problems. In addition, some problems are really just guidance on proving
theorems; I expect these problems to be done as well. If you meet these
requirements, you’ll earn an A in the course.
1
Divisibility
1.1
Definition
Let m, n 2 Z. We say that m divides n if
n = md
for some d 2 Z.
We write m|n to indicate that m divides n.
Problem 1 Show that if m and n are both positive and m|n then m  n.
Problem 2 Investigate the possibility and consequences of allowing m or n
to be zero in the definition of m|n.
Theorem 1 (The Division Algorithm) Let m, n 2 Z with m > 0. Then
there are unique q, r 2 Z such that
n = mq + r
and
0  r < m.
Problem 3 Let m and n be as in Theorem 1.
(a) Show that the set Q = {k | k 2 Z, n
(b) Show that q = min(Q) and r = n
of the theorem.
mk
0} is nonempty.
mq satisfy the requirements
(c) Verify the uniqueness.
Problem 4
(a) Show that if m|x and n|y then mn|xy.
(b) A number n is even if 2|n. Show that if n2
8|(n2 1).
1 is even, then
If d|n and 1 < d < n, then we say that d is a proper divisor of n.
Problem
5 Show that if n has a proper divisor, then it has a proper divisor
p
d  n.
1.2
Greatest Common Divisors
We say that an integer d is a common divisor of m and n if d|m and d|n.
Clearly the set
D(m, n) = {d | d is a common divisor of m and n}
is a nonempty set containing 1.
Problem 6 Under what conditions is D(m, n) bounded above?
When D(m, n) is bounded above, it has a greatest upper bound, which
is the greatest common divisor of m and n. This is denoted (m, n).
Problem 7 Let m, n 2 Z, not both zero.
(a) Show that (m, n) = ( m, n) and (m, n) = (n, m).
(b) Show that (m, n) = (m, n + m).
(c) Deduce (m, n) = (m, n
dm) for any d 2 Z.
The observation of Problem 7(c) is the basis for an algorithm to find the
greatest common divisor for any two numbers.
Problem 8 Suppose 0 < m  n. Show that there are integers a and b with
(m, n) = (a, b) and 0  a, b < n.
The process you invented to prove Problem 8 clearly cannot be applied
more than n times. At the stopping point you will have produced an equation
of the form (m, n) = (d, 0) for some number d > 0. Since (d, 0) = d, you’ve
found the greatest common divisor of m and n.
Theorem 2 Let m, n 2 Z, not both zero. Then
(m, n) = min (N \ {mx + ny | x, y 2 Z}) .
Problem 9 Let m, n 2 Z, not both zero, let L = N \ {mx + ny | x, y 2 Z},
and write d = (m, n).
(a) Show that d 2 L.
Hint Show that in Problem 8 both the numbers a and b are linear
combinations of m and n.
(b) Show that if ` 2 L, then d  `.
(c) Prove Theorem 2.
Problem 10
(a) Show that if d is any common divisor of m and n, then (m, n)|d.
(b) Show that (m, n)|(m, nx) for any x.
These ideas generalize to greatest common divisors for more than two
numbers.
Project 11 Let n1 , n2 , . . . , nr be nonzero integers.
(a) Formulate the definition of the greatest common divisor of n1 , n2 , . . . , nr .
It is denoted (n1 , n2 , . . . , nr ).
(b) Show that (n1 , n2 , . . . , nr ) = (n1 , n2 , . . . , nr 2 , (nr 1 , nr )).
(c) Develop an algorithm to compute the greatest common divisor of
r integers.
(d) Show that (n1 , n2 , . . . , nr ) can be written as an integer linear combination of n1 , n2 , . . . , nr , and deduce that it is precisely the smallest positive such linear combination.
Problem 12 Let m, n and x be integers and let d be a common divisor of
m and n.
(a) Show that (mx, nx) = (m, n)x.
(b) Show that ( md , nd ) = d1 (m, n).
(c) Show that d|(m, n).
(d) Suppose m and n are nonzero integers. Show that (m, m + n)
divides n.
1.3
Relatively Prime Numbers
Two integers m, n 2 Z are said to be relatively prime if their greatest
common divisor is 1, that is if (m, n) = 1.
