Math 445 Homework 7 Solutions
25. Show that if p is an odd prime and a is a primitive root mod p, then
a
= 1 .
p
p 1
a
a 2 (mod p) . Since a is a primitive root mod p, ordp (a) =
p
p 1
p 1, so x = a 2 1 (mod p), since p1 < p 1. but x2 = ap1 1 (mod p) so,
Math 445 Homework 6 SOlutions
21. Show that if an integer n can be expressed as the sum of the squares of two rational
numbers
c
a
(*) n = ( )2 + ( )2 ,
b
d
then n can be expressed as the sum of the squares of two integers.
(Hint: Not directly! Show that
Math 445 Homework 5
Due Friday, October 10
17. Show that if a2 + 2b2 = c2 and p|(a, c) for some prime p, then p|b.
(Note: p = 2 is treated dierently.)
If p|(a, c), then a = px, c = py for some integers x, y , so a2 + 2b2 = c2 implies
2b2 = c2 a2 = (p2 )(y
Math 445 Homework 4 solutions
13. Show that if n|m, and (10, m) = 1, then the period of the decimal expansion of 1/n
divides the period of the decimal expansion of 1/m .
Translating this into the language of orders, if n|m and (10, m) = 1, then we wish to
Math 445 Homework 3 Solutions
9. Our description of RSA assumed that for n = pq , that (a, n) = 1 . But we dont
control a, the sender does! Show that in any event, the RSA algorithm works even if
(A, n) > 1 :
Show that if n = pq is a product of distinct p
Math 445 Homework 1 solutions
1. (NZM, Problem 1.3.27) Show that if n is not prime, then n|(n 1)! .
If nisnt prime, then n = ab , with 1 < a b < n . Then a and b are both among the factors
of (n 1)! . So if they are dierent, then ab|1 (a 1)a(a +1) (b 1)b(