PRACTICE PROBLEMS:
PRIMITIVE ROOTS AND ARITHMETIC FUNCTIONS
Problem 1.
(a) Find a primitive root mod 13.
(b) Find all primitive roots mod 13.
Solution.
(a) We need to find a ∈ Z with order ϕ(13) = 12, so it suffices to find a ∈ Z coprime to 13
with a4 6≡ 1 mod 13 and a6 6≡ 1 mod 13. We compute 24 ≡ 3 and 26 ≡ −1, so 2 works.
(b) From a homework ord(αe ) = ord(α)/ gcd(e, ord(α)). Thus, the primitive roots are 2, 25 , 27 , 211 , and
indeed we know there must be ϕ(12) = 2 · 2 = 4 elements of order 12. Evaluating the powers, these
are 2, 6, 11, 7.
Problem 2. What is the order of 6 mod 19?
Solution. Let d = ord19 (6) be the order of 6 mod 19. ϕ(19) = 18 = 2 · 32 , so d = 2, 3, 6, 9, 18. 66 ≡ (−2)3 ≡
−8, so d = 9, 18 and 69 ≡ (−8) · (−2) · 6 ≡ −18 ≡ 1. Thus, d = 9.
Problem 3. Find a primitive root mod 11100 .
Solution. We know for an odd prime p that: (a) there is some a ∈ Z that is a primitive root mod p; (b) if
a ∈ Z is a primitive root mod p, either a or a + p is a primitive root mod p2 ; (c) if a ∈ Z is a primitive root
mod p2 then it is a primitive root mod pe for all e ≥ 2. Thus, we compute: (a) 2 is primitive mod 11 since
22 6≡ 1 mod 11 and 25 6≡ 1 mod 11; (b) ord112 (2) is either 10 or 110, and since 210 ≡ 128 · 8 ≡ 7 · 8 6≡ 1 mod 112 ,
2 is a primitive root mod 112 . 2 is therefore a primitive root mod 11e for all e ≥ 1.
Problem 4. Suppose we have distinct α, β ∈ (Z/n)∗ . Show that ord(αβ)| lcm(ord(α), ord(β)). Must we
have equality?
Solution. Let d = lcm(ord(α), ord(β)). Certainly αd = 1 = β d , so (αβ)d = 1 and thus ord(αβ)|d. If α is not
of order 2, then β = α−1 6= α, and ord(αβ) = 1.
Problem 5. Find all solutions to x5 ≡ −1 mod 112
Solution. Substituting x = 2d , and using the fact that −1 = 255 (which we know because 255 must have
order 2) the congruence is equivalent to 5d ≡ 55 mod 110, so we must have d ≡ 11 mod 22, and the 5
solutions mod 110 are 11, 11 + 22, 11 + 2 · 22, 11 + 3 · 22, 11 + 4 · 22 = 11, 33, 55, 77, 99. Thus, the solutions to
x5 ≡ −1 mod 112 are 211 , 233 , 255 , 277 , 299 . If we really wanted to we could evaluate these to be 112, , −1, Problem 6. Find all solutions to x4 ≡ 5 mod 2100
Solution. Recall that 5 has order 2e−2 mod 2e , and that every invertible residue class mod 2e can be written
as ±5d . Therefore substituting x = ±5d , the equation implies 4d ≡ 1 mod 8, which has no solutions.
Problem 7. Find all solutions to x2 ≡ 16 mod 77 (you may use the fact that 3, 2 are primitive roots mod 7, 11
respectively).
Solution. This is equivalent to solving x2 ≡ 16 mod 7 and x2 ≡ 16 mod 11. Both of these have the solution
±4, so there are 4 solutions mod 77 given by solving x ≡ ±4 mod 7 and x ≡ ±4 mod 11. x ≡ ±4 mod 77
work, and so do x ≡ ±18 mod 77 since 18 ≡ 4 mod 7 and 18 ≡ −4 mod 11. Note we didn’t even have to use
primitive roots explicitly.
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Problem 8. How many solutions does x11 ≡ −1 mod 23100000000 have?
Solution. There are clearly solutions because x ≡ −1 is a solution. The number of solutions is then
gcd(11, ϕ(n)) for n = 23100000000 , and since ϕ(23100000000 ) = 2399999999 · 22, there are 11 solutions. Note we
could have easily seen from our criterion that there are solutions since (−1)ϕ(n)/ gcd(11,ϕ(n)) ≡ 1.
Problem 9. Show that there are no solutions to xn ≡ −1 mod 2e for e ≥ 3 and n even.
Solution. We know every invertible residue mod 2e is congruent to ±5d for a unique choice of ± and d with
0 ≤ d < 2e−2 . Setting x = ±5d and plugging in, we must have 5nd ≡ −1 mod 2e . By uniqueness, we can
never write −1 as +5f , so there cannot be any solutions.
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Problem 10. Suppose we have two functions f, g : N → C with f (n) = d|n g(d), and suppose further that
f (n) is zero if n is divisible by a nontrivial square (i.e. m2 for m > 1). Show that g(n) is zero for n divisible
by a nontrivial cube.
Solution. By Möbius inversion,
g(n) =
X
f (d)µ(e)
de=n
d,e∈N
If n is divisible by m3 for m > 1, then for any factorization de = n, either m2 |d or m2 |e in which case either
f (d) = 0 or µ(e) = 0.
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Problem 11. Define a function f : N → C implicitly by n2 = d|n f (d). What is f (pe q f ) for p, q distinct
primes?
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2
e
e−1 2
Solution. By Möbius inversion, f (n) =
) (−1) + (pe )2 =
d|n d µ(n/d), so evaluating f (p ) = (p
2e−2 2
f
p
(p − 1), and similarly for f (q ). The Dirichlet product of two multiplicative functions is multiplicative,
so
f (pe q f ) = f (pe )f (q f ) = p2e−2 q 2f −2 (p2 − 1)(q 2 − 1)
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P
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Problem 12. Recall that τ (n) = d|n 1. Express d|n τ (d)ϕ(n/d) in terms of and σ(n) = d|n d.
Solution. This is easiest using the notion of Dirichlet product. Let u(n) = 1 for all n ∈ N and N (n) = n.
Then τ = u ∗ u and we know N = u ∗ ϕ, so τ ∗ ϕ = u ∗ u ∗ ϕ = u ∗ N = σ.
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k
k
Problem 13. Recall that σk (n) = d|n d for k ∈ Z. Show that n σ−k (n) = σ(n).
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Solution. We know if g(n) is multiplicative then f (n) = d|n g(d) is too. Thus, σk (n) is multiplicative for
all k, as is ρk (n) = nk σ−k (n). We just need to show that ρk and σk agree on prime powers:
ρk (pe ) = (pe )k 1 + p−k + (p2 )−k + · · · + (pe )−k = (pe )k + (pe−1 )k + (pe−2 )k + · · · + 1 = σk (pe )
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