00 E 8 0 5 | Mon 01:37 93% iviwoods × 傑里科的書×螞蟻【全新X 全文:鸚鵡螺 × B【電子書】×【樂天Kobo] Answered: X F DC bartleby.com/questions-and-answers/if-p-7-mod-8-where-p-is-prime-show-th: ± 0 : + Q SEARCH ASK VX MATH SOLVER → Steps Part 1: If p = 7 (mod 8) where p is prime, show that p divides 2^((p-1)/2)-1 My approach was correct. When p = 7 (mod 8), we have p=8k +7 and (p-1)/2=4k+3. *how come By Fermat's Little Theorem: 2^(p-1) = 1 (mod p) This means 2^(8k+6) = 1 (mod p) Since (2^(4k+3))^2=2^(8k+6), we have (2^(4k+3))^2 = 1 (mod p) TL
00 E 8 0 5 | Mon 01:37 93% iviwoods × 傑里科的書×螞蟻【全新X 全文:鸚鵡螺 × B【電子書】×【樂天Kobo] Answered: X F DC bartleby.com/questions-and-answers/if-p-7-mod-8-where-p-is-prime-show-th: ± 0 : + Q SEARCH ASK VX MATH SOLVER → Steps Part 1: If p = 7 (mod 8) where p is prime, show that p divides 2^((p-1)/2)-1 My approach was correct. When p = 7 (mod 8), we have p=8k +7 and (p-1)/2=4k+3. *how come By Fermat's Little Theorem: 2^(p-1) = 1 (mod p) This means 2^(8k+6) = 1 (mod p) Since (2^(4k+3))^2=2^(8k+6), we have (2^(4k+3))^2 = 1 (mod p) TL
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 13E
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E 8 0 5 | Mon 01:37
93%
iviwoods × 傑里科的書×螞蟻【全新X 全文:鸚鵡螺 × B【電子書】×【樂天Kobo] Answered: X
F
DC bartleby.com/questions-and-answers/if-p-7-mod-8-where-p-is-prime-show-th: ± 0 :
+
Q SEARCH
ASK
VX MATH SOLVER
→ Steps
Part 1: If p = 7 (mod 8) where p is prime, show that p
divides 2^((p-1)/2)-1
My approach was correct. When p = 7 (mod 8), we have
p=8k +7 and (p-1)/2=4k+3.
*how come
By Fermat's Little Theorem: 2^(p-1) = 1 (mod p)
This means 2^(8k+6) = 1 (mod p)
Since (2^(4k+3))^2=2^(8k+6), we have (2^(4k+3))^2 = 1
(mod p)
TL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fa74a7fc4-965e-4912-be41-b0a4757a2f71%2Fets1eee_processed.png&w=3840&q=75)
Transcribed Image Text:00
E 8 0 5 | Mon 01:37
93%
iviwoods × 傑里科的書×螞蟻【全新X 全文:鸚鵡螺 × B【電子書】×【樂天Kobo] Answered: X
F
DC bartleby.com/questions-and-answers/if-p-7-mod-8-where-p-is-prime-show-th: ± 0 :
+
Q SEARCH
ASK
VX MATH SOLVER
→ Steps
Part 1: If p = 7 (mod 8) where p is prime, show that p
divides 2^((p-1)/2)-1
My approach was correct. When p = 7 (mod 8), we have
p=8k +7 and (p-1)/2=4k+3.
*how come
By Fermat's Little Theorem: 2^(p-1) = 1 (mod p)
This means 2^(8k+6) = 1 (mod p)
Since (2^(4k+3))^2=2^(8k+6), we have (2^(4k+3))^2 = 1
(mod p)
TL
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