Math327-Homework 1
Ali Akçay
October 26, 2023
Question 1.
(a) Prove that a positive integer is not expressible as a sum of consecutive integers if and only if it
is not a power of 2.
(b) Prove that an odd integer n>1 is prime if and only if it is not expressible as a sum of three or
more consecutive integers.
Proof.
(a) Let m be a power of 2 and say that it is a sum of consecutive integers. Also say that
there are l summands. Meaning m = (n + 1) + (n + 2) + ... + (n + l)
Case 1 - l is even: Then m is divisible by (n + 1)+(n + l) Which is an odd integer since
l + 1 is odd. Which is a contradiction.
(l + 1)
Which means l divides m but
Case 2 - l is odd: Then we see that m = nl + l
2
since l is odd this is a contradiction.
(b) Let p be a prime number. if p is equal to 2 or 3 then the statement is clearly true. If not,
say p = (n + 1) + (n + 2) + ... + (n + l) where l is at least 3, similar to (a), if l is odd then
l divides p but l<p so contradiction. If l is even (n + 1)+(n + l) divides p which is again
strictly smaller then p, since there must be at least 4 summands.
Conversely suppose n>1 is a non-prime odd integer and let p be the smallest prime
dividing n. Write n = pk, we have
p−1
p−1
n =(k −
) + ... + (k − 1) + k + (k + 1) + ... + (k +
)
2
2
Where there are at least 3 summands since p ≥ 3, and also the beginning term can not
be 0 as p ≤ k.
Question 2. Let a and b be positive integers such that (1 + ab) (a2 + b2 ). Show that the integer
a2 + b2
must be the square of an integer.
1 + ab
1
1
Let a and b two positive integers satisfying the equation and a ≥ b. Suppose for
a2 + b 2
contradiction that k =
is not a perfect square. Replace a with x in the equation and
1 + ab
obtain
x2 − kbx + b2 − k = 0
Proof.
We see that this is a second order rational coefficient polynomial equation and since
it already has a solution in rationals it also has a second solution in rationals, call it c. c
b2 − k
satisfies c = kb − a which shows it is an integer, c =
which shows it is non-zero since
a
c 2 + b2
could
k is not a perfect square by assumption. Also if c were negative then 1 + bc =
k
not have been satisfied since the right hand side is positive. This shows us that we have
x2 + y 2
obtained another positive 2-integer solution (c, b) to
= k. Now we have,
1 + xy
a≥b≥
b2
b2 − k
b2 − k
>
≥
=c
b
b
a
So c is strictly less then both a and b. Replacing b with c, a with b and carrying on
the same calculations we see that we get an infinite, strictly decreasing, positive series of
integers a + b, b + c, ... . But this is impossible since any strictly decreasing series of positive
integers must terminate. Contradiction.
Question 3. Let σ−1 (n) denote the sum of the reciprocals of the divisors of n.
P
(a) Express σ−1 (n) = ... by filling in the dotted places. Explain whether σ−1 (n) is a multiplica...
tive function of n or not.
P
n
(b) Evaluate σ−1 (d)µ( )
d
d|n
P
(c) Evaluate
σ−1 (n) = Main term + O(Error) as x →
− ∞, by finding the main term and the
n≤x
error term as functions of x.
Proof.
P1
1
. Since f (n) = is a multiplicative function and σ−1 (n) is defined as the
n
d|n d
1
sum of the f (d) = for d dividing n it is also multiplicative.
d
P1
(b) This is just Mobius inversion formula applied to σ−1 (n) =
thus the result gives us
d
(a) σ−1 (n) =
d n
f (n) =
1
.
n
1
This solution is what remains in my mind from a solution that I was exposed long before I have started
taking this course. I have not used any aid after the homework was assigned.
2
(c)
X
σ−1 (n) =
n≤x
XX 1
n≤x d|n
d
by letting dm = n for a divisor d of n the sum is equal to
XX 1 X1 X
X1 x
X 1
X1 x
X1
=
⌊ ⌋=
( + O(1)) = x
O(1)
1=
+
2
d
d
d
d
d
d
d
d
x
x
d≤x
d≤x
d≤x
d≤x
d≤x
d≤x
m≤ d
m≤ d
=x
∞
X
X1
π2
1
1
)
+
O(
)
=
x
+ O(log x)
+
xO(
2
d
x
d
6
d=1
d≤x
x
x
x
x
Here we have used ⌊ ⌋ = + O(1) which is clear as and ⌊ ⌋ differ from each other at
d
d
d
d
most by a factor of d and it is a constant independent of x. Also
∞ 1
P 1
P 1
P
π2
O(1) = O(
) = O(log x). Lastly
=
. All of which were previously done
2
6
d≤x d
d≤x d
d=1 d
in the class.
