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\begin{document}
\begin{center}
{\large Rational Points of Order Four on Elliptic Curves }\\
{\bf Leighton Cowart, adviser Kevin Iga}\\
{\footnotesize November 2002}
\end{center}
Recall that an elliptic curve $C$ is a set of points $(x,y)$ that satisfy the relation
\[y^{2} = f(x) = x^{3} + ax^{2} + bx + c\]where $a$, $b$, and $c$ are integers and the complex roots of $f(x)$ are distinct. In this work we use $\rr$ to refer to the real numbers, $\qq$ to refer to the rationals, and $\zz$ to refer to the integers. We also use a superscript $+$ to denote the positive elements of these sets ($\qq^{+}$ refers to the positive rationals, for example). The primary results of this paper follow.\\
\noindent
{\bf The following are necessary conditions for rational points of order four to occur on $C$:}\\
\noindent
1. The root $(\alpha, 0)$ of the component of $C$ containing infinity must be rational.\\
\noindent
2. There exists $r \in \qq^{+}$ such that $r = \sqrt{f'(\alpha)}$.\\
\noindent
3. At least one of the following holds:
(i) There exists $p \in \qq^{+}$ such that $3a + \alpha + 2r = p^2$.
(ii) There exists $q \in \qq^{+}$ such that $3a + \alpha - 2r = q^2.$\\
\noindent
{\bf Suppose the three conditions stated above hold. Then $(r - \alpha, \pm rp)$ are rational points of order 4 on $C$ if 3(i) holds, and $(-r - \alpha, \pm rq)$ are rational points of order 4 on $C$ if 3(ii) holds.}
\section{$C$ as a Group and Points of Finite Order}
An addition operation may be defined on $C$ in such a way that the operation on $C$ (together with a point at infinity) forms a group. We will denote the elliptic curve addition operation with the symbol $\bigoplus$. Treating $C$ under elliptic curve addition as a group, we note that the point at infinity satisfies the requirements for the group identity, and we call it $O$ (or sometimes the {\bf origin}). Later in this work, if we speak of $C$ as a set, we refer to the points $(x,y) \in C$ together with the point at infinity; if we refer to $C$ as a group, we mean the set $C$ together with the point at infinity all under the operation $\bigoplus$. \\
For some point $P \in C$, consider $P \bigoplus P$. We call this result $2P$, and refer to finding $2P$ as {\bf duplication}. We also give $2P \bigoplus P$ the name $3P$. Note that since $\bigoplus$ is associative, $2P \bigoplus P = P \bigoplus 2P$ so that it doesn't matter what order we take these terms in. More generally, we recursively define $nP$ for $n \in \zz^{+}$ to be $((n-1)P) \bigoplus P$.\\
We deal with the duplication of $P$ often enough that it is useful to have an explicit formula for $2P$. We therefore cite the following. With $P = (x,y)$ and $2P = (x_{0},y_{0})$: \[x_{0} = \frac{x^{4} - 2bx^{2} -8cx + b^{2} - 4ac}{4y^{2}}\]
\[y_{0} = \frac{(3x^{2} + 2ax + b)x_{0}}{2y} - \frac{3x^{3} + 2ax^{2} + bx}{2y} + y \] The former is derived in I.4 of [1], and the latter follows from the definition of $\bigoplus$ with a little algebra.
