Math 7 Spring 2017
TA: Serin Hong
PROBLEM SET 3 SOLUTIONS
1. Solve the following system of congruences:
x ≡ 2 (mod 5)
x ≡ 5 (mod 7)
x ≡ 3 (mod 11)
Solution. There are many ways of solving this system. We will present a solution using the constructive
proof of the Chinese Remainder Theorem.
The product 7 · 11 = 77 ≡ 2 (mod 5) has multiplicative inverse 3 (mod 5). Similarly, the products
5 · 11 = 55 ≡ 6 (mod 7) and 5 · 7 = 35 ≡ 2 (mod 11) respectively have multiplicative inverse 6 (mod 7)
and 6 (mod 11). Hence the solution is given by
x ≡ 2 · (7 · 11) · 3 + 5 · (5 · 11) · 6 + 3 · (5 · 7) · 6 = 2742 ≡ 47 (mod 385).
2. (a) Show that if a and b are each a sum of two squares, so is ab. In other words, show that for any
x, y, u, v ∈ Z, (x2 + y 2 )(u2 + v 2 ) is a sum of two squares. (Hint: ±2xyuv).
Solution. For the proof, we simply observe the following identity:
(x2 + y 2 )(u2 + v 2 ) = x2 u2 + y 2 u2 + x2 v 2 + y 2 v 2
= (x2 u2 + y 2 v 2 + 2xyuv) + (y 2 u2 + x2 v 2 − 2xyuv)
= (xu + yv)2 + (yu − xv)2 .
(b) Prove that if every prime divisor of a number n ∈ N is a sum of two squares, so is n.
Solution. This is an immediate consequence of part (a).
Math 7 Spring 2017
TA: Serin Hong
Dirichlet products of arithmetic functions.
Any f : N −→ C is called an arithmetic function. If f, g are two such functions, their Dirichlet product
p = f ∗ g is defined by the formula
p(n) =
X
f (d)g
n
d
d|n
.
You can assume the following relations, which are stragith-forward to verify:
f ∗ g = g ∗ f,
f ∗ (g ∗ h) = (f ∗ g) ∗ h.
For each of the following problems, you can assume the ones before it as fact.
3. Let f be an arithmetic function.
(a) Verify that e ∗ f = f ∗ e = f , where e : N −→ C is
e(n) =



1
if n = 1


0
if n > 1
.
Solution. Since we know that e ∗ f = f ∗ e, it suffices to prove that e ∗ f = f . In fact, we have
(e ∗ f )(n) =
X
e(d)f
d|n
n
d
= e(1)f (n) = f (n)
for all n ∈ N
where the second equality follows since e(d) = 0 for d > 1.
(b) Prove that if f (1) 6= 0, there exists a unique arithmetic function f −1 such that
f ∗ f −1 = f −1 ∗ f = e.
(Hint: Write the formula for (f ∗ f −1 )(k) and use it to prove that the values f −1 (r), for r < k,
determines f −1 (k) uniquely. Then apply induction.)
Solution. We will present a recursive formula for the values of f −1 . For every integer k we have
e(k) = (f
−1
∗ f )(k) =
X
f
−1
d|k
k
(d)f
.
d
When k = 1, this simplifies to 1 = (f −1 ∗ f )(1) = f −1 (1)f (1), so we find that
f −1 (1) =
1
.
f (1)
(1)
Math 7 Spring 2017
TA: Serin Hong
For k > 1, (1) yields
0=
X
f −1 (d)f
d|k
k
.
d
We can write this as
0=f
−1
(k)f (1) +
X
f
−1
d|k
d<k
k
(d)f
d
which yields
f −1 (k) = −
1 X −1
k
f (d)f
.
f (1)
d
d|k
d<k
This gives a recursive formula for an arithmetic function f −1 satisfying f −1 ∗ f = e. Note that we
also have f ∗ f −1 = e, as f ∗ f −1 = f −1 ∗ f . Uniqueness is also clear since the proof shows that any
such arithmetic function should satisfy this recursive formula.
αr
1
The Moebius function µ is defined as follows: If n has prime factorization n = pα
1 · · · pr ,
µ(n) =



(−1)r


0
Let
if α1 = α2 = · · · = αr = 1
.
otherwise
1 denote the constant function 1 = 1.
4. Prove that
1 = µ−1 .
αr
1
(Hint: Suppose n > 1 has prime factorization n = pα
1 · · · pr . Express the formula for (1 ∗ µ)(n) as a
sum over subsets of {p1 , · · · , pr }. Show the expression is equal to the binomial expansion of (1 − 1)r . )
Solution. It suffices to prove that µ ∗ 1 = e. For each n ∈ N, we have
(µ ∗ 1)(n) =
X
µ(d)1
d|n
n
d
=
X
µ(d).
(2)
d|n
β1
αr
βr
1
Let n = pα
1 · · · pr be a prime factorization of n. Then every divisor d of n is of the form d = p1 · · · pr
where 0 ≤ βj ≤ αj for j = 1, 2, · · · , r. Since µ(d) = 0 if βj > 1 for some j, we have µ(d) 6= 0 if and only if d
divides p1 · · · pr . Hence we may write (2) as
(µ ∗ 1)(n) =
X
d|p1 ···pr
µ(d).
(3)
Math 7 Spring 2017
TA: Serin Hong
Now if d divides p1 · · · pr , we have µ(d) = (−1)s where s is the number of distinct prime factors of d. For a
r
fixed s, the number of d’s with µ(d) = (−1)s is
. Hence we have
s
X
µ(d) =
r X
s=0
d|p1 ···pr



1
r
(−1)s = (1 − 1)r =

s

0
if r = 0
otherwise
where the middle equality follows from the binomial theorem. Note that r = 0 if and only if n = 1. We thus
obtain from (3) the desired identity µ ∗ 1 = e.
5. Prove the Moebius inversion formula: If f is an arithmetic function, and
g(n) =
X
f (d),
d|n
then
f (n) =
X
g(d)µ
n
d|n
d
.
(Hint: Write g as a Dirichlet product, apply Problem 4.)
Solution. We may write g = f ∗ 1 since
g(n) =
X
f (d) =
d|n
X
f (d)1
d|n
n
d
= (f ∗ 1)(n)
for each n ∈ N.
Then we have f = f ∗ e = f ∗ (1 ∗ µ) = (f ∗ 1) ∗ µ = g ∗ µ, where the first two equalities respectively follow
from Problem 3 (a) and Problem 4. This gives the desired identity.
6. Show that ϕ ∗ 1 = N , where ϕ is the Euler ϕ-function and N (n) = n.
(Hint: Show this is a property of ϕ proved in class.)
Solution. For each n ∈ N we have
(ϕ ∗ 1)(n) =
X
d|n
ϕ(d)1
n
d
=
X
ϕ(d) = n
d|n
where the last equality follows from a property of ϕ proved in class. Hence we find ϕ ∗ 1 = N , as desired.
Math 7 Spring 2017
TA: Serin Hong
7. Prove that
ϕ(n) =
X n
µ
d.
d
d|n
(Hint: Apply the Moebius inversion formula.)
Solution. This is an immediate consequence of the Moebius inversion formula (Problem 5) applied to f = ϕ
and g = N .