Number Theory and Cryptography
Homework 4 (due 2/17) (with solution for problem
5)
1. a) Prove that 5 is a generator of Z×
23 .
b) Find all generators of Z×
,
and
express
each of them as a power of 5.
23
2. Prove that 11 is a generator of Z×
73 .
3. Let p be a prime with p = 1 (mod 4), and suppose a ∈ Z×
p is a generator.
p−1
a) Prove that x = a 4 has order exactly 4.
b) Prove that x and −x both satisfy the equation x2 = −1 in Zp .
c) Find two solutions of the equation x2 = −1 in Z73 . (Hint: Use problem 2.)
2
4. We say x ∈ Z×
n is a square if there is some t ∈ Z with t = x (mod n).
×
Suppose we have two elements x, y ∈ Zn such that x is a square and xy is a
square. Prove that y must be a square as well.
5. Let a, b ∈ Z×
n be two elements with orders j, k.
a) Prove that the order of ab divides jk.
b) Prove that if (j, k) = 1 then ab has order exactly jk.
Solution: For brevity we set h = ordn (ab).
a) We have (ab)jk = ajk bjk = (aj )k (bk )j = (1)k (1)j = 1 in Zn . Since we
m
prove in class that (for any c ∈ Z×
n ) c = 1 in Zn exactly when ordn (c) divides
jk
m, so (ab) = 1 implies that h divides jk.
b) From part a), we know that h|jk, which means in particular that h ≤ jk.
Therefore, it’s enough to prove that under the assumption (j, k) = 1, we have
jk ≤ h.
First we show that j|hk and k|hj. By the result recalled in our solution for
part a), we know that j|hk if and only if ahk = 1 in Zn . To show that ahk = 1,
we note that bk = 1, so bhk = 1h = 1, and therefore
ahk
=
ahk · 1
=
=
ahk · bhk
(ab)hk
=
=
((ab)h )k
(1)k
=
1.
Likewise, swapping the roles of a, b and j, k shows that bhj = 1 and therefore
that k|hj.
Since j|hk, and (j, k) = 1 by assumption, we deduce that j|h. (This is
problem 1 on HW3.) Switching j and k, we also get that k|h. In particular,
1
h is a common multiple of j and k. Since lcm(j, k) is defined as the smallest
common multiple, we now see that lcm(j, k) ≤ h. But by problem 4 on HW2,
jk
we have lcm(j, k) = (j,k)
= jk
1 = jk, so we deduce that jk ≤ h as desired.
2