ELEMENTARY NUMBER THEORY PROBLEM SHEET 4
Exercise 1. Suppose a is a positive integer which is coprime to 10. Show that
a2001 ≡ a (mod 1000)
(In other words: the last three decimal digits of a2001 are the same as those of a.)
Solution 1. We have φ(1000) = 400 and 2001 ≡ 1 (mod 400). So we have a400 ≡ 1
(mod 1000) by the Fermat–Euler theorem. So a2001 ≡ a1 (mod 1000).
×
Exercise 2. Find the order of each element of Z×
9 and Z10 .
Solution 2. We have Z×
9 = {[1], [2], [4], [5], [7], [8]}. The order of an element is a
divisor of φ(9) = 6. So the possibilities for the orders are 1, 2, 3 and 6. We have
o([1]) = 1, o([2]) = 6 (as 23 and 22 are not congruent to 1 mod 9), o([4]) = 3 (since
4 = 22 ), o([5]) = 6 (since 5 is an inverse to 2 modulo 9), o([7]) = 3 (since 7 ≡ 42
(mod 9)), o([8]) = 2.
One systematic way of finding the orders is to find i ∈ 0, 1, . . . , 5 for each congruence class [a] such that a ≡ 2i (mod 9). Then the order of a modulo 9 is
φ(9)/ gcd(i, φ(9)).
×
Now we do the Z×
10 example: we have Z10 = {[1], [3], [7], [9]} and the order of an
element is a divisor of φ(10) = 4. So the possibilities are 1, 2, 4. We have o([1]) = 1,
o([3]) = 4 (since 32 ≡ −1 (mod 10)), o([7]) = 4, and o([9]) = 2.
Exercise 3. If a is a primitive root modulo an odd prime number p, show that ap
is not a primitive root modulo p2 .
2
Solution 3. We have aφ(p ) = ap(p−1) ≡ 1 (mod p2 ) by the Fermat–Euler theorem.
So ap has order modulo p2 a divisor of p − 1 (in fact it must be equal to p − 1 since
ap ≡ a (mod p) is a primitive root modulo p).
Exercise 4.
(1) Find a primitive root modulo 17.
(2) Find a primitive root modulo 5n for all n.
(3) Find all the primitive roots modulo 5.
(4) Find all the primitive roots modulo 25.
Solution 4.
(1) We have φ(17) = 16. So a is a primitive root modulo 17
if and only if a8 6≡ 1 (mod 17). We have 24 = 16 ≡ −1 (mod 17) so
28 ≡ 1 (mod 17). So 2 is not a primitive root. Next we try 3. We have
34 = 81 ≡ −4 (mod 17). So 38 ≡ 16 ≡ −1 (mod 17). So 3 is a primitive
root modulo 17.
(2) First we note that 2 is a primitive root modulo 5 (since 22 ≡ −1 (mod 5)).
Next we note that 24 = 16 6≡ 1 (mod 25). So 2 is also a primitive root
modulo 25. Now a result from lectures (if a is a primitive root modulo p2
for an odd prime p, then a is a primitive root modulo pn for all n ≥ 2)
shows that 2 is a primitive root modulo 5n for all n ≥ 1.
(3) Since 32 ≡ −1 (mod 5), 3 is a primitive root. −1 is not, so the primitive
roots are the integers congruent to 2, 3 (mod 5).
(4) Here are two ways of doing this: First, we know that 2 is a primitive root
modulo 25. So the primitive roots modulo 25 are given by the congruence
classes of 2i for i coprime to φ(25) = 20. In particular, the primitive
Date: Tuesday 28th February, 2017.
1
2
ELEMENTARY NUMBER THEORY PROBLEM SHEET 4
roots are given by φ(20) = 8 congruence classes modulo 25. You can then
explicitly work out all these powers of 2 if you like.
Another way is as follows: we know that the primitive roots are given
by 8 = φ(φ(25)) congruence classes modulo 25. A primitive root modulo
25 has to be a primitive root modulo 5, so the candidates are the integers
congruent to 2, 3 mod 5. This gives 10 congruence classes in Z×
25 . So we just
need to exclude two of these congruence classes. The previous exercise tells
us that 25 ≡ 7 (mod 25) and 35 ≡ −7 (mod 25) are not primitive roots. So
the primitive roots modulo 25 are all the integers x ≡ 2, 3 (mod 5) except
integers which are congruent to ±7 (mod 25).
