# Pell's equation and Rational points on elliptic curve

## Abstract

In the quest of solving a problem of finding natural numbers which are simultaneously triangular and square, we landed in the field of Pell's Equations and its different types. Closer investigations lead to interesting observation regarding its nature and solutions. Elementary modifications in coefficients resulted in a few more observations. Then, we looked into Negative Pell's Equations. At one point, from finding the integral solution for a generalized Diophantine biquadratic equation, we shifted our interest to rational solutions for such equations. Thus, we ended up in cultivating rational points on unit circle, which eventually was reduced to geometry of cubic curves. We introduced the idea of projective geometry, with constant comparison between points on Euclidean (affine) plane and projective plane. We tried constructing a group out of the rational points on the cubic curve, defining composition as group operation. It is assigned the nature of an additive group. Next, we tried addressing Mordell’s theorem (naively). For this, we reduced generalized cubic curves to Weierstrass Normal form. Explicit formulae for group laws were also cultivated with the help of examples of elliptic curves.

**Keywords**: Negative Pell’s equation, projective geometry, Mordell’s theorem, Weierstrass Normal form, elliptic curve

## INTRODUCTION

## Background

Like Diophantine equations, one of its special type, namely Pell's equation, has also been of great interest to mathematicians around the globe since long. It is heard that Euler, after a recursory reading of Wallis's *Opera Mathematica*,* *mistakenly attributed the first serious study of non-trivial solutions of such equations to John Pell. However there is no evidence that Pell, professor of University of Amsterdam, had ever considered solving such equations; while many other noted mathematicians like Fermat, Archimedes, Theon of Smyrna, Diophantus, Brahmagupta, Bhaskaracharya is found to have put immense effort and interest in developing the theory for such equations, for integral solutions of the variables. The standard form of the Pell-equation is $x^2\;-\;dy^2\;=\;1$ where $d$ is a square free natural number. We aim to find the solution for $x$ and $y$ in $\mathbb{Z}$; when it is of the form $x^2\;-\;dy^2\;=\;a$, it is called Pell-type equation.

## Rationale

## Statement of the Problem

We initially started with an elementary problem in number theory which goes like:

The first two numbers that are both triangular and square simultaneously are 1 and 36. Find the next one and, if possible the one after that. Can you figure out an efficient way to find triangular-square numbers? Do you think that there are infinitely many such numbers?

Addressing the problem:

We know, the triangular numbers can be represented as $\frac{n(n+1)}2$ and the square number as $m^2$, for some $n,m\in\mathbb{N}$

According to the condition,

Now, without loss of generality, we may consider $n$ to be odd i.e. $n\;=\;2k-1,\;k\in\mathbb{N}$

Putting in equation Equation 2, we get,

Now, if the product of two distinct coprime positive integers is a perfect square then each of them individually should be perfect squares. Hence, let $\;k=x_1^2$and $(2k-1)\;=\;x_2^2$ we get,

This is of the form $2y^2\;-\;x^2\;=\;1$. Thus, it got reduced to our Pell's Equation.

## Objective

- To study Pellian equations
- To study rational points of known order on non-singular elliptic curve

## PELL'S EQUATIONS

The equation of the form $x^2\;-dy^2\;=\;1\;$ is called a Pell's Equation.

Why does the definition of Pell's equation assume $d$ to be square-free?** **

Suppose, d is a square number, i.e. let $d=c^2$ , $c \in \mathbb{Z}$. Thus the equation can be factorized

as $(x+cy)(x-cy)=1$ but this is impossible for integral x and y, as two integers

multiplied cannot give 1, if not each equal to 1.

Since, we are unable to find a factorization in $\mathbb{Z}$ , we would look for a factor in $\mathbb{Z}[\sqrt{d}]$ where d is non-zero square-free integer.

Our equation can be factorised as $(x+y\sqrt{d})(x-y\sqrt{d})=1$. The set of numbers of the

form $(x+y\sqrt{d})$with $x,y \in \mathbb{Z}$, in denoted as $\mathbb{Z}[\sqrt{d}]$. In fact, $(\mathbb{Z}[\sqrt{d}], +, .)$ forms a ring.

