[Number Theory] How do you solve question 2? thanks 1. Show that 25 is a strong pseudoprime base 7, i.e., passes Miller's test base 7.2. Use the method of squaring to compute51711 mod 1911.3. (i) For a prime p suppose that n = 2² - 1 is not a prime.Show that n is a pseudoprime base 2.(ii) Show every composite Fermat number Fm4. Show that if=n22m + 1 is a pseudoprime base 2.a²p 1a² - 1where a > 1 is an integer and p is an odd prime withpła(a²-1),then n is a pseudoprime base a.Hint: First show that 2p | (n − 1). Then consider an−¹ – 1.Note that this shows that there are infinitely many pseudoprimes base a for any a.

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Answered: 1. Show that 25 is a strong pseudoprime…

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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A prime of the form 2n − 1 is called a Mersenne prime. (e) More generally, prove that if n is a composite number, then 2n − 1 is not prime. Thus, all Mersenne primes have the form 2p − 1 with p a prime number.
Show that the result of the previous problem does not provide a sufficient condition for the existence of a quadratic residue of n; in other words, find relatively prime integers a and n, with a^((phi(n)/2)) ≡ 1 (mod n), for which the congruence x^(2) ≡ a (mod n) is not solvable.
I know that using a corollary that the congruence x2 ≡ 0 (mod p) that the only solutions are + and - 1 but thats when p is prime. How would i go about proving this... Prove that if n ≡ 2(mod 4), then n can't be written as am for any integer m with m > 1.
Find all values of n with 0 ≤ n ≤ 35 such that the congruence 24x ≡ n (mod 36) has a solution. Showall the necessary work to obtain your answers.
Prove, that if 2m + 1 is a prime number, then m must be a power of 2 (m = 2n, for some n ∈ N).Prime numbers in the form p = 2m + 1 are called Fermat's prime numbers.
We stated that the number 561 is a Carmichael number, but we never checked that a561 ≡ a (mod 561) for every value of a.(a) The number 561 factors as 3 · 11 · 17. First use Fermat’s little theorem to prove thata561 ≡ a (mod 3), a561 ≡ a (mod 11), and a561 ≡ a (mod 17)for every value of a. Then explain why these three congruences imply that a561 ≡ a (mod 561) for every value of a.(b) Mimic the idea used in (a) to prove that each of the following numbers is aCarmichael number. (To assist you, we have factored each number into primes.)(i) 1729 = 7 · 13 · 19(ii) 10585 = 5 · 29 · 73(iii) 75361 = 11 · 13 · 17 · 31(iv) 1024651 = 19 · 199 · 271
(This exercise constructs another proof of the infinitude of pimes) Show that the integer Qn = n! + 1, where n is a positive integer, has a prime divisor greater than n. Conclude that there are infinitely many primes.
Find the sequence of pseudorandom numbers generated by the linear congruential method with modulus m = 9, multiplier a = 89, increment c = 2, and seed x0= 5. xn+1 = (axn + c) mod m.
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If 2n +1 is prime for some n ≥ 1, prove that n is a power of 2.(Primes of the form 2n +1 are called Fermat primes.)
a) Show that 2340 ≡ 1 (mod 11) by Fermat’s little theorem and noting that 2340 = (210)34.b) Showthat 2340 ≡ 1 (mod 31) using the fact that 2340 = (25)68 = 3268.c) Conclude from parts (a) and (b) that 2340 ≡ 1 (mod 341).

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