Exercise 5.
(1) Find all integer solutions to x6 ≡ 4 (mod 23).
(2) Find all integer solutions to x4 ≡ 4 (mod 99). (Hint: Use the Chinese
Remainder Theorem and primitive roots modulo 9 and 11)
Solution 5.
(1) First we find a primitive root modulo 23. We have φ(23) =
22 = 2 · 11. So a, coprime to 23, is a primitive root if and only if a11 and a2
are not congruent to 1 mod 23. We have 211 ≡ 2(25 )2 ≡ 2(9)2 ≡ 2 · 12 ≡ 1
(mod 23). So (−2)11 ≡ −1 (mod 23) and −2 is a primitive root modulo
23. We have 4 = (−2)2 . So we need to solve (−2)6i ≡ (−2)2 (mod 23),
or in other words, 6i ≡ 2 (mod 22), or 3i ≡ 1 (mod 11). We have [3]−1
11 =
[4]11 so we get i ≡ 4 (mod 11). So the solutions are x ≡ (−2)4 , (−2)15
(mod 23). We have (−2)15 ≡ −(−2)4 , since (−2)11 ≡ −1 (mod 23). We
have (−2)4 = 16 ≡ −7 (mod 23). So finally we get the answer x ≡ ±7
(mod 23).
(2) First we separately solve x4 ≡ 4 (mod 9) and x4 ≡ 4 (mod 11). 2 is a
primitive root modulo 9. The equation 24i ≡ 22 (mod 9) is equivalent to
4i ≡ 2 (mod 6) which is equivalent to 2i ≡ 1 (mod 3) which is solved by
i ≡ 2 (mod 3). So the solutions to x4 ≡ 4 (mod 9) are given by x ≡
22 , 25 ≡ ±4 (mod 9).
Exercise 6. Suppose n = pq, with p and q distinct prime numbers. What is the
maximum order of an element of Z×
n ? (Hint: Use the Chinese Remainder Theorem)
Solution 6. We have xi ≡ 1 (mod n) if and only if xi ≡ 1 (mod p) and xi ≡ 1
(mod q). This happens whenever i is a common multiple of p − 1 and q − 1. So
we have xlcm(p−1,q−1) ≡ 1 (mod n) for all x coprime to n. Conversely, if we take x
congruent to a primitive root modulo p and a primitive root modulo q, then x has
order exactly lcm(p − 1, q − 1). So the maximum order is lcm(p − 1, q − 1).
Exercise 7. (*) Suppose m, n are coprime positive integers. Show that 2m − 1 and
2n − 1 are coprime.
Solution 7. Let d = gcd(2m − 1, 2n − 1). The order of 2 modulo d divides m, since
2m − 1 ≡ 0 (mod d). Similarly, the order of 2 modulo d divides n. So the order
of 2 modulo d is a common divisor of m, n. Since m, n are coprime, the order of 2
modulo d must be 1. So 2 ≡ 1 (mod d). This implies that d = 1.
Exercise 8. (**)
n−2
(1) Let n ≥ 2. Show that the order of 5 in Z×
.
2n is 2
×
(2) Show that Z2n is isomorphic, as a group, to a product of cyclic groups
C2 × C2n−2 of order 2 and 2n−2 .
(3) Use the Chinese Remainder Theorem to establish the fact quoted in lectures
i
i
that Z×
m is cyclic if and only if m = 1, 2, 4, p or 2p for some odd prime p
and positive integer i.
ELEMENTARY NUMBER THEORY PROBLEM SHEET 4
3
n−2
Solution 8.
