{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw1 - Solutions to Assignment 1 September 9 2009 Section...

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

Solutions to Assignment 1 September 9, 2009 Section 1.2 1a. Apply Euclidean Algorithm: 7469 = 3 × 2464 + 77 2464 = 32 × 77 So (7469 , 2464) = 77. 3a Note that 423 / 9 = 47 , 198 / 9 = 22. So it suffice to find integer solution to the equation 47 x + 22 y = 1 Apply Euclidean Algorithm: 47 = 2 × 22 + 3 22 = 7 × 3 + 1 3 = 3 × 1 Working back (or you can also use matrices): 1 = 22 - 7 × 3 = 22 - 7 × (47 - 2 × 22) = - 7 × 47 + 15 × 22 So a solution is x = - 7 , y = 15 (you may obtain other solutions). 14. Any odd number n is of the form 4 k + 1, or 4 k + 3. If n = 4 k + 1, then n 2 = (4 k + 1) 2 = 16 k 2 + 8 k + 1 = 8(2 k 2 + k ) + 1 Similarly, if n = 4 k + 3, then n 2 = 16 k 2 + 24 k + 9 = 8(2 k 2 + 3 k + 1) + 1 So 8 | ( n 2 - 1) in all cases. 1

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

View Full Document
15. Write x = 2 m + 1 , y = 2 n + 1. Then x 2 + y 2 = (2 m + 1) 2 + (2 n + 1) 2 = (4 m 2 + 4 m + 1) + (4 n 2 + 4 n + 1) = 4( m 2 + n 2 + m + n ) + 2 So x 2 + y 2 is even, but not divisible by four. In fact, it’s an integer of the form 4 k + 2. Section 1.3 24. Argue by contradiction. Suppose n is composite, and all the prime factors of n are > n 1 / 2 . Write n = ab , with a, b 6 = 1. The prime factors of a, b are those of n , hence are > n 1 / 2 . In particular, a, b > n 1 / 2 . But then n = ab > n 1 / 2 × n 1 / 2 = n a contradiction. Hence some prime factor of
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

hw1 - Solutions to Assignment 1 September 9 2009 Section...

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

View Full Document
Ask a homework question - tutors are online