[Back] For a finite field elliptic curve we have for a curve of \(y^2 = x^3 + ax +b \pmod p\) and for a defined prime number (\(p\)). If we have a point \(P\), we can then calculate \(2P\) (and use this to find \(nP\) - where \(n\) is the number of times we add \(P\)) [Calculating nP] In this case we will add two points on the elliptic curve together to get a resultant point:

## ECC Point Addition Calculator |

## Coding

In this case we calculate \(x^3+ax+b \pmod p\). We can add two points \((x_1,y_1)\) and \((x_2,y_2)\):

Let \(s = (y_1-y_2)/(x_1-x_2)\)

Then:

\(x_2 = s^2 - x_1 - x_2\)

\(y_2 = s(x_1 - x_2) - y_1\)

import sys a=0 b=7 p=37 x1=6 x2=8 if (len(sys.argv)>1): x1=int(sys.argv[1]) if (len(sys.argv)>2): x2=int(sys.argv[2]) if (len(sys.argv)>3): p=int(sys.argv[3]) if (len(sys.argv)>4): a=int(sys.argv[4]) if (len(sys.argv)>5): b=int(sys.argv[5]) def modular_sqrt(a, p): """ Find a quadratic residue (mod p) of 'a'. p must be an odd prime. Solve the congruence of the form: x^2 = a (mod p) And returns x. Note that p - x is also a root. 0 is returned is no square root exists for these a and p. The Tonelli-Shanks algorithm is used (except for some simple cases in which the solution is known from an identity). This algorithm runs in polynomial time (unless the generalized Riemann hypothesis is false). """ # Simple cases # if legendre_symbol(a, p) != 1: return 0 elif a == 0: return 0 elif p == 2: return p elif p % 4 == 3: return pow(a, (p + 1) / 4, p) # Partition p-1 to s * 2^e for an odd s (i.e. # reduce all the powers of 2 from p-1) # s = p - 1 e = 0 while s % 2 == 0: s /= 2 e += 1 # Find some 'n' with a legendre symbol n|p = -1. # Shouldn't take long. # n = 2 while legendre_symbol(n, p) != -1: n += 1 # Here be dragons! # Read the paper "Square roots from 1; 24, 51, # 10 to Dan Shanks" by Ezra Brown for more # information # # x is a guess of the square root that gets better # with each iteration. # b is the "fudge factor" - by how much we're off # with the guess. The invariant x^2 = ab (mod p) # is maintained throughout the loop. # g is used for successive powers of n to update # both a and b # r is the exponent - decreases with each update # x = pow(a, (s + 1) / 2, p) b = pow(a, s, p) g = pow(n, s, p) r = e while True: t = b m = 0 for m in xrange(r): if t == 1: break t = pow(t, 2, p) if m == 0: return x gs = pow(g, 2 ** (r - m - 1), p) g = (gs * gs) % p x = (x * gs) % p b = (b * g) % p r = m def legendre_symbol(a, p): """ Compute the Legendre symbol a|p using Euler's criterion. p is a prime, a is relatively prime to p (if p divides a, then a|p = 0) Returns 1 if a has a square root modulo p, -1 otherwise. """ ls = pow(a, (p - 1) / 2, p) return -1 if ls == p - 1 else ls def egcd(a, b): if a == 0: return (b, 0, 1) else: g, y, x = egcd(b % a, a) return (g, x - (b // a) * y, y) def modinv(a, m): g, x, y = egcd(a, m) if g != 1: print "x-point is not on the curve. Please select another point." sys.exit() else: return x % m print "a=",a print "b=",b print "p=",p print "x-point=",x1 print "x-point=",x2 z=(x1**3 + a*x1 +b) % p y1=modular_sqrt(z, p) z=(x2**3 + a*x2 +b) % p y2=modular_sqrt(z, p) print "\nP1\t(%d,%d)" % (x1,y1) print "P2\t(%d,%d)" % (x2,y2) s=(y2-y1)* modinv(x2-x1,p) x3=(s**2-x2-x1) % p y3=((s*(x2-x3)-y2)) % p print "P1+P2\t(%d,%d)" % (x3,y3)

And a sample run:

a= 0 b= 7 p= 37 x-point= 6 x-point= 8 P1 (6,1) P2 (8,1) P1+P2 (23,36)

## Presentation

The following is a presentation: