10 Problem 1. The following assertions may be true or false, depending on the choice of the integers a, b 0. a "

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1 Math 4161 Dr. Franz Rothe December 9, FALL\4161_fall13f.tex Name: Use the back pages for extra space Final Problem 1. The following assertions may be true or false, depending on the choice of the integers a, b 0. a " gcd(a, b) "a and a " gcd(a, b) and and b are relatively prime" b are relatively prime" gcd(a, b) b are relatively prime" gcd(a, b) Give an example of two numbers a and b for which the two first lines are both false. Which one of the three assertions is for any integers a, b 0 always true. 1

2 Problem 2. Let a, d, m be integers and d 2, m 2. We check whether the terms a, a + d, a + 2d,... of a (long) arithmetic sequence are divisible by integer m. Under which condition is it true that every m-th term of the sequence is divisible by m? Give several counterexample with d = 4, m = 6 and d = 6, m = 4 to confirm that your condition is necessary. 2

3 Proposition 1 (The Euclidean Property). Let a, b, c be integers. divides the product ab and gcd(c, a) = 1, then c divides b. If a number c Problem 3. Complete the standard proof of the Euclidean property for integers. Standard proof. Since c divides the product ab, there exists an integer q such that? By the extended Euclidean algorithm, there exist integers s, t such that? Multiplying both sides of the last formula by b, and using ab = qc yields b = sab+tcb. A bid of calculation allows to factor out c. One can see that as follows:? Hence c divides b, as to be shown. 3

4 Problem 4. Let p be a prime and 1 k p 1 an integer. Complete the following proof that p divides the binomial coefficient ( p k). In the definition of the binomial coefficient, we factor the top ( ) p p (p 1) (p k + 1) = = p b k 1 k k! with b = (p 1) (p k + 1). I have used the standard abbreviation k! for the product? which occurs in the bottom. It is known that the given binomial coefficient is?. Hence the bottom k! divides the product? We also know that gcd(p, k!) = 1 since the number p is? factors in the product k! are? than p. and all the We can now use the Euclidean property stated in the last problem. Since k! divides the product p b and gcd(p, k!) = 1, we conclude that? Hence is divisible by the prime p. ( ) p = p b k k! = p b k! 4

5 Problem 5. For which values of n is it true that 13 divides n Determine these values in the range 0 n < 12. The remaining values are easily determined by a calculation modulo 13. What are all solutions of 13 n

6 Problem 6. Find some solutions of the equation φ(n + 2) = φ(n) + 2, in the hope to get infinitely many solutions. Here are three possible ideas: Assume that p and p + 2 are twin primes. What do choose for n? Assume that p and q = 2 p 1 are prime. What can you choose for n? Assume that p and q = 2p + 1 are prime. What can you choose for n? Remark. Each of these ideas would now imply existence of infinitely many solutions of the equation φ(n + 2) = φ(n) + 2, if one would only know that there exist infinitely many either twin primes, or Mersenne primes or Sophie Germain primes. 6

7 Problem 7. Prime factor the number 40! and explain the method, credited to Legendre. 7

8 Problem 8. Prove that in the sequence of primes there exist arbitrary long gaps. In other words for any given natural number n, there exists a natural numbera such that all numbers a + 2, a + 3,..., a + n are composite. 8

9 Problem 9. One finds in the literature the assertion that Chinese generals used to count the number of their soldiers using the Chinese remainder theorem. They gave military commands to line up in several squares: Line up 7 by 7. Line up 11 by 11. and each time asked the remaining soldiers to report back. The general needs only to count these remaining soldiers. Suppose that after lining up in 7 by 7 squares, there are soldiers remaining, after lining up 11 by 11, there are 8 soldiers remaining. Set up the simultaneous congruences to determine the number of soldiers. Solve these congruences and determine the number of soldiers. 9

10 Problem. Solve the simultaneous congruences Complete and explain the method below. g 2 (mod 15) g 3 (mod 7) We want to calculate with a sequence of congruences modulo the product m = 15 7 = 5, and use the same operations as done by the Euclidean algorithm calculating the greatest common divisor of? and? q i s i r i congruence (g 2) 15(3 2) 0 7 7(g 2) 1 1(g 2) 0 0(g 2) One gets?, which can easily be checked. The very last line give the? of the simultaneous congruences exists. under which the solution

11 Problem 11. Let p, q, r be three different odd primes and m = pqr. What is the maximal possible order modulo m that any integer a can have. Given are a primitive root a modulo p, a primitive root b modulo q, and a second primitive root c modulo r. Set up the Chinese Remainder Problem to determine an integer g that has maximal possible order modulo m. 11

12 Problem 12. Put m = 5 in the previous problem, as a numerical example. I take the primitive roots from a table: 2 is a primitive root modulo 3, and 2 is a primitive root modulo 5, finally 3 is a primitive root modulo 7. What is λ(5). Set up the Chinese Remainder Problem to determine an integer g that has maximal possible order modulo 5. 12

13 Problem 13. Complete the following table of the values of the Euler totient function φ(m) and φ(φ(m)) for m = 1, 2,..., 20. x φ(x) φ(φ(x)) exists primitive root? (no) yes yes yes yes yes yes no

14 Problem 14. How many primitive roots are there modulo 157, are how many primitive roots are there modulo 237. Why does there occur a difference? 14

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