Question 4. Let θ ∈ R and 0 < θ < 1. Define gn as:
(
0 if ⌊nθ⌋ = ⌊(n − 1)θ⌋
gn =
1 otherwise
Prove that:
g1 + g2 + . . . + gn
=θ
n→∞
n
P
Proof. We have gn = ⌊nθ⌋ − ⌊(n − 1)θ⌋ So that
gn = ⌊N θ⌋ = N θ + O(1) Thus we have
lim
1≤n≤N
g1 + g2 + . . . + gn
nθ + O(1)
= lim
=θ
n→∞
n→∞
n
n
lim
Question 5. Show that
P
n≤x
P
x
µ(n)
µ(n)⌊ ⌋ = 1. Hence prove that
≤ 1.
n≤x
n
n
Proof. Let N = ⌊x⌋
X
X X
XX
X
x
N
l
µ(n) ⌊ ⌋ =
µ(n)⌊ ⌋ =
µ(n) =
µ( ) =
ν(l) = 1
n
n
m
N
n≤x
n≤N
n≤N
l≤N
l≤N
X
m|l
m≤ n
For the other part, we write
X µ(n)
n≤x
n
1
=
x
x
µ(n)
n
n≤x
X
!
1
=
x
3
X
x
x
µ(n)⌊ ⌋ +
µ(n){ }
n
n
n≤x
n≤x
X
!
P
x
x
x
Where { } is the fractional part of . From the first part we have n≤x µ(n)⌊ ⌋ = 1.
n
n
n
We claim,
X
µ(n) ≤ x − 1
n≤x
for x ≥ 2.
Let N = ⌊x⌋ and by induction on N . For N = 2 it is clearly true. for any N > 2
X
X
µ(n) + µ(N ) ≤ N − 2 + 1 = N − 1
µ(n) ≤
n≤N −1
n≤N
.
x
Since we always have { } < 1, we obtain,
n
X
x
µ(n){ } ≤ x − 1
n
n≤x
for x ≥ 2. And
X µ(n)
n≤x
n
1
=
x
X
x
x
µ(n){ }
µ(n)⌊ ⌋ +
n
n
n≤x
n≤x
!
X
≤
1
(1 + x − 1) = 1
x
The result clearly is also true for x < 2
n
P
Question 6. Show that
e2πia/n = µ(n) for all n ∈ Z.
a=1,(a,n)=1
Proof. We begin by showing that left hand side is a multiplicative function, meaning if we
let (m, n) = 1,
n
X
a=1,(a,n)=1
2πia/n
e
m
X
e
2πik/m
mn
X
=
k=1,(k,m)=1
e2πij/mn
j=1,(j,mn)=1
If we choose a term from the left hand side we have
e2πia/n e2πik/m = e2πi(am+kn)/nm
Since (a, n) = 1 and (k, m) = 1 by Chinese remainder theorem there is a unique j mod mn
which satisfies these congruences and also congruent to (am + kn) mod mn. Using this
we have obtained j ≡ am + kn mod mn which satisfies (mn, j) = 1, j ≤ mn. Also from
this calculation it is clear that this mapping is injective.
e2πia/n e2πik/m = e2πij/nm
Conversely if we choose a term from the right hand side
e2πij/mn
4
We can write, again by Chinese remainder theorem, j ≡ xm + ny mod mn, where x ≡ j
mod m and y ≡ j mod n are integers so we can rewrite it as
e2πij/mn = e2πiax/n e2πiy/m
Again this mapping is injective because of the uniqueness of the solutions between modulus’ that the Chinese remainder theorem grants.
Since this function is multiplicative as shown above, we will only compute it for prime
powers and show that it agrees with µ(n). Firstly for any integer n we have
n
X
e
2πia/n
n
X
=
2πia/n
e
−
a=1
a=1,(a,n)=1
n
X
e2πia/n
a=1,(a,n)̸=1
if we choose a prime p this becomes
p
X
p
X
e2πia/p =
a=1
a=1,(a,p)=1
p
X
e2πia/p −
e2πia/p = 0 − 1
a=1,(a,p)̸=1
This is because all nth roots of unity for any integer n add up to 0 as they satisfy the polynomial xn −1 = (x−1)(xn−1 +xn−2 +...+x+1) and evaluating the polynomial at x = e2πi/n
yields the result.
If p is a prime and k > 1 is an integer
k
p
X
a=1,(a,pk )=1
k
2πia/pk
e
=
p
X
k
2πia/pk
e
p
X
−
a=1
pk−1
e
2πia/pk
=0−
X
k−1
e2πia/p
= 0.
a=1
a=1,(a,pk )̸=1
Thus the function is multiplicative and agrees with µ(n) on prime powers.