There are lots of great things to say about relatively prime numbers.
Problem 13 Show that if (m, n) = 1 and m|nx, then m|x.
Hint What is (mx, nx)?
Problem 14
(a) Show that if g = (m, n), then
m
g
and
n
g
are relatively prime.
(b) Show that if (m, x) = 1 = (n, x), then (mn, x) = 1.
(c) Show that (m, n) divides (ma, nb) for any m, n, a, b 2 Z.
Problem 15 Suppose (m, n) = 1. Show that for any x 2 Z
(mn, x) = (m, x)(n, x).
? Problem
16 Show that if m and n are relatively prime and d|mn then
there is a unique factorization d = d1 d2 with d1 |m and d2 |n.
Hint This seems pretty simple, but my proof of this is kind of complicated.
Don’t get discouraged if you can’t find a short proof. Start by making an
intelligent guess as to the values of d1 and d2 .
Exercise 17 How much of Problem 16 remains true if you don’t require m
and n to be relatively prime?
1.4
Prime Numbers
Prime numbers are often defined to be those numbers that have no proper
divisors, and this is not wrong. However in algebra it has emerged over
time that the main property of prime numbers is not what their divisors are,
but their properties as divisors. We follow this point of view and say that a
nonzero nonunit p 2 Z is prime if whenever p|mn, then either p|m or p|n (or
both). If a nonzero nonunit n 2 Z is not prime, then it is called a composite
number.
It is of course not hard to prove that the other definition is equivalent.
Problem 18 Let p 2 Z be prime.
(a) Show that if d|p, then either d = 1 or d = ±p. (and conversely).
(b) Show that for any x, either x is relatively prime to p or else p|x.
(c) Show that (p, x) is either p or 1, depending on whether or not p|x.
Euler defined the ‘totient function’
: N ! N by the rule
(x) = #{y 2 N | y  x, (x, y) = 1}.
It tells you how many numbers less than x are relatively prime to x.
Problem 19 Let p be prime and n 2 N. What is (pn )?
Prime Factorization. Part of the importance of prime numbers is that
every integer can be factored, uniquely, as a product of primes.
Theorem 3 (Fundamental Theorem of Arithmetic) Every integer n > 1 can
be written as a product of positive prime numbers, and that factorization is
unique, except possibly for the order in which the factors are written.
No fair
using
prime
factorizations!
Problem 20 Show that every integer n > 1 can be written as a product of
prime numbers.
Hint Suppose there were counterexamples, and study the smallest one.
Could it be prime? Could it be composite?
Problem 21
(a) Now suppose you have an integer n with two prime factorizations:
p 1 p 2 · · · p r = n = q1 q2 · · · q s
where you (initially) have no information about the numbers r
or s or the relationship between the prime numbers {pi } and the
prime numbers {qj }. Show that p1 = qj for some j.
(b) Finish the proof of Theorem 3.
Problem 22 Give another proof of Theorem 3 using Problems 20 and 16.
Theorem 4 (Euclid) There are infinitely many prime numbers.
Problem 23 Prove it!
1.5
More Problems
Problem 24
(a) Let p and q be prime. What is (pq)? What is (pa q b )?
(b) Show that there are infinitely many prime numbers of the form
4n + 1 with n 2 Z.
(c) Show that if mx|my then x|y.
(d) Suppose x and y are both odd (not divisible by 2). Show that
x2 + y 2 is not a square.
(e) Write d(n) = #{positive divisors of n}. Show that if m and n are
relatively prime then d(mn) = d(m)d(n).
(f) Suppose I tell you that n has an odd number of divisors. What
can you deduce about the number n?
Project 25 A common multiple of integers n1 , n2 , . . . , nr is an integer
m such that nj |m for j = 1, 2, . . . , r. The least common multiple is the
smallest positive common multiple; it is sometimes denoted [n1 , n2 , . . . , nr ].
(a) Show that [n1 , n2 , . . . , nr ] divides every other common multiple of
n 1 , n2 , . . . , n r .