We know, the conjugate the conjugate elements of an algebraic element $\alpha$ , over a field extension $\mathbb{Z}$, are the roots of the minimal polynomial $m_{\mathbb{Z},\alpha}(x)$ of $\alpha$ over $\mathbb{Z} \implies m_{\mathbb{Z},\alpha}(\alpha) =0$. If $z= x+y\sqrt{d}$, then its conjugate is defined and denoted as $\overline z = x- y\sqrt{d}$ , and its norm as $N(z)=z\overline z = x^2 -dy^2 \in \mathbb{Z}$.

The norm and conjugates are multiplicative in $\mathbb{Z}$ , i.e.,

- $N(z_1z_2)=N(z_1)N(z_2)$
- $\overline{z_1z_2}=\overline z_1 \overline z_2$

## Solution to Pell's Equation

Let us denote the Pell's equation as $N(z)=1$ with minimal solution as $(u_0,v_0)$. Minimal solution is trivially (1,0) in such cases.

**Theorem. **

Given d>0, the equation

has infinitely many solutions in positive integers.

The general solution is given by $(u_n,v_n)_{n\geq0}$ ,

where $(u_1,v_1)$ is its least non-trivial solution.

*Proof. *

*Existence of fundamental solution* $(u_1,v_1)$:

Let, $c_1>0, c_1 \in \mathbb{Z}$ , we have to show that $\exists\; t_1,w_1 \geq 1$ such that

Considering, $l_k=\left[k\sqrt {d}+1\right],\;k=0,\dots,c_1$ yields $0\leq l_k-k\sqrt d\leq1,\;k=0,\dots,c_1$ and, since, $\sqrt d$ is an irrational number, ${l}_{{k}^{\text{\'}}}\ne {l}_{{k}^{"}}$ whenever ${k}^{\text{\'}}\ne {k}^{"}$ . As there are $c_1$ intervals in $(\frac{p-1}{c_1}, \frac{p}{c_1}), p=1, ... , c_1, c_1 +1$ numbers of the form $l_k - k\sqrt{d}, k=1, ... , c_1$, by Pigeonhole Principle, there exists $i,j,p \in \{0,1,2,..., c_1\},i \neq j, p \neq 0$ such that

Since the inequality leads to $\left|(l_i-l_j)-(j-i)\sqrt d\right|<\frac1{c_1}$, and putting $\left|(l_i-l_j)\right|=t_1\;\mathrm{and}\;\left|j-i\right|={\mathrm w}_1$ . So, we get the previous expression $|t_1− w_1\sqrt{d}|< \frac{1}{c_1}$ and $w_1 \leq c_1$. Now multiplying this inequality by $|t_1+w_1\sqrt{d}| < 2w_1\sqrt{d}+1$ we get,

Choosing a positive integer $c_2>c_1$ such that $\left|t_1-w_1d\right|>\frac1{c_2}$ , we obtain positive integers $t_2,\;c_2$ with the properties

and

By continuing this procedure, we find a sequence of distinct pairs $(t_n,w_n)_{n\geq1}$ satisfying the inequalities

Thus the interval $(-2d-1,2d+1)$ contains a non-zero integer k such that there exists a subsequence of $(t_n,w_n)_{n\geq1}$ satisfying the equation $t^2-w^2d=k$. This subsequence contains at least two pairs of $(t_p,w_p),\;(t_q,w_q)$ for which $t_p\equiv t_q(mod\left|k\right|)$

$w_p\equiv w_q(mod\left|k\right|)$, and $t_pw_q-t_qw_p\neq0$, otherwise $t_p=t_q$and $w_p=w_q$ , in contradiction with $\left|t_p-t_q\right|+\left|w_p-w_q\right|\neq0$, refer Equation 11.

