(10) Prove that for any prime number P>3. p² = 1 (mod 24) (Hint: chinese remainder theorem tells us need to show SP² = 1 (mod 3) 2p² = 1 (mod 8) we only The hard part is to show p² = 1 (mod 8), which means 81 p2-1 = P (P-1) (P+) we know that must be odd. Hence, P two integers. P+1 are " So it is enough to show 21욎. 왤).

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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(10) Prove that for any prime number P>3.
p² = 1 (mod 24)
(Hint: chinese remainder theorem tells us
SP² = 1 (mod 3)
p² = 1 (mod 8)
need to show
we only need to
p²=1 (mod 8), which
we know that
P
PH
two integers.
The hard part is to show
means S1 p²2-1
81 =
(P-1) (PH).
must be odd. Hence, P-1,
So it is enough to
show
2 |
P-1
F
are
PH).
Transcribed Image Text:(10) Prove that for any prime number P>3. p² = 1 (mod 24) (Hint: chinese remainder theorem tells us SP² = 1 (mod 3) p² = 1 (mod 8) need to show we only need to p²=1 (mod 8), which we know that P PH two integers. The hard part is to show means S1 p²2-1 81 = (P-1) (PH). must be odd. Hence, P-1, So it is enough to show 2 | P-1 F are PH).
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