(b) Show that n1 n2 = [n1 , n2 ](n1 , n2 ) for any nonzero n1 , n2 2 Z.
(c) Generalize (a) to larger collections of integers.
2
Number Theory and Algebra
In this section we’ll introduce congruence modulo n. The congruence classes
of integers modulo n are a ring under the product and sum the receive from
the integers, and we’ll use the algebra of these rings to prove many interesting
number-theoretical results.
2.1
Modular Arithmetic
For each nonzero integer n we define an equivalence relation on Z by the rule
x⌘y
mod n
if and only if
n|(x
y)
called congruence modulo n.
Problem 26 Verify that ⌘ mod n is in fact an equivalence relation.
The key to the power of congruence modulo n is that it respects addition
and multiplication.
Problem 27 Let x0 , x1 , y0 , y1 and n 2 Z. Assume x0 ⌘ x1
y0 ⌘ y1 mod n.
(a) Show that x0 + x1 ⌘ y0 + y1
(b) Show that x0 · x1 ⌘ y0 · y1
mod n and
mod n.
mod n.
Problem 28 Suppose ax ⌘ ay mod n. What can you deduce about the
relationship between x and y?
Problem 29
(a) Show that if p is prime, the congruence x2 ⌘ 1 mod p has only
two solutions (mod p), namely x ⌘ 1 mod p and x ⌘ 1 mod p.
(b) Does this fail if you try to solve x2 ⌘ 1 mod n with nonprime
n? Can you find specific examples? Can you describe a general
pattern of examples?
Thinking in Terms of Algebra. If X is a set with an equivalence relation
⇠, the ⇠-equivalence class of x 2 X is the set
[x] = {y 2 X | x ⇠ y}.
The defining properties of equivalence relations force it to be true that two
classes are either disjoint or equal. Thus an equivalence relation gives rise to
a set of ⇠-equivalence classes:
X/ ⇠ = {[x] | x 2 X}.
Furthermore, this new set is related to the original set X by a function
q : X ! X/ ⇠ defined by the rule q(x) = [x].
In the special case of congruence mod n, we write Z/n for the set of
congruence classes.
Problem 30 Show that there is exactly one way to define a sum and multiplication on Z/n making Z/n into a ring so that the quotient function
q : Z ! Z/n
is a homomorphism of rings.
From now on we will work with this ring structure on Z/n without any
further comment.
Problem 31
(a) Show that if x ⌘ y mod m then (x, m) = (y, m).
(b) Show that [x] 2 Z/n is invertible if and only if (x, n) = 1.
(c) Show that #(Z/n)⇥ = (n).
Problem 32 Under what conditions on n is the ring Z/n a field?
2.2
Orders of Integers in Modular Arithmetic
Remember that the order of an element g of a group G is the least n 2 N
such that g n = 1, the (multiplicative) identity element of the group. When
we study rings, we have two di↵erent groups, namely the additive group R
(given by forgetting how to multiply) and the multiplicative group R⇥ . So
every element x 2 R has an additive order, meaning its order as an element
of the additive group R. If x 2 R⇥ , then x also has a multiplicative order,
meaning its order in the multiplicative group R⇥ .
Additive Order. It is comparatively easy to understand the additive order
of integers in Z/n.
Problem 33 Let x 2 Z. What is the additive order of [x] in Z/n?
Multiplicative Order. Here we have a variety of interesting results, but
none quite as sweeping as what you’ve proved for additive order.
? Problem 34 (Euler)
Show that if (n, x) = 1, then x (n) ⌘ 1 mod n.
Hint Translate this question into an algebraic question about the ring Z/n.
It follows that if [x] 6= 0 2 Z/n, then the multiplicative order of [x] divides
(n). The special case when p is prime is known as Fermat’s theorem; it
says that if p is prime and p - x then xp 1 ⌘ 1 mod p.
1)! ⌘
Theorem 5 (Wilson) If p is prime, then (p
Problem 35
1 mod p.
(a) Prove Wilson’s Theorem.
(b) What is (n
2.3
1)! congruent to modulo n when n is not prime?
