HW5_solns

# HW5_solns - MATH 3360 Sketch of solutions for Assignment 5...

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

MATH 3360 Sketch of solutions for Assignment 5 9C.49 Since ϕ (18) = 6, we have 5 6 = 5 · 5 5 1(mod 18). Answer: 5 5 . 9C.50 [graded out of 2 points] f 1(mod ϕ ( m )) f = ( m ) + 1 for some k Z . Thus, a f = a ( m )+1 = ( a ϕ ( m ) ) k · a 1 · a (mod m ) by Euler. 9C.62 [graded out of 5 points] We are given that the order of 2 is q - 1, so 2 q - 1 1(mod q ). We also have ( q, 2) = 1, so 2 is a unit and by Euler 2 ϕ ( q ) 1(mod q ). Since q - 1 is the order of 2, we must have q - 1 ϕ ( q ). By deﬁnition of ϕ , we must also have ϕ ( q ) q - 1, so ϕ ( q ) = q - 1 q is prime. Note that 2 10 1(mod 341). 10A.7 [graded out of 2 points] m = pq , ϕ ( m ) = ( p - 1)( q - 1) = pq - p - q + 1 = m - p - m/p + 1 ( m ) = pm - p 2 - m + p p 2 + ( ϕ ( m ) - m - 1) p + m = 0. Also note that m and ϕ ( m ) are both symmetric in p and q . Thus, a quadratic equation x 2 + ( ϕ ( m ) - m - 1) x + m = 0, has both p and q as its solution. 11B.11

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.

{[ snackBarMessage ]}

### Page1 / 2

HW5_solns - MATH 3360 Sketch of solutions for Assignment 5...

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

View Full Document
Ask a homework question - tutors are online