Let $t_0=t_pt_q-dw_pw_q$ and $w_0=t_pw_q-t_qw_p$ then $t_0^2-w_0^2d=k^2$

On the other hand, as the solutions of Pell's equation Equation 7 are of the form Equation 15, we can approximate the value of $\sqrt{d}$, d being square-free. If $(u_k,v_k)$ is one of the solutions of Equation 7 for some $k \in \mathbb{N}$, then

The pair $(u,v)$ where $u=\frac{t_0}{\left|k\right|}$ and $v=\frac{w_0}{\left|k\right|}$ are non-trivial solution to Pell equation.

Let $(u_1,v_1)$ be the least such solution with u (and implicitly v) minimal. Now we show that the pair $(u_n,v_n)$ defined by Equation 7 satisfies Pell's equation Equation 7 by principle of mathematical induction on n. If $(u_1,v_1)$ is a solution to the equation Equation 7, then $(u_n,v_n)$ is also a solution for $n >1, n\in \mathbb{N}$. We proceed with assumption $(u_n,v_n)$ is a solution, need to show for n+1.

i.e. the pair $(u_{n+1},v_{n+1})$ is also a solution to the Pell's equation.

Thus, for all positive integer n,

Equation 17 also yields the trivial solution $(u_0,v_0)=(1,0)$

Let $z_n=u_{n\;}+v_n\sqrt d=\;(u_1+v_1\sqrt d)^n$ and note that $z_0<z_1<z_2<\dots<z_n<\dots$

We will prove now that all solutions to the are of the form $u_n+v_n\sqrt d=(u_1+v_1\sqrt d)^n$. Indeed, if the had a solution $(u,v)$ such that $u+v\sqrt d$ is not of the form of Equation 15 then $z^m<z<z^{m+1}$ for some integer *m*.

Then $1<\;(u+v\sqrt d)(u_m-v_m\sqrt d)<(u_1+v_1\sqrt d)$ and therefore $1<\;(uu_m-dvv_m)\;+(u_mv-uv_m)\sqrt d<(u_1+v_1\sqrt d)$.

On the other hand, $(uu_m-dvv_m)^2-d(u_mv-uv_m)^2=(u^2-dv^2)(u_m^2-dv_m^2)=1$, i.e., $(uu_m-dvv_{m\;},\;u_mv-uv_m)$is a solution of smaller than $(u_1,v_1)$ in contradiction with the assumption that $(u_1,v_1)$ is the minimal non-trivial solution.(q.e.d)

## Remarks

The Equation 8 and can be expressed in matrix form as the following:

where $\begin{pmatrix}u_n\\v_n\end{pmatrix}=\begin{pmatrix}u_1&dv_1\\v_1&u_1\end{pmatrix}^n\begin{pmatrix}u_0\\v_0\end{pmatrix}$.

If we take $\begin{pmatrix}u_1&dv_1\\v_1&u_1\end{pmatrix}=\begin{pmatrix}a_n&b_n\\c_n&d_n\end{pmatrix}$ then it is understood that ${a}_{n},{b}_{n},{c}_{n},{d}_{n}$ are linear combination of ${\lambda}_{1}^{n},{\lambda}_{2}^{n}$ where ${\lambda}_{1},{\lambda}_{2}$ are the eigenvalues of the matrix $\begin{pmatrix}u_1&dv_1\\v_1&u_1\end{pmatrix}$.

and

On the other hand, as the solutions of Pell's equation Equation 7 are of the form Equation 15, we can approximate the value of $\sqrt{d}$, d being square-free. If $(u_k,v_k)$ is one of the solutions of Equation 7 for some $k \in \mathbb{N}$, then

Thus, it follows that

Thus, we may also write

In other words, the fraction $\frac{u_k}{v_k}$ approximates to $\sqrt d$ with an error term less than $\frac1{v_k^2}$.

Equation of type $a{x}^{2}-b{y}^{2}=1$

Considering a general equation of the form

where $a,b\in \mathrm{\mathbb{Z}}$. If $\u2206=4ab>0$, then we may reduce it to Pellian equation by continued fraction method. Detailed discussion on this method can be found in Andreescu, et al, 2015.

Observation:

*If* $ab=k^2$ , *where k is an integer greater than 1, then the Equation 22 does not have solutions in positive integers*.