P
Question 7. Show that for n > 1 we have
1 = 2ω(n)−1 (ω(n): the number of distinct prime
d|n
µ(d)=1
factors of n)
Proof. We have
X
1=
d|n
µ(d)=1
X
X
µ(d) +
d|n
1
d|n
µ(d)=−1
Where the first summand is ν(n) and since n > 1 it is equal to 0. Thus we conclude
X
X
1=
1
d|n
µ(d)=1
d|n
µ(d)=−1
Which implies the number of integers d dividing n such that µ(d) = 1 is equal to those
µ(d) = −1. Therefore we have
X
d|n
µ(d)=1
1=
1X
2ω(n)
µ(d)2 =
= 2ω(n)−1
2
2
d|n
5
The second equality holds because if the number of distinct divisors of n is m then
there are precisely 2m distinct square free divisors of n.
Question 8. Express the sum
P µ2 (d)σ(d)
for n ∈ Z+ as a product. Evaluate this sum.
ϕ(d)
d|n
µ2 (d)σ(d)
is multiplicative, since the sum is defined over the divisors
Proof. The function
ϕ(d)
of n, the sum is also a multiplicative function of n. Therefore we compute for prime powers
pk . We have
X µ2 (d)σ(d)
d|pk
ϕ(d)
X µ2 (pi )σ(pi )
µ2 (p)σ(p)
p+1
2p
=
=
+1=
+1=
i
ϕ(p )
ϕ(p)
p−1
p−1
0≤i≤k
With this we obtain,
X µ2 (d)σ(d)
d|n
ϕ(d)

=
Y
p|n,p:prime
2p
1
= 2ω(n) n 
p−1
n

Y
p|n,p:prime
n
p 
= 2ω(n)
p−1
ϕ(n)
Question 9.
a
a
(i.e., 0 < < 1, a, b > 0) can be written as a sum of distinct
b
b
1
unit fractions (fractions of the form with c ∈ Z+ , c > 1).
c
(b) Find the largest proper fraction which can be written as a sum of four or fewer unit fractions.
(a) Show that any proper fraction
Proof.
a
(a) By induction on the height of the rational number H( ) = max{|a|, |b|} = b. Suppose
b
a
a
the statement holds for any rational number with (a, c) = 1 where 3 < c < b. Pick
c
b
with (a, b) = 1.If a = 1 we are done so suppose not. Then there is an integer 1 < l < b such
a
1
la − 1
that al ≡ 1 mod b. Now consider − =
. Since b divides la − 1 which is strictly
b
lb
lb
la − 1
la − 1
1
1
1
larger then 1,
has height l which is less then b. If we write
=
+ + ... +
lb
lb
x1 x2
xs
a
1
1
1
1
1 1 2
1 1
then =
+
+ ... +
+ . The base cases are , , = +
b
x1 x2
xs lb
2 3 3
6 2
(b) We start by observing that there is a canonical candidate. For length 1 clearly it is just
1
1 1
1 1 1
, for length 2, + , for length 3, + + , for the length 4 we continue our path and
2
2 3
2 3 7
1 1 1
1
1 1 1
1
1805
claim that the answer is + + +
= + + +
=
.
2 3 7 lcm(2, 3, 7) + 1
2 3 7 43
1806
6
1 1 1 1
1
+ + + be another unit fraction sum of length 4. If is not present there then
a b c d
2
57
1
1 1 1 1
which is much smaller then our result, so
the largest sum we get is + + + =
3 4 5 6
60
2
has to be there. Then we are looking for the maximum of a unit fraction sum of length 3
1
1
that is less then by eliminating from both sums.
2
2
1 1
1
451
1
Our candidate is + +
=
. So suppose is not there, then we observe if all
3 7 43
903
3
1 1 1
1 1 1
73
451
1
<
the unit fractions are higher or equal to we have + + ≤ + + =
6
b
c d
6 7 8
168
903
1
then there has to be 1 or 2 fractions less or equal to .
5
1 1
1 1
1
209
We choose from , . If we choose both the largest we get is + +
=
. Which
4 5
4 5 21
401
1 1
1
is roughly 0.4976 but this is less then our guess + +
which is roughly 0.4994.
3 7 43
1
1
1 1 1
1
If we only choose , then the second one has to be or because if not + + ≈ 0.48.
4
6
7
4 8 9
1 1
1
1 1
1
Other two cases are + +
≈ 0.493 and + +
≈ 0.492.
4 6 13
4 7 10
1
1
1 1 1
If we choose then the second one has to be otherwise + + ≈ 0.491. In the
5
6
5 7 8
1
1 1 1
case of we have + + ≈ 0.491.
6
5 6 8
1
Now we want to show has to be present. We are looking for the maximum of a unit
7
1
fraction sum of length 2 that is less then . If we choose both of them smaller or equal to
6
1 1
1
1
25
1 1
50
1
then + ≤
+
=
< +
=
≈ 0.166. So we have to choose at least one
12
c d
12 13
156
7 43
301
1 1 1 1
1
1
1 1
1
1
from the set { , , , }. The cases are +
= 0.165, +
≈ 0.163,
+
= 1.625,
8 9 10 11
8 25
9 19
10 16
1
1
+
≈ 0.162
11 14
1 1 1
1
After making sure , , have to be in the sum
comes by itself.
2 3 7
43
Let
7