We say that a point $P = (x,y) \in C$ is of {\bf finite order} if there exists a $k \in \zz^{+}$ such that $kP = O$, and we further say that $P$ {\bf has order $k$} or {\bf is of order $k$} if $k$ is the smallest positive integer satisfying $kP = O$.\\
\section{Real and Rational Points}
As might be expected, a point $(x, y) \in C$ is called {\bf real} if both $x$ and $y$ are real, and {\bf rational} if both $x$ and $y$ are rational. An interesting thing to investigate is what the set of real points on $C$ looks like. It turns out that the set of real points on the component of $C$ containing infinity is isomorphic to the circle group $S^{1}$ (i.e., the set of complex numbers of magnitude 1 under multiplication). The group of real points on $C$ under addition turns out to be isomorphic to $S^{1}$ if $C$ is connected, and isomorphic to $S^{1} \times \zz_{2}$ if $C$ is disconnected. This is explained on p. 42 of [1].\\
We now turn to real points of finite order. Knowing what we do about the group structure of the real points on $C$, we may say that the points on $C$ of order dividing $k$ are isomorphic to $\zz_{k}$ if $C$ is connected. If $C$ is disconnected, the real points of order dividing $k$ are isomorphic to $\zz_{k}$ if $k$ is even and isomorphic to $\zz_{k} \times \zz_{2}$ if $k$ is odd. (The real points of order $k$ correspond to the $k$th roots of unity.) Naturally, the rational points on $C$ of order dividing $k$ form a subgroup of the real points of order $k$.\\
What happens with a specific value of $k$, for instance $k = 2$? To find points $P$ of order 2, we want to have $2P = O$. In other words, the line tangent to $P$ intersects $C$ at infinity. It is geometrically evident that this line is vertical, and that a vertical tangent happens only at the zeroes of $C$. But $(x,y) \in C$ is a zero if and only if $x$ is a zero of $f(x)$. Thus, the real points of order two are of the form $(\gamma, 0)$ where the $\gamma$ are the roots of $f(x)$. The rational points of order 2 are easy to find: they correspond to the rational roots of $f(x)$.\\
We could go through a similar process to investigate points of order 3, specifically points $P$ such that $2P = -P$, and in fact Silverman and Tate do so in [1]. But here we are more interested in points of order 4.\\
A point $P$ of order 4 has the property that $4P = 2(2P) = O$, or equivalently that $2P$ is a point of order 2. We want to find out if there are rational points of order 4 on $C$. Since the real points of order 4 on $C$ form a group, we know that there exist real points of order 4: it is just a question of seeing how many (if any) of the real points are rational.
\begin{theorem}The following are necessary conditions for rational points of order 4 to occur on an elliptic curve $C$:\\
\noindent
1. The root $(\alpha, 0)$ of the component of $C$ containing infinity must be rational.\\
\noindent
2. There exists $r \in \qq^{+}$ such that $r = \sqrt{f'(\alpha)}$.\\
\noindent
3. At least one of the following holds:
(i) There exists $p \in \qq^{+}$ such that $3a + \alpha + 2r = p^2$.
(ii) There exists $q \in \qq^{+}$ such that $3a + \alpha - 2r = q^2.$
\end{theorem}
\section{Criterion 1: Rational root of the component containing infinity}
It is evident from the duplication formulae that if a point $P$ is rational, $2P$ is rational. Therefore if we want rational points of order 4, there had better be a rational point of order 2, which corresponds to a rational root $\alpha$ of $f(x)$.
But not just any rational root will do: we show that it must be the component of $C$ containing infinity. \\
This is obvious when $C$ is connected and there is only one real root. When $f(x)$ has three real roots,the points of order dividing four form a group isomorphic to $G = \zz_{4} \times \zz_{2}$. Consider the points of order 4 in $G$: $(1,0)$, $(3,0)$, $(1,1)$, and $(3,1)$. Each of these points added to itself (by the usual addition operation in $G$) is $(2,0)$. Because of the isomorphism we may say that all real points of order 4 on $C$ map to a single point of order 2. Call this point $(\alpha, 0)$. If there are to be any rational points of order 4, $\alpha$ must be rational. \\
A geometric argument shows that $(\alpha, 0)$ must be the root of the component of $C$ containing infinity. Consider a nonvertical line intersecting a root $(\beta, 0)$ of the component not containing infinity. This line will hit the component not containing infinity in exactly one other place, thus hitting the component with infinity in exactly one place---so for all real points $P'$ on $C$, $(\beta, 0) \not= 2P'$. Thus we have shown the necessity of our first criterion.
