^2$ would then give $2 \equiv (\lambda c)^2$ (mod~$5$);
but this is a contradiction, since $2$ is not a quadratic residue
mod~$5$ [to see this, check that: $0^2 \equiv 0,\ 1^2 \equiv 1,\
2^2 \equiv 4,\ 3^2 \equiv 4,\ 4^2 \equiv 1$ (mod~$5$)].
This contradiction shows that our assumption `$5$ does not divide~$a$'
must be false, and so we conclude that: $5|a$.
\par Since $5|a$ and $2a^2 + 5b^2 = c^2$, we can deduce
that $5|c^2$ and so $5|c$. Therefore $5^2|a^2$ and $5^2|c^2$
and so $5| 5b^2$; but only $5^1$ divides into $5$,
and so $5| b^2$, giving $5|b$. We now have that
$5$ divides all of $a,b,c$, contradicting the fact that
$\hbox{gcd}(a,b,c) = 1$. This proves that no such $a,b,c$ exist.
\medskip
\noindent {\bf 12.} Define $S(n) = \{ x \in {\Bbb N} :
1 \leqslant x \leqslant n \hbox{ and gcd}(x,n) = 1\}$,
so that $\phi(n) = \# S(n)$.
\par
For any~$p$, to get $S(p^r)$ we remove all multiples of~$p$
from the set $\{ 1,\ldots , p^r \}$, and so
$\phi (p^r) = \# S(p^r)$ is $p^r - {1\over p}p^r$; that is: $p^r - p^{r-1}$.
\par
For two
distinct primes $p_1,p_2$, note that $S(p_1p_2)$ can be obtained
by removing from the set $\{ 1,\ldots p_1p_2 \}$ all multiples
of $p_1$ (of which there are ${1\over p_1}p_1p_2 = p_2$)
and all multiples 
of $p_2$ (of which there are ${1\over p_2}p_1p_2 = p_1$). But
note that the element $p_1p_2$, which is a multiple of both $p_1$ and
$p_2$, should only be counted once. Hence the total number of
elements removed is: $p_2 + p_1 - 1$, so
$\phi (p_1p_2) = p_1p_2 - (p_2 + p_1 - 1) = (p_1-1)(p_2-1)$.
\par Applying Euler's Theorem, that $a^{\phi(n)} \equiv 1$~(mod~$n$)
when $\hbox{gcd}(a,n) = 1$, gives as follows.
\par
$2^{12} \equiv 2^{\phi(13)} \equiv 1$ (mod~$13$), since 
$\hbox{gcd}(2,13) = 1$.
$3^{12} \equiv 3^{\phi(13)} \equiv 1$ (mod~$13$), since
$\hbox{gcd}(3,13) = 1$.
Hence, $3^{24} \equiv (3^{12})^2
\equiv 1^2 \equiv 1$ (mod~$13$).
Similarly, $3^{12000} \equiv (3^{12})^{1000}
\equiv 1^{1000} \equiv 1$ (mod~$13$),
and $3^{12002} \equiv (3^{12})^{1000}\cdot 3^2
\equiv 1^{1000}\cdot 9 \equiv 9$ (mod~$13$).
\par
Since $(4,35)=1$ and $\phi(35) = \phi(5)\phi(7) = 4\cdot 6 = 24$,
we have $4^{24} \equiv 1$ (mod~$35$). So,
$4^{48} \equiv (4^{24})^2 \equiv 1^2 \equiv 1$ (mod~$35$), also.
Similarly, $4^{48000001} \equiv 4^{48000000}\cdot 4
\equiv (4^{24})^{2000000}\cdot 4 \equiv 1^{2000000}\cdot 4 \equiv 4$
(mod~$35$).
\par
We cannot do the same thing for $7^r$, since $(7,35) \not= 1$.
Note that $7^5 \equiv 7$ (mod~$35$) [this is due to the
fact that $7^5 \equiv 7$ modulo each of $5$ and $7$],
and so $7^{24} \equiv (7^5)^4\cdot 7^4 \equiv 7^4\cdot 7^4
\equiv 7^5\cdot 7^3 \equiv 7\cdot 7^3 \equiv 21$ (mod~$35$).
Similarly,
$7^{48}  \equiv (7^5)^9\cdot 7^3 \equiv 7^9\cdot 7^3
\equiv (7^5)^2 \cdot 7^2 \equiv 7^2 \cdot 7^2 \equiv 21$ (mod~$35$).
Similarly,
$7^{48000000} \equiv 21$ (mod~$35$), so that
$7^{48000001} \equiv 21\cdot 7 \equiv 7$ (mod~$35$).
\vfil \eject \end
