Theorem 1. Let x, y, z be positive integers where x is odd and y is even. Then x, y, z is a primitivePythagorean triple if and only if there are positive integers u and v such thatv², y= 2uv,z = u? + v?(1)where (u, v) = 1, u > v and exactly one of u and v is even.Theorem 2 (Euler's Theorem). Let m be a natural number and let a be an integer such that(a, m) = 1. Then aº(m) = 1 (mod m).Theorem 3. Let m e N and let a € Z be such that (a, m) = 1. Let n be a natural number. Thena" = 1 (mod m) if and only if om(a) |n.(a, b). If d{ c, then theTheorem 4. Suppose that a and b are non-zero integers and let d =equationax + by = c(2)does not have any integer solutions. If d|c, then this equation has an infinite number of solutions.In fact, if ro and yo is one integer solution to equation (2), then all such solutions have the formx = xo +tay = Yo – twhere t varies over the integers.Definition 5. Let m e N and let a e Z be such that (a, m) = 1. The order (mod m) of a,denoted by om(@), is the least d e N such that ad = 1 (mod m).Definition 6. Let m be a natural number and let a be an integer such that (a, m)Om (a) = 0(m), then a is said to be a primitive root (mod m).= 1. IfCorollary 7. Let m e N. If a primitive root (mod m) exists, then there are exactly o(0(m))incongruent primitive roots (mod m).Theorem 8. Let p be a prime number and let d e N be such that d|(p – 1). Then there areexactly ø(d) many incongruent integers of order d modulo p.Corollary 9. Let p be a prime number. Then there are exactly ø(p–1) many incongruent integersof order p – 1 modulo p; that is, there are exactly ø(p – 1) many primitive roots (mod p).Lemma 10. Let p >1 be a prime number and let a > 1. Then$(p*) = pª – pª=1 = p° ( 1 –Theorem 11. Let a, b, n E Z. If n and a are relatively prime and n| (ab), then n|b.Theorem 12. Let a, b, n be integers where a |n and b|n. If (a, b) = 1, then (ab) |n.Lemma 13 (Euclid's Lemma). Let a and b be natural numbers and let p be a prime. If p|(ab),then p|a or p|b.Corollary 14. Let a be a natural number and p be a prime. If p|a², then p|a.Theorem 15. Let a, r, m be integers. Then (a, m) = 1 and (r, m) = 1 if and only if (ar, m) = 1.Theorem 16. Let a, b, m be integers where m > 1. Suppose that the congruence a = b (mod m)holds. Then (a, m) = 1 if and only if (b, m) = 1.Corollary 17. Let a, x, b, m be integers where m > 1. If (b, m) = 1 and ax = b (mod m), then(а, т) — 1 and (х, m) — 1. Let m e N and a e Z. Suppose that 3|o(m) and that a is aprimitive root (mod m). Prove that a is not a primitive root (mod m).[You must clearly write and justifysteps in your proof.]

Answered: Image /qna-images/answer/5190b396-bbc7-40f1-bd12-afc7fba0531e.jpg

Answered: Let m e N and a E Z. Suppose that 3|(m)…

Let m e N and a E Z. Suppose that 3|(m) and that a is a primitive root (mod m). Prove that a is not a primitive root (mod m). [You must clearly write and justify the steps in your proof.]

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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Please give detailed steps. I’m really trying to understand this problem. Much appreciated!
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…
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