\section{Criterion 2: Rational Square Root of $f'(\alpha)$}
Knowing from our prior discussion that $(\alpha, 0)$ corresponds to a rational root $\alpha$ of $f(x)$, we now translate the curve to make calculations easier. Let $C'$ consist of points $(x,y)$ that satisfy $Y^2 = F(x) = f(x + \alpha)$; simplification shows this to be equal to $x^{3} + Ax^{2} + Bx$, where $A = 3a + \alpha$ and $B = 3\alpha^{2} + 2a\alpha + b$. Note that $B = f'(\alpha).$ \\
The primary value of this exercise has been to relocate the important point $(\alpha, 0)$ so that for points $P$ of order 4, $2P = (0,0)$. (Recall that there is a one-to-one correspondence between rational points on $C$ and $C'$.) Given this, we can use the duplication formula with $x_{0} = 0$ to solve for the $x$-coordinates of the points of order 4. \\
Recall the duplication formula for $x$-coordinates: \[x_{0} = \frac{x^{4} - 2bx^{2} - 8cx + b^{2} - 4ac)}{4y^{2}}\] With $x_{0} = 0$ and $c = 0$, we get $0 = x^{4} - 2Bx^{2} + B^{2} = (x^{2} - B)^{2}$; thus $x = \pm \sqrt{B}$. Thus another necessary condition for rational points of order 4 is that $B$ be a perfect square in $\qq$. Recall that $B = f'(\alpha)$; so $f'(\alpha)$ must be a perfect square.\\
%Observe further that $B \not= 0$, otherwise $F(x) = x^{3} + Ax^{2} = (x^{2})(x + A)$ would have nondistinct roots---a contradiction since we assumed that $C$ is an elliptic curve. (Remember that an elliptic curve translated along the x-axis is also an elliptic curve.) This is omitted from our list of requirements for rational points of order four to appear, since it is guaranteed by a prior assumption, but we will use this fact later.
\section{Criterion 3: More conditions on $A$ and $B$}
Given the $x$-coordinates of our points $P$, there are two ways to find the $y$-coordinates. We let $Y^2 = F(\pm \sqrt{B})$ and solve for $Y$, leaving it to the reader to show that substituting $\sqrt{B}$ and $-\sqrt{B}$ for $x$ in the duplication formula for $y$-coordinates gives the same answers for y.\\
Case 1: $x = \sqrt{B}$.
Then $Y^{2} = (\sqrt{B})^{3} + A(\sqrt{B})^2 + B(\sqrt{B}) = B(A + 2\sqrt{B})$, and \par $Y = \pm\sqrt{B}\sqrt{A + 2\sqrt{B}}$.\\
Case 2: $x = -\sqrt{B}$.
Then $Y^{2} = (-\sqrt{B})^{3} + A(-\sqrt{B})^2 + B(-\sqrt{B}) = B(A - 2\sqrt{B})$, and \par $Y = \pm\sqrt{B}\sqrt{A - 2\sqrt{B}}$.\\
We have therefore found four complex points of order 4 which always occur when criteria 1 and 2 are satisfied:
\par
$(\sqrt{B}, \sqrt{B}\sqrt{A + 2\sqrt{B}})$
\par
$(\sqrt{B}, -\sqrt{B}\sqrt{A + 2\sqrt{B}})$
\par
$(-\sqrt{B}, \sqrt{B}\sqrt{A - 2\sqrt{B}})$
\par
$(-\sqrt{B}, -\sqrt{B}\sqrt{A - 2\sqrt{B}})$\\
We now examine under what conditions these points are both real and rational. Observe that the first two are always real, and that the second two will be real when $A > 2\sqrt{B}$. \\
Under what conditions are these points rational? Assuming that all prior necessary conditions are satisfied (in particular, $\sqrt{B}$ is rational), we conclude that the first two points will be rational iff $A + 2\sqrt{B}$ is a perfect square in $\qq$, and the second two will be rational iff $A - 2\sqrt{B}$ is a perfect square in $\qq$. Thus if neither $A + 2\sqrt{B}$ nor $A - 2\sqrt{B}$ is a perfect square (i.e., neither 1.3(i) nor 1.3(ii) holds), there will be no rational points of order 4. This concludes our list of necessary conditions. $\Box$
\section{Sufficient Conditions \& Explicit Description}
\begin{theorem}
Suppose the three conditions stated above hold. Then \break $(r - \alpha, \pm rp)$ are rational points of order 4 on $C$ if 1.3(i) holds, and \break $(-r - \alpha, \pm rq)$ are rational points of order 4 on $C$ if 1.3(ii) holds.
\end{theorem}
Suppose 1.1, 1.2, and 1.3(i) hold---then $r = \sqrt{B}$ is rational and $A + 2r = p^2$ for $p \in \qq$, so that $(\sqrt{B}, \pm \sqrt{B}\sqrt{A + 2r}) = (r, \pm rp)$ are rational points. The calculations above demonstrate that they are in fact points of order four.\\
Suppose 1.1, 1.2, and 1.3(ii) hold---then $r = \sqrt{B}$ is rational and \break $A - 2r = q^2$ for $q \in \qq$, so that $(-\sqrt{B}, \pm \sqrt{B}\sqrt{A - 2r}) = (-r, \pm rq)$ are rational points. As before, the calculations above demonstrate that they are indeed points of order four.\\
Therefore we have shown that the conditions listed in Theorem 1 are not only necessary but sufficient for rational points of order 4 to occur on $C$. However, more than that, we have provided their locations explicitly---on $C'$, our shifted curve. Recall that on $C'$, $Y^2 = F(x) = f(x + \alpha)$, so on $C$, $y^2 = f(x) = F(x - \alpha)$. We therefore shift the points we've found by $-\alpha$ along the $x$-axis, which gives us an explicit description of our points on $C$. $\Box$
\section{Examples}
The following examples may prove instructive; they illustrate the necessity of each of the conditions listed above. Verification is left to the reader.\\
$y^{2} = x^{3} - 2$ has no rational points of order four. Its real root is irrational.\\
Use Eisenstein's criterion and the discriminant to generate a polynomial with three irrational real roots\\
$y^{2} = x^{3} + x^{2} - 2x - 2 = (x + 1)(x^{2} - 2)$ has no rational points of order four. It has a rational root at $x = -1$, but not in the component containing infinity.\\
$y^{2} = x^{3} + 3x^{2} + 2{x} = x(x + 1)(x + 2)$ has no rational points of order four. All its roots are rational, but $\sqrt{f'(\alpha)} = \sqrt{f'(0)} = \sqrt{2} \not\in \qq$.\\
$y^{2} = x^{3} + 5x^{2} + x$ has no rational points of order four. $\alpha = 0$ and $\sqrt{f'(\alpha)} = 1$, but $\sqrt{A - 2\sqrt{B}} = \sqrt{3} \not\in \qq$ and $\sqrt{A + 2\sqrt{B}} = \sqrt{7} \not\in \qq$.\\
$y^{2} = x^{3} + 3x^{2} + x$ has two rational points of order four, $(-1, -1)$ and $(-1, 1)$. $\alpha = 0$, $\sqrt{f'(\alpha)} = 1$, and $\sqrt{A - 2\sqrt{B}} = 1 \in \qq$, though \break $\sqrt{A + 2\sqrt{B}} = \sqrt{5} \not\in \qq$.\\
$y^{2} = x^{3} + 7x^{2} + x$ has two rational points of order four, $(1, -1)$ and $(1, 1)$. $\alpha = 0$, $\sqrt{f'(\alpha)} = 1$, and $\sqrt{A + 2\sqrt{B}} = 3 \in \qq$, though $\sqrt{A - 2\sqrt{B}} = \sqrt{5} \not\in \qq$.\\
$y^{2} = x^{3} + 5x^{2} + 4x$ has four rational points of order four, $(-1, -1)$, $(-1, 1)$, $(1, -1)$, and $(1, 1)$. $\alpha = 0$, $\sqrt{f'(\alpha)} = 2$, $\sqrt{A + 2\sqrt{B}} = 3 \in \qq$, and $\sqrt{A - 2\sqrt{B}} = 1 \in \qq$.\\
\section{Bibliography}
[1] Joseph Silverman \& John Tate, Rational Points on Elliptic Curves, Springer-Verlag 1992\\
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