(4) Consider the primes p = 13 and q = 19. Let e = 43.(a) Compute pq and (p − 1)(q − 1).Prove that gcd(e, (p − 1)(q − 1)) = 1.Compute the smallest positive integer d such thatde = 1 (mod (p − 1)(q − 1))Equivalently,43d = 1 (mod 216)

To compute and .Given and The value of and is calculated as follows: and .

Answered: Consider the primes p = 13 and q = 19.…

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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C
Can you help me to show that the systhem has a unique solution.
Does x° = a (mod p) always have a solution for every value of a, whenever p is prime?
answer no.2
ne Zt. Solve the congruence x = n" (mod p), where p is a prime of the form 4n ± 1,
b) Let p≥ 3 be prime. Show that the only solutions to x² =1 (mod p) are x = ±1 (mod p). Hint: Apply part a) to (x+1)(x-1)
For each part, use the data provided to find values of a and b satisfying a2 ≡ b2 (mod N), and then compute gcd(N, a − b) in order to find a nontrivial factor of N, as we did in Examples 3.37 and 3.38. (c) N = 198103 11892 ≡ 27000 (mod 198103) and 27000 = 23 · 33 · 53 16052 ≡ 686 (mod 198103) and 686 = 2 · 73 23782 ≡ 108000 (mod 198103) and 108000 = 25 · 33 · 53 28152 ≡ 105 (mod 198103) and 105 = 3 · 5 · 7 Here is an example of a similar problem solved in the book: We factor N = 914387 using the procedure described in Table 3.4. We first search for integers a with the property that a2 mod N is a product of small primes. For this example, we ask that each a2 mod N be a product of primes in the set {2, 3, 5, 7, 11}. Ignoring for now the question of how to find such a, we observe that 18692 ≡ 750000 (mod 914387) and 750000 = 24 · 3 · 56 19092 ≡ 901120 (mod 914387) and 901120 = 214 · 5 · 11 33872 ≡ 499125 (mod 914387) and 499125 = 3 · 53 · 113 None of the numbers on the right is a…
(10) Prove that for any prime number P>3. p² = 1 (mod 24) (Hint: chinese remainder theorem tells us SP² = 1 (mod 3) p² = 1 (mod 8) need to show we only need to p²=1 (mod 8), which we know that P PH two integers. The hard part is to show means S1 p²2-1 81 = (P-1) (PH). must be odd. Hence, P-1, So it is enough to show 2 | P-1 F are PH).
1+n 3n+1 For each natural number n, Let - Find UA, and 4. n n EN EN If R={(x, y) & RxR: x² = 3 +5y}, S={(x,y) & R×R: y³ = 3x²+5}, Then find RoS and SoR Solve the simultaneous congruences: x = 3 (mod 2) x = 5 (mod 7) (x = 11 (mod 13)
solve no.3
solve no.1

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Consider the primes p = 13 and q = 19. Let e = 43. (a) Compute pq and (p − 1)(q − 1). Prove that gcd(e, (p − 1)(q − 1)) = 1. Compute the smallest positive integer d such that de 1 (mod (p − 1)(q − 1)) 43d 1 (mod 216) Equivalently,