Divisibility Rules
If we want to test if 5|x or if 7|x, etc, we can simply divide and look at the
remainder, but in practice1 it is nice to have simple rules that reduce the
work required. Since integers in our world are generally presented in decimal
notation, it is nice to have simple rules that take advantage of that special
form.
Problem 36 Let n be a nonnegative integer written in decimal notation
n = dr 10r + dr 1 10r
1
+ · · · + d1 10 + d0 .
with 0  di < 10 for each i.
(a) Show that n ⌘ d0 + d1 + · · · + dr
(b) Show that n ⌘ d0
1
+ dr
d1 + · · · + ( 1)r 1 dr
mod 9.
1
+ ( 1)r dr
mod 11.
Derive the resulting rules for testing divisiblity by 9 and 11.
Problem 37 Now write
n = dr 1000r + dr 1 1000r
1
1
+ · · · + d1 1000 + d0 .
In my case, when taking a baby for a walk in the stroller and factoring all the house
numbers as I go.
with 0  di < 1000 for each i. Show that
n ⌘ d0
d1 + · · · + ( 1)r 1 dr
1
+ ( 1)r dr
mod 7.
Show that the same result is true mod 11 and 13.
We’ve seen that for p = 3, 11, testing for divisibility by p can be reduced to
checking divisibility for 1-digit (decimal) numbers. Divisibility by p = 7, 13
can be reduced to checking divisibility for 3-digit (decimal) numbers.
? Problem 38
True or false: For each prime p testing divisibility by p can
be reduced to testing divisibilty of m-digit numbers by p for some m. Explain
your answer carefully (i.e., prove or counterexamp).
2.4
Divide and Conquer
There is a basic fact of algebra that enables us to, very often, simplify our
work in solving problems involving congruences. To state it properly, we
need to introduce the product of rings. If R and S are rings (commutative,
with 1), their product is the ring
R ⇥ S = {(r, s) | r 2 R, s 2 S}.
The algebraic structure is just the original sum and product from R in the
first coordinate and the sum and product from S in the second coordinate.
Problem 39 Show that (r, s) 2 R ⇥ S is a unit if and only if r 2 R is a unit
and s 2 S is a unit.
Theorem 6 Suppose n = ab with (a, b) = 1. Then the function
p : Z/n ! Z/a ⇥ Z/b
given by reduction in the two coordinates is an isomorphism of rings.
Problem 40 Suppose n = ab with (a, b) = 1.
(a) Show that #(Z/a ⇥ Z/b) = n = #(Z/n).
(b) Determine the kernel of p.
(c) Prove Theorem 6.
We’ll make use of this fact over and over. Here’s an example of how it
can be used.
Problem 41
(a) Show that if (a, b) = 1, then (ab) = (a) (b).
(b) Since you know (pa ) for p prime, you can deduce a formula for
(n) in terms of its prime factorization. Write it down.
(c) Can it happen that (ab) = (a) (b) when a and b are not relatively prime?
2.5
Polynomials Versus Functions
We have abstract polynomials
R[x] = {polynomials with coefficients in R}.
We also have the set F (R, R) of all functions f : R ! R. We are used to
confusing polynomials with functions, and we will continue to do so in this
course after we determine the precise nature of the confusion.
Problem 42 Let R be a ring and let Q be a finite set.
(a) Show that pointwise addition and multiplication of functions gives
F (Q, R) the structure of a ring.
(b) Now consider the ‘indicator functions’
x
:Q!R
given by
x (y)
=
⇢
1 if x = y
0 if x 6= y,
defined for each x 2 Q. Show that every element of F (Q, R) is an
R-linear combination of the functions { x | x 2 Q}.
Each abstract polynomial p gives rise to a function fp : R ! R, given by
the formula
fp (↵) = the element of R that results from replacing x with ↵ in p
The rule p 7! fp defines a ‘functioning’ function
: R[x] ! F (R, R).
? Problem 43
(a) Show that
: R[x] ! F (R, R) is a ring homomorphism.
(b) Show that : Z/p[x] ! F (Z/p, Z/p) is surjective if p is prime.
What can you say about the kernel of ?
Project 44
(a) Is surjective for R = Z/n with n not prime? Can you give a
specific counterexample?
(b) Describe