Elliptic Curve (When , Why , Where ??)

Hi Folks ! This blog is regarding Elliptic curves. Hope after reading this blog you will be able to know about the basics of elliptic curves. 

Una and Jammu are the two best friends. One day Una saw the equation  of elliptic curve. Now, Una became curious regarding elliptic curve means what is elliptic curve and how they are generated and why they are generated ?. 😓

So he call to his friend Jammu.

Una : Hey Jammu ! How are you ??

Jammu : I'm good Una. Thanks for asking. 

Una : Hey Jammu ! you know about my nature 😁 so I don't want to waste your time. Yesterday, I saw the equation of elliptic curve. so can you please tell me about elliptic curve means how they generated and why they generated ? 

Jammu : Hahahah !!! Ok, Sure Una. Have you know about Tunnell's theorem ??

Una : No. 😌.

Jammu : Don't Panic Una. See my below text. 

Finding a simple test to determine whether or not a given integer say ‘n’ is the area of some right triangle all of whose sides are rational numbers.

Una : Ohh Sorry Jammu ! I forget, I can also relate this tunnell's theorem with Congruent numbers.

A natural number ‘n’ is called congruent number if there exist a right triangle with all three sides are rational and area n.

am I right Jammu ?

Jammu : Yes Una you are right. 👍. Hey Una, now come to your basic mathematics. 

Una : Oo Wow . Mathematics !!! My Favourite One. 

Jammu : okok. Let me check your knowledge. Do yon know Fermat's Last Theorem. 

Una : Of course. 

According to Fermat’s Last theorem.

                Xn + Yn = Zn

Where n >2, and X,Y,Z ∉ N  

Symbol N denotes set of natural number {1,2,3,…..}.

Jammu : Great Una ! 👏 . Now it's my time, I will give you the basic knowledge regarding right triangle and it's mathematics. 

Assume X, Y,and Z be the sides of right triangle where Z be the longer side. So  according to Pythagorean theorem :

                 X2 + Y2 = Z2                                                                                        (1) 

(Hey Una ! Don’t Forget X,Y,Z ∉ N or X,Y,Z ∈ Q where Q be the set of rational numbers. }

          

We also Know that the area of triangle is :

                ½XY = n or XY=2n                                                                             (2)

 

If we add or subtract four times the (2) with (1) then :

                (X+Y)2 = Z2 + 4n                                                                                (3)

                (X-Y)2 = Z2 – 4n                                                                                 (4)

 

If we multiply equation (3) with equation (4).

                 ((X2-Y2)/4)2 = (Z/2)4 – n2                                                                  (5)

 

This shows that the equation u4 – n2 = v2 has a rational solution, namely u = Z/2 and v = (X2-Y2)/4.

We next multiply through by u2 to obtain u6 – n2u2 = (uv)2.

If we x = u2 = (Z/2)2 and further y = uv = (X2-Y2)Z/8, then we have a pair of rational numbers (x,y) satisfying the cubic equation.

                             Y2 = x3 – n2x

or I can also write it as y2 = x3 + Ax + b. (also known as weierstrass equation)

Una :  Hey Wow I got the result but here u , and v are not defined what is this 😩.

Jammu : hahaha. 😀You are so curious Una but that's great Ok. Now come to little bit knowledge regarding Euclid's big problem. (Dear Una I will mail you one link. I hope this link will also very beneficial for you as for reference. (https://www.youtube.com/watch?v=6Lm9EHhbJAY) .

1. Consider two points, we know that from that two points we can draw a line and that line can be extended. Here A and B are the two points and we connect A and B and from A we draw an arc and from B we draw an another arc then we find the point of intersection at C and we know that this intersection is perpendicular (as shown in Fig. 1). We also deduce that it can make a triangle.   


Fig. 1 (Line with two points)

Now come to next methodology. Take two positive integers a and b with a>b draw the line in uv plane through the point (-1,0) with slope b/a. Assume (u,v) be the second point of intersection of this line with the unit circle (as shown in Fig. 2). It can be easily said that.

Fig. 2 : Circle. 

 u = (a2-b2)/(a2+b2)

 v = 2ab/(a2+b2)

Then, the integer X = a2-b2, Y = 2ab, Z = a2+b2 are the sides of right triangle; the fact that X2+Y2 = Z2 follows because u2+v2=1

Una : Hurrah !!! Now My all doubts related to Elliptic curve are clear. Thanks Jammu. 

Jammu. Ok Una. Good Luck !!! If you have still any doubt then you can write in the Comment section. 

Thanks to Author Neal Koblitz (Introduction to Elliptic curves and modular form Second Edition.). 

Hope you Enjoy the Journey of Elliptic curve. 😊


 *************************************************************************************************************

Comments

Post a Comment

Popular posts from this blog

Homomorphic Encryption: A Basic Idea

Image
Homomorphic encryption has the same aim as other encryption-decryption mechanisms: maintaining confidentiality, security, privacy, etc.  In this blog, we will see homomorphic encryption and its type.  Homomorphic Encryption (HE):  It is an encryption that computes operations over encrypted data without decrypting it (or without the need for a secret key).  Mathematically speaking in a client-server architecture: client has message 'm' and function f. The task of client is to compute f(m). Due to computational expensiveness (maybe one of the reason) client sends encryption of message (say Enc(m)) and 'f' to the server. Server will compute function over encrypted message, that is, f(Enc(m)), and return the result to the client, that is, Enc(f(m)). Client will decrypt it Dec(Enc(f(m)), that is, f(m). Hence, client achieves f(m).  Let's understand HE with our daily life (Tea making process): Figure 1. Here m is message, Enc is encryption, f is function, and Eval in...
Read more

Fast Base Conversion and Its Application

Image
Fast Base Conversion (FBC) is a technique used to convert a number from one modulus space to another. For example, if we have 14 m o d     77 14 \mod 77  and want to convert it to x m o d     390 x \mod 390 , FBC is widely used. Mathematically, it can be represented as follows: Here, the above equation represents an integer x x  in modulus q q , and FBC converts this integer into the modulus space t t . The modulus q q  is the product of small coprime moduli (i.e., q = q 1 q 2 … q k q = q_1 q_2 \dots q_k ​ ), and x x  is an integer such that x ∈ [ 0 , q ) x \in [0, q) . We also note that q q  and t t  are coprime numbers. Here, we remark that it is not necessary for the above equation to produce the exact result for x x . Instead of x m o d     t x \mod t , it will provide x ′ m o d     t x' \mod t . Thus, the output of the above equation is as follows: FBC ( x , q , t )  =  x  +  A q  mod  t , where A ∈ [ 0 , k − 1 ] A \in [0,...
Read more

Brakerski-Fan-Vercauteren (BFV) Homomorphic Encryption

Image
Homomorphic encryption means computation over encrypted data without secret key.  How this is possible  ?? 😴 In this blog, we will explore the types of Homomorphic Encryption and demonstrate how computations can be performed on encrypted data without the secret key. In 2009, Gentry proposed Fully Homomorphic Encryption. In 2012, Brakerski proposed a leveled homomorphic encryption scheme that works over integer arithmetic, which later became known as the BFV scheme. BFV is an encryption scheme. Thus, as an encryption scheme, it is expected to be secure. To date, under certain conditions, it is assumed that the scheme is safe, and its security is based on the Learning With Errors (LWE) assumption.. Learning-with-error (LWE): Given a pair of a, b and the task is to recover secret key (s), such that b = a.s  + e  Here e is error mostly picked from gaussian distribution, and a, s are selected from Rq (known as ring module R) The above problem is HARD. The proof is not si...
Read more