135_7_-_Solutions

# 135_7_-_Solutions - Assignment 7 – Solutions 1 Prove...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment 7 – Solutions 1. Prove, without using a calculator, that 641 divides the Fermat number F (5) = 2 2 5 + 1 . Solution: We have to prove that 2 2 5 +1 ≡ 0 (mod 641); i.e. 2 32 ≡ - 1 (mod 641). We therefore need to compute large powers of 2 modulo 641. We note that 641 = 640+1 = 2 7 · 5+1, so 2 7 · 5 ≡ - 1 (mod 641) (2 7 · 5) 4 ≡ (- 1) 4 (mod 641) 2 28 · 625 ≡ 1 (mod 641) 2 28 (- 16) ≡ 1 (mod 641) , since 625 ≡ - 16 (mod 641)- 2 32 ≡ 1 (mod 641) 2 32 ≡ - 1 (mod 641) . Therefore, 641 | 2 32 + 1 . 2. Find 27 7 9 (mod 11). Solution1: By Fermat’s little thereom, since 11 6 | 27, 27 10 ≡ 1 (mod 11). Consider 7 9 (mod 10). Note that 7 2 ≡ 49 ≡ - 1 (mod 10) We have 7 9 ≡ (7 2 ) 4 · 7 ≡ (- 1) 4 · 7 ≡ 7 (mod 10) Thus 7 9 = 10 q + 7, for some integer q. The original formula 27 7 9 ≡ 27 10 q +7 ≡ 27 10 q · 27 7 ≡ 27 7 (mod 11) Since 27 ≡ 5 (mod 11), 27 7 9 ≡ 27 7 ≡ 5 7 ≡ 3 (mod 11) . Solution2: Since 27 ≡ 5 (mod 11) we have 27 7 9 ≡ 5 7 9 (mod 11). We make a list of powers of 5 modulo 11....
View Full Document

## This note was uploaded on 07/24/2011 for the course MATH 135 taught by Professor Andrewchilds during the Winter '08 term at Waterloo.

### Page1 / 3

135_7_-_Solutions - Assignment 7 – Solutions 1 Prove...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online