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(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).

(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).

Advanced Engineering Mathematics
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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(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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