PROBLEMS 8,1 (a) modulo 17. (b) modulo 19. (c) modulo 23.
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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Question 1 8.1
![of 9, these the 2 and 5.
1. Find the of the integers 2, 3, and 5:
(b) The odd of the nº + 1 are of form 8k + 1. o
(c) The odd of the n² that are from 3 are of
4 can be illustrated by taking a = 2 andn=9. Because ø(9) = 6,
1. 2, 4, 5, 7, 8, and we see that
prime to 9 are
2' =2
23 = 8
22 = 4
ve of the corollary, there are exactly o((9)) = ¢(6) = 2 primitive roots
25 =
7.
5 w 26 = 1 (mod 9)
PROBLEMS 8.1
Find the order of the integers 2, 3, and 5:
by
(a) modulo 17.
(b) modulo 19,
(c) modulo 23.
2 Establish each of the statements below:
(0) If a has order hk modulon, then a" has order k modulo n.
b) If a has order 2k modulo the odd prime p, then ak = -1 (mod p).
() If a has order n – 1 modulo n, then n is a prime.
3. Prove that ø(2" – 1) is a multiple of n for any n > 1.
[Hint: The integer 2 has order n modulo 2" – 1.]
4. Assume that the order of a modulo n is h and the order of b modulo n is k, Show that the
order of ab modulo n divides hk; in particular, if ged(h, k) = 1, then ab has order hk.
5. Given that a has order 3 modulo p, where p is an odd prime, show that a +1 must have
order 6 modulo p.
Hint: From a² +a +1 = 0 (mod p), it follows that (a + 1) = a (mod p) an
(a + 1)³ = –1 (mod p).]
) The odd prime divisors of the integer n² + 1 are of the form 4k + 1.
nt: n' = -1 (mod p), where p is an odd prime, implies that 4|$(p) I
Theorem 8.1.]
fo odd prime divisors of the integer n2 +n+1 that are different from 3 are of
6k + 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2Fc3681411-98bd-445c-95ac-a6c156d8817c%2Ft6pimmp.jpeg&w=3840&q=75)
Transcribed Image Text:of 9, these the 2 and 5.
1. Find the of the integers 2, 3, and 5:
(b) The odd of the nº + 1 are of form 8k + 1. o
(c) The odd of the n² that are from 3 are of
4 can be illustrated by taking a = 2 andn=9. Because ø(9) = 6,
1. 2, 4, 5, 7, 8, and we see that
prime to 9 are
2' =2
23 = 8
22 = 4
ve of the corollary, there are exactly o((9)) = ¢(6) = 2 primitive roots
25 =
7.
5 w 26 = 1 (mod 9)
PROBLEMS 8.1
Find the order of the integers 2, 3, and 5:
by
(a) modulo 17.
(b) modulo 19,
(c) modulo 23.
2 Establish each of the statements below:
(0) If a has order hk modulon, then a" has order k modulo n.
b) If a has order 2k modulo the odd prime p, then ak = -1 (mod p).
() If a has order n – 1 modulo n, then n is a prime.
3. Prove that ø(2" – 1) is a multiple of n for any n > 1.
[Hint: The integer 2 has order n modulo 2" – 1.]
4. Assume that the order of a modulo n is h and the order of b modulo n is k, Show that the
order of ab modulo n divides hk; in particular, if ged(h, k) = 1, then ab has order hk.
5. Given that a has order 3 modulo p, where p is an odd prime, show that a +1 must have
order 6 modulo p.
Hint: From a² +a +1 = 0 (mod p), it follows that (a + 1) = a (mod p) an
(a + 1)³ = –1 (mod p).]
) The odd prime divisors of the integer n² + 1 are of the form 4k + 1.
nt: n' = -1 (mod p), where p is an odd prime, implies that 4|$(p) I
Theorem 8.1.]
fo odd prime divisors of the integer n2 +n+1 that are different from 3 are of
6k + 1.
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