Prof. Assume that Equation 22 has a solution (x, y), where x, y are positive integers. Then $ax^2-by^2=1$ , and clearly a and b are relatively prime. From the condition $ab=k^2$ it follows that $a=k_1^2$and $b=k_2^2$ for some positive integer $k_1$and $k_2$. The

relation $k_1^2x^2-k_2^2y^{2\;}\;=\;1$ can be written as $(k_1x+k_2y)(k_1x-k_2y)=1$. This is not possible simultaneously for positive integral x,y,k_{1},k_{2} as atleast one term in the expression should be less than 1.

From this, we may easiely deduce the **Pell's Resolvent*** *for equations of the form of Equation 22, as

Suppose, the Equation 22 has solution in positive integers, then by the afore mentioned procedure, the general solution $(x_n,y_n)_{n\geq0}$ in terms of smallest solution $(x_0,y_0)$ , is given as

and

where $(u_n,v_n)_{n\geq0}$ is the general solution to the Pell's Resolvent.

An easy cross-verification would be the following:

Remarks:

(1) Relation between the fundamental solution $(u_1,v_1)$ of Pell's Resolvent and the smallest solution $(x_0,y_0)$ to Equation 22 , will be

(2) Comparing Equation 20 & 21 and Equation 27 & 28, we get

and

(3) Matrix form of the solution is the following:

(4) The next result is further generalisation:

If $1<a<b$ are integers such that *ab* is square-free, then at most one of the two equations $ax^2-by^2=\pm1$ is solvable. Elaborate discussion can be found in Nagell, et al, 1954.

**Algebraic property of the Solution to Equation** $ax^2-by^2=1$:

Defining $F(x_n,y_n):=(x_0\sqrt a+y_0\sqrt b)^{\;2}=(ax_0^2+by_0^2)\;+\;2x_0y_0\sqrt{ab}$ for a solution $(x_{n}\sqrt a+y_n\sqrt b)$of Equation 22 , then the following holds:

**Properties:**

- $F(-x,-y)=F(x,y)$
- $F(x,-y)=F\overline{(x,y)}$ , the conjugate of
*F(x,y)* - $F((x,y).(r,s))=F(x,y).(r,s)^2$ or $F((x,y).(r,s))=F(x,y).(r+s\sqrt{ab})^2$

## The Negative Pell's Equation

Unlike the equation $x^2-dy^2=1$ which is usually solvable for square-free d, the equation

is solvable only for some particular values of d.

In Olds, et al, 2009, it is explicitly shown that if r is the period of the expansion of $\sqrt d$ in continued fractions, then, if r is even, the Equation 31 has no solution. If r is odd, then $x=h_{nr-1}$and $y=k_{nr-1}$ give all positive solutions to Equation 31 for n=1,3,5,...

Equations of the form Equation 31 are called *N**egative Pell's Equation. *

The solution in general and in matrix representation can be found as before.

## Theorem

Let p be a prime ≥ 3. The Negative Pell’s Equation

is solvable in positive integers if and only if $p\equiv1(mod\;4)$.

* Proof. *Let us suppose that the Equation 32 is solvable. Then there are positive integers $u,v$ such that $u^2-pv^2=-1$i.e., $u^2-(-1)=pv^2$. Reading this equation modulo p, we get $u^2-(-1)\equiv{0 (\mod p)} \implies -1 \equiv{u^2 (\mod p)}$.

Considering $\left(\frac ap\right)$ be the Legendre symbol, by definition we know,

Then, by Quadratic Reciprocity, we have $\left( \frac {-1}{p} \right)=+1$ . Hence, $\left(\frac{-1}p\right)=(-1)\frac{p-1}2 \implies p\equiv1(mod\;4)$.

Let $(u_0,v_0)$ be the fundamental solution to the Pell’s resolvent $u^2-pv^2=1$.

Then $u_0^2-1=pv_0^2$, and $u_0$ cannot be even, for in this case we should have ${-1}\equiv p(mod\;4)$.

Hence $u_0$is odd and the numbers $u_0 -1$and $u_0+1$ have the greatest common divisor 2.

Therefore $u_0\pm1=2 \alpha ^2$and $u_0\mp1=2p\beta^2$, where α and β are positive integers such that $v_0=2\alpha\beta$.

By elimination of $u_0$ we get $\pm1= \alpha^2-p\beta^2$. Since $\beta<v_0$ , we cannot have the upper sign.

Thus, ${-1}=\alpha^2-p\beta^2$ proves the theorem.(q.e.d)

## RATIONAL POINTS ON ELLIPTIC CURVES

The Diophantine quadratic equation

with integral coefficients *a, b, c, d, e, f* reduces in its main case to a Pell-type equation. The general method of reduction is the following:

The Equation 34 represents a conic in the Cartesian plane, therefore solving Equation 34 in integers means finding all lattice points situated on this conic. We will solve the Equation 34 by reducing the general equation of the conic to its canonical form. We introduce the discriminant of the Equation 34 by $\Delta=b^2-4ac$. When $\Delta<0$, the conic defined by Equation 34 is an ellipse and in this case the given equation has only a finite number of solutions. If Δ = 0, then the conic given is a parabola. If $2ae − bd = 0$, then the Equation 34 becomes $(2ax+by+d)^2=d^2-4af$ and it is not difficult to solve. In the case $2ae-bd\neq0$, by performing the substitutions $X= 2ax+by+d$ and $Y=(4ae-2bd)y+4af-d^2$, the Equation 34 reduces to $X^2+Y=0$ which is also easy to solve. The most interesting case is $\Delta>0$, when the conic defined by Equation 34 is a hyperbola. Using a sequence of substitutions, the Equation 34 reduces to general Pell-type Equation:

We say that the conic curve given by Equation 34 is *rational *if the coefficients are rationals instead of integers. Similarly, a point in the X-Y plane is called a *rational point* if both its coordinates $(x,y)$ are rational numbers; a line a *rational line* if the equation of the line can be written with rational numbers; that is, if its equation is $ax+by+c=0$ with *a,b,c* rational.

- Now what about the intersection of a rational line with a rational conic? Will it be true that the points of intersection are rational. Using analytic geometry to find the coordinates of these points, a quadratic equation for the x-coordinate of the intersection can be obtained. In such case, the quadratic equation will have rational coefficients. In other words, the two points of intersection will be rational if and only if the roots of that quadratic equation are rational; in general, they might be quadratic irrationalities.

- Let us suppose that one of the rational point O on our rational conic is known to us, then can we get all of them very simply? More generally, on a given rational conic whether there are any rational points? One solution is just to draw some rational line and project the conic onto the line from this point O. (Projection is by the tangent line to the conic at O.) A line meets a conic in two points, so for every point P on the conic, there will be a point Q on the line, and vice verse. We get a one-to-one correspondence between the points on the conic and points on the line. Now, the rational points on the line are easily described in terms of rational values of some parameter.

Let us exercise it for a unit circle with the equation $x^2+y^2=1$. We choose the point (-1,0) and project it onto the y-axis and let (0,t) be a point on the circle. The equation of the line joining (-1,0) and (0,t) will be $y=t(x+1)$ . So, the point of intersection (x,y) will be given as $1-x^2=y^2=t^2(1+x^2)$.

Solving x and y in terms of t, we get, the familiar rational parametrization of the circle as

We now extend this to cubics.

Let $ax^3+bx^2y+cxy^2+dy^3+ex^2+fxy+gy^2+hx+iy+j=0$ be the equation of a general cubic. We say that a cubic is *rational* if the coefficients of its equation are rational numbers. We cannot use the geometric principle that worked so well for conics because a line generally meets a cubic in three points. And if we have one rational point, we cannot project the cubic onto a line, because each point on the line would then correspond to two points on the curve.

But there is a geometric principle we can use. If we can find two rational points on the curve, then we can generally find a third one. Namely, draw the line connecting the two points you have found. This will be a rational line, and it meets the cubic in one more point. To find three intersections of a rational line with a rational cubic, we find that we come out with a cubic equation with rational coefficients. If two of the roots are rational, the third must also be so.

