-
Thông tin
-
Quiz
Lifting the Exponent - Lý thuyết số - Toán Kinh Tế | Đại học Tôn Đức Thắng
Question 11. Let a > b > 1be positive integers, and let bbe odd. Let nbe a positive inte-ger such that bn|an−1. Show that n·ab≥3n.Question 12. This exercise aims to prove that the group Z∗pnis cyclic for all odd primes pandnatural numbers n, assuming Z∗pis cyclic. Tài liệu được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Toán Kinh Tế (C01120) 73 tài liệu
Đại học Tôn Đức Thắng 3.5 K tài liệu
Lifting the Exponent - Lý thuyết số - Toán Kinh Tế | Đại học Tôn Đức Thắng
Question 11. Let a > b > 1be positive integers, and let bbe odd. Let nbe a positive inte-ger such that bn|an−1. Show that n·ab≥3n.Question 12. This exercise aims to prove that the group Z∗pnis cyclic for all odd primes pandnatural numbers n, assuming Z∗pis cyclic. Tài liệu được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!
Môn: Toán Kinh Tế (C01120) 73 tài liệu
Trường: Đại học Tôn Đức Thắng 3.5 K tài liệu
Thông tin:
Tác giả:



Tài liệu khác của Đại học Tôn Đức Thắng
Preview text:
Lecture 3: Lift That Exponent December 13
Lecturer: Shourya Pandey Scribe: Shourya Pandey The Lemma
Definition: For an integer x and a prime p, vp(x) denotes the higher power of the prime p dividing
x. We define vp(0) = ∞ for all primes p.
Lift The Exponent Lemma: Let n be a positive integer and let p = 2 be a prime. Suppose x
and y are integers such that neither x nor y is divisible by p, but p | x − y. Then
vp(xn − yn) = vp(x − y) + vp(n)
What about the case of p = 2? Let n be a positive even integer and let x, y be odd integers. Then
vp(xn − yn) = vp(x − y) + vp(x + y) + vp(n) − 1
Also, if 4 | x − y then for any positive integer n, we have
vp(xn − yn) = vp(x − y) + vp(n)
The theorem is simple enough, so let’s tackle some questions right away! 1 Easier Problems
Question 1. Let a, n be two positive integers, and let p be an odd prime such that ap ≡ 1 (mod pn) Prove that a ≡ 1 (mod pn−1) Is the above true if p = 2?
Question 2. Let p be a prime and let a, n be positive integers. Show that if an = 2p + 3p then n = 1.
Question 3. Find all positive integral solutions to (n − 1)! + 1 = nm Question 4. Evaluate v 19911992 1991(1990 + 199219911990). 1 2 Medium Problems
Question 5. Let a, b be distinct reals such that
a − b, a2 − b2, a3 − b3, · · ·
are integers. Show that a and b are integers.
Question 6. Find all positive integers n for which 2n + 1 n2 is a positive integer.
Question 7. Does there exist a positive integer n such that n has exactly 2000 prime divisors, and n | 2n + 1?
Question 8. Let n > 1 be a positive integer. Show that the sequence (ak)k∈N defined as ⌊ nk ⌋ ak = k
has infinitely many odd natural numbers. 3 Harder Problems
Question 9. Find all triples (x, y, p) consisting of two positive integers x and y, and a prime p,
such that both xp−1 + y and yp−1 + x are powers of p.
Question 10. Find all triples (a, b, c) of positive integers for which ab − c, bc − a, ca − b are all powers of 2.
Question 11. Let a > b > 1 be positive integers, and let b be odd. Let n be a positive inte-
ger such that bn | an − 1. Show that n · ab ≥ 3n.
Question 12. This exercise aims to prove that the group Z∗ is cyclic for all odd primes p and pn
natural numbers n, assuming Z∗ is cyclic. p
1. Make sure you can recall the definitions of Zm, Z∗ , primitive roots (or generators), and m cyclic groups. 2
2. Assume you know there exists an element g such that 1, g, g2, · · · , gp−2 are all distinct mod-
ulo p. Such a g is called a primitive root (or a generator) modulo p. Then, how many
1 ≤ i ≤ p − 1 exist such that gi is also a generator? Note that for gi to be a generator, it is
necessary and sufficient for 1, gi, g2i, · · · , g(p−2)i to be distinct modulo p.
3. The answer to the previous question is ϕ(p − 1) = ϕ(ϕ(p)). Now, we try to find a generator
modulo p2. Let g be a generator modulo p. Let h be the order of p modulo p2. Prove that
either h = p − 1 or h = p(p − 1).
4. Let g be a generator modulo p, where 0 < g < p. Then note that each of the numbers
g, g + p, g + 2p, · · · , g + (p − 1)p will be generators modulo p as well. Prove that gp−1, (g +
p)p−1, · · · , (g + (p − 1)p)p−1 are all distinct modulo p2.
5. Conclude from the previous exercise that among the numbers g, g+p, g+2p, · · · , g+(p−1)p,
exactly one of the numbers has order p−1 modulo p2, and the rest have order p(p−1) modulo p2.
6. Conclude that not only does there exist a number with order p(p − 1) modulo p2, but there
are, in fact, (p − 1) · ϕ(p − 1) = ϕ(ϕ(p2)) such numbers.
7. Let n ≥ 3 be a positive integer. Prove that if g is a generator modulo pn, then it is a generator
modulo p2. Conversely, show that if g is a generator modulo p2, then it is a generator modulo
pn. This shows that pn has pn−2 · (p − 1) · ϕ(p − 1) = ϕ(ϕ(pn)) generators modulo pn. This is where you might use LTE.
Question 13. From the previous exercise, we got that the set of numbers coprime to pn form a
cyclic group, where p is an odd prime and n is a positive integer.
1. Prove that for any odd prime p and natural number n, the group Z∗2 is cyclic. ·pn
2. Prove that the groups Z∗2 and Z∗4 are cyclic.
3. Prove that for m > 1, the group Z∗ is cyclic if, and only if, m = 2, 4, pn, or 2 · pn, for some m
odd prime p and some natural number n. 3