Hw3 - (2.1) (2.1) p. 106 #20. a) Since , we can apply...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (2.1) (2.1) p. 106 #20. a) Since , we can apply Euler's Theorem, p. 81, to conclude that . Hence is an with the property that . Since is, by definition, the smallest such value of , it follows that . b) = = = = 1 mod n. c) 1 = = = = . d) Since by definition is the smallest positive exponent of to give 1, and since , it follows from c) that is zero. e) Taking , and dividing by , we see from part d) that the remainder is zero. Hence | . p. 106 #21. isprime(601); true a) If r | 600 then the factorization of r contains only powers of the primes 2, 3, and 5. If r < 600, then at least one of the exponents of 2, 3, or 5 in the factorization of r must be less than the corresponding exponent in the factorization of 600. The maximal such elements are 300 where the exponent of 2 drops by 1; 200, where the exponent of 3 drops by 1; and 120, where the exponent of 5 drops by 1 from the exponents in the factorization in 600. b) 601 is prime, hence = 600. By #20, | 600. If < 600, then must divide one of the maximal proper divisors, 300, 200, or 120. c) By #20 b), if ord[601](7) divided 300, then would be 1 mod 601, a contradiction. Similarly for the other two cases. d) Since by c) does not divide any of the proper subdivisors of , but by #20 e) must divide , then . e) If for some , so that 2 is not primitive, and t is any positive integer, then = = = , so is also not primitive. Hence, there is no point in testing for primitivity powers of integers where the original integer is known not to be primitive. We cannot exclude 6, even if 2 and 3 are known not to be primitive, since the (4.1) (4.1) exponents at which the 2 and the 3 become congruent to 1 may be different, so that the product of the powers itself is different than 1 at each stage. Once you know the order of 2 and the order of 3, however, you can immediately discern the order of 6, since it is the least common multiple of the orders of 2 and of 3. In fact, since the lcm(n,m) is often greater than both n and m, it is possible that the composite 6 will have maximal order and 6 will be a primitive root of unity although neither 2 or 3 is primitive....
View Full Document

This note was uploaded on 02/14/2012 for the course MATH 470 taught by Professor Staff during the Spring '08 term at Texas A&M.

Page1 / 8

Hw3 - (2.1) (2.1) p. 106 #20. a) Since , we can apply...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online