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(Modification of exercise 36 in section 2.5 of Rosen.) The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p 71, q = 53 and form n = 71 · 53 = 3763. [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 19. This will work since gcd(19, (p – 1)(q– 1)) = gcd(19, 3640) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E3D4,F=5,G=6,H=7,1=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 1549 mod 3763. This is not as hard as it seems and you might consider using fast modular multiplication. 19 The canonical representative of 15° mod 3763 is 7

(Modification of exercise 36 in section 2.5 of Rosen.) The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p 71, q = 53 and form n = 71 · 53 = 3763. [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will know n but p and q will be kept private.] Now we choose our public key e = 19. This will work since gcd(19, (p – 1)(q– 1)) = gcd(19, 3640) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E3D4,F=5,G=6,H=7,1=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. Thus we need to calculate 1549 mod 3763. This is not as hard as it seems and you might consider using fast modular multiplication. 19 The canonical representative of 15° mod 3763 is 7

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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(Modification of exercise 36 in section 2.5 of Rosen.) The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p 71 · 53 = 3763. 71, q = 53 and form n = [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will known but p and q will be kept private.] Now we choose our public key e = 19. This will work since gcd(19, (p – 1)(a – 1)) = gcd(19, 3640) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. 19 Thus we need to calculate 15 mod 3763. This is not as hard as it seems and you might consider using fast modular multiplication. 19 The canonical representative of 15 mod 3763 is 7
Transcribed Image Text:(Modification of exercise 36 in section 2.5 of Rosen.) The goal of this exercise is to work thru the RSA system in a simple case: We will use primes p 71 · 53 = 3763. 71, q = 53 and form n = [This is typical of the RSA system which chooses two large primes at random generally, and multiplies them to find n. The public will known but p and q will be kept private.] Now we choose our public key e = 19. This will work since gcd(19, (p – 1)(a – 1)) = gcd(19, 3640) = 1. [In general as long as we choose an 'e' with gcd(e,(p-1)(q-1))=1, the system will work.] Next we encode letters of the alphabet numerically say via the usual: (A=0,B=1,C=2,D=3,E=4,F=5,G=6,H=7,l=8, J=9,K=10,L=11,M=12,N=13,0=14,P=15,Q=16,R=17, S=18,T=19,U=20,V=21,W=22,X=23,Y=24,Z=25.) We will practice the RSA encryption on the single integer 15. (which is the numerical representation for the letter "P"). In the language of the book, M=15 is our original message. The coded integer is formed via c = Me mod n. 19 Thus we need to calculate 15 mod 3763. This is not as hard as it seems and you might consider using fast modular multiplication. 19 The canonical representative of 15 mod 3763 is 7
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