Geometrically, we can define some composition laws: Starting with two points P and Q, we draw the line through P and Q and let P * Q denote the third point of intersection of the line with the cubic. Even if we only have one rational point P, we can still generally get another. By drawing the tangent line to the cubic at P, we are essentially drawing the line through P and P. The tangent line meets the cubic twice at P, and the same argument will show that the third intersection point is rational. Then we can join these new points up and get more points. So if we start with a few rational points, then by drawing lines, we generally get lots of others.

## Mordell's Theorem

If a non-singular plane cubic curve has a rational point, then the group of rational points is finitely generated.

In order to prove the above theorem, we would require some more insights into the geometry of curves. In general, two cubic curves meet in nine points. Stating Bezout's theorem naively,

- Let C, C
_{1}, and C_{2}be three cubic curves. Suppose С goes through eight of the nine intersection points of C_{1}and C_{2}. Then С goes through the ninth intersection point.

Let F_{1}(x,y) = 0 and F_{2}(x,y) = 0 be the cubic equations giving C_{1} and C_{2}. We can then find cubics going through the eight points by taking linear combinations $\lambda_1F_1+\lambda_2F_2$. Because the cubics going through the eight points form a one dimensional family, and because the set of cubics $\lambda_1F_1+\lambda_2F_2$ is a one dimensional family, we see that the cubic С has an equation $\lambda_1F_1+\lambda_2F_2=0$ for a suitable choice of $\lambda_1,\lambda_2$. Now about the ninth point - Since that ninth point is on both C_{1}and C_{2}, we know that F_{1}(x,y) and F_{2}(x,y) both vanish at that point. It follows that $\lambda_1F_1+\lambda_2F_2$ also vanishes there, and this means that С also contains that point.

Reformulating Mordell's theorem, if we have any two rational points on a rational cubic, say P and Q, then we can draw the line joining P to Q, obtaining the third point which we denoted P * Q. The set of all rational points on the cubic can be considered and assigned a law of composition. Cultivating upon the algebraic structure of the set along with the composition laws whether a group or not, we define the identity O as the zero element. (We will denote the group law by "+" because it is going to be a commutative nature.) The rule is as follows:

To add P and Q, take the third intersection point P*Q, join it to O, and then take the third intersection point to be P + Q. Thus by definition, P + Q = O*(P*Q).

- What is the negative of the point Q?

The negative of Q is the reflected point; i.e.if Q=(x,y) then —Q=(x,—y). To verify the negative, let us add Q and —Q. To do this we take the third intersection of the line through Q and —Q, which is S; and then join S to О and take the third intersection point S * O. But the line through S and O, because it is tangent to the cubic at O, meets the cubic once at S and twice at O. So the third intersection is the second time it meets O. Therefore, Q+(—Q) = O.

To prove the associativity: Let P, Q, R be three points on the curve. We want to prove that (P + Q)+R = P + (Q + R). To get P+Q, we form P*Q and take the third intersection of the line connecting it to O. To add P+Q to R, we draw the line from R through P+Q, and that meets the curve at (P + Q) * R. To get (P + Q)+ R, we would have to join (P + Q) * R to О and take the third intersection. To show (P + Q) + R = P + (Q + R), it will be enough to show that (P + Q) * R = P * (Q + R). We show it pictorially.

Thus, we have successfully formed a group structure on the set of rational points. Now we would reduce it to Weierstrass Normal Form. On other words, we need to show that every cubic is birationally equivalent to a cubic of this type. Using a little bit of projective geometry, we consider a rational point О on C, and Z = 0 to be the tangent line to С at O. This tangent line intersects С at one other point, and we take the X = 0 axis to be tangent to С at this new point. Finally, we choose Y = 0 to be any line (other than Z = 0) which goes through 0. After choosing axes in this fashion, if we consider $x=\frac {X}{Z}$ and $y=\frac {Y}{Z}$, then we get some linear conditions on the form the equation will take in these coordinates. This is called a **projective transformation**.

Skipping the algebra, we get to the end equation given as:

Calling the variable xy as y, we get,

Replacing $y-\frac{(ax+b)}{2}$ another linear transformation which amounts to completing the square on the left-hand side of the equation, we obtain $y^2=cubic\; in\; x$. So we do finally get an equation in Weierstrass form. To get rid of the $x^2$ term in the cubic, replace x by x — $\alpha$ for an appropriate choice of $\alpha$.

A cubic equation in normal form looks like

Assuming that the (complex) roots of f(x) are distinct, such a curve is called an * elliptic curve*. (More generally, any curve birationally equivalent to such a curve is called an elliptic curve.)

Now we are going to look deeper into the explicit formula of group laws.

Homogenising the cubic equation $y^2=x^3+ax^2+bx+c$ as $x=\frac {X}{Z}$ and $y=\frac{Y}{Z}$ yielding $Y^2Z= X^3+ aX^2Z+bXZ^2+cZ^3$

- How do we add two points P and Q on a cubic equation in Weierstrass form? First we draw the line through P and Q and find the third intersection point P * Q. Then we draw the line through P * Q and 0, which is just the vertical line through P*Q. A cubic curve in Weierstrass form is symmetric about the x axis, so to find P + Q, we just take P * Q and reflect it about the x axis. This procedure is illustrated in the following figure-

## Explicit formulae for group addition operation

Let, $P_1\equiv(x_1,y_1), P_2\equiv(x_2,y_2)\;$ then, $\; P_1*P_2 \equiv (x_3,y_3) \\ P_1+P_2=(x_3,-y_3)$. This is how we quantitively find the points. But, in order to prove Mordell's theorem, we would need some more rigor!

## Heights and descents

Defining **height** of a rational point $x=\frac m n$ reduced to its lowest terms, as

Thus, height of any rational number is a positive integer, with the following property:

**Finiteness property of heights.** The set of all rational numbers whose height is less than a fixed positive integer is a finite set.

Suppose, $y^2=f(x)=x^3+ax^2+bx+c$ be the non-singular cubic curve C with integer coefficient a,b,c; and if P=(x,y) is a rational point on the curve, then we defined height of P as simply the height of its x-coordinate

The property $H(P+Q)=H(P)H(Q)$ shows that the height has somewhat multiplicative structure. But, as we have always wanted to have an additive structure, we reduce H to "small h", which is defined as $h(P)=log\;H(P)$. Parallely, h is always a non-negative real number.

Note: "h" of rational points on curve C follow finiteness property; i.e. if M be a postive number, then the set ${P∈C(Q):h(P)≤M}$ is a finite set. Again, H(O)= 1 is defined this way.

Our ultimate aim is to show that the group of rational points $C\left(\mathbb{Q}\right)$ is finitely generated. We would need following four lemmata for the finite generation argument:

**Lemma I .**For every real number M, the set $\{P\in C(\mathbb{Q}):h(P)\leq M\}$ is finte set.**Lemma II.**Let P_{0}be a fixed rational point on C. There exists a constant K_{0}depending on P_{0}_{}and on a,b,c so that $h(P+P_0)\;\leq\;2h(P)+K_0\;\forall\;P\in C(\mathbb{Q})$ .**Lemma III.**There is a constant K depending on a,b,c such that $h(2P)\geq2h(P)-K\; \forall\; P\in C(\mathbb {Q})$.**Lemma IV.**The index $(C(\mathbb{Q}) : 2C(\mathbb{Q}))$ is finite.

Lemma I is well-explained already. Now, suppose, we are given an commutative group T , written additively, and a height function h as

from T to non-negative real integers. Now we state that T should be finitely generated.** **This statement** **along with the lemmata is called the** Descent Theorem, **where T is** ** $C(\mathbb {Q})$** . **

## The Height of P+P_{0}

We first make a remark that, for a rational point P(x,y) on the curve C, then x and y can be expressed as:

for integers m,n,e>0 and (m,e)=(n,e)=1. Let us take $x=\frac m M$ and $y=\frac n N$ in lowest term with M>0 and N>0. Putting in curve, we get,

Since,^{} $N^2$