(1) We have φ(2n ) = 2n−1 . So we need to show that 52
≡1
n
2n−3
n
2n−2
(mod 2 ) and (if n ≥ 3) 5
6≡ 1 (mod 2 ). In fact we have a
≡1
(mod 2n ) for every odd integer a and n ≥ 3. Let’s prove this by induction on
n−2
n. For n = 3 we can check by hand. If it’s true for n, then we have a2
=
n−1
1 + 2n b for some integer b. So a2
= (1 + 2n b)2 = 1 + 2n+1 b + 22n b2 ≡ 1
(mod 2n+1 ), and it’s also true for n + 1.
n−3
Now we show that 52
6≡ 1 (mod 2n ). If you apply the binomial
n−3
2
theorem to (1 + 4)
you’ll see that it looks reasonable to guess that
n−3
52
≡ 1 + 2n−1 (mod 2n ). We are going to prove this by induction.
n−3
Again, for n = 3 it’s easy to check. Now suppose 52
= 1 + 2n−1 + 2n b
for some integer b. Squaring both sides gives
n−2
52
= (1 + 2n−1 )2 + 22n b2 + 2n+1 b(1 + 2n−1 ) ≡ (1 + 2n−1 )2
(mod 2n+1 )
We have (1+2n−1 )2 = 1+2n +22n−2 and since n ≥ 3 we have 2n−2 ≥ n+1,
so (1 + 2n−1 )2 ≡ 1 + 2n (mod 2n+1 ). This gives the required inductive step.
(2) We are going to show that Z×
2n is a product of its cyclic subgroups, h[−1]i
and h[5]i generated by −1 and 5 respectively.
In other words, we want to show that the group homomorphism
h[−1]i × h[5]i → Z×
2n
(x, y) 7→ xy
is an isomorphism.
We have shown that the cyclic subgroup generated by [5] has order 2n−2 .
So the product of the orders of the two cyclic subgroups is equal to the order
of Z×
2n . So it suffices to check that the group homomorphism is injective (an
injective homomorphism between two finite groups of the same cardinality
is an isomorphism). The kernel of the homomorphism is non-trivial if and
only if [−1] is contained in the cyclic subgroup generated by [5]. To show
that it is not, we consider the square roots of 1 in h[5]i. Since this is a cyclic
n−3
group of order 2n−2 the only square roots of 1 are given by [1] and [5]2 .
n−3
So if [−1] is in the subgroup generated by [5], we have [−1] = [5]2 . But
n−3
we have already shown that 52
≡ 1 + 2n−1 (mod 2n ), and this is not
congruent to −1. So [−1] is not in the subgroup generated by [5] and we
are done.
Qr
(3) Suppose n = 2a i=1 pai i , with pi distinct odd primes and ai > 0. By the
Chinese Remainder Theorem, we have
∼
Z×
n =
r
Y
i=1
Z×
a
p i
i
if a = 0, and
∼ ×
Z×
n = Z2a ×
r
Y
i=1
Z×
a
p i
i
if a > 0. We have to decide if these groups are cyclic. First suppose that
there are at least two distinct odd prime factors of n, p1 6= p2 . Then every
a
a
×
lcm(φ(p1 1 ),φ(p2 2 ))
element of the group Z×
= [1]. Since
a × Z a2 satisfies [a]
p1 1
p2
p1 , p2 are odd, φ(pa1 1 ) and φ(pa2 2 ) are both even, so this lcm is less than
×
×
φ(pa1 1 pa2 2 ). So Z×
a × Z a2 is not cyclic. This implies that Zn is not cyclic.
p 1
p
1
2
So, for Z×
n to be cyclic, we have at most one odd prime factor of n. So we
can write n = 2a pb , with p an odd prime. If a = 0 then Z×
n is cyclic. If
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ELEMENTARY NUMBER THEORY PROBLEM SHEET 4
b = 0 then it follows from the previous part that n = 2a is cyclic if and
only if a ≤ 2. Finally, we assume that a, b are both > 0. So
∼ Z×a × Z×b
Z× =
n
2
p
we must have a ≤ 2, otherwise one of the factors in the product is not
×
× ∼
b
cyclic, so Z×
n is not cyclic. We have Z22 × Zpb = C2 × Cφ(pb ) and φ(p ) is
even, so this group is not cyclic (by the earlier lcm argument). We conclude
∼ ×
that a = 0, 1. In these cases, Z×
n = Zpb which is cyclic. Putting everything
together, we have shown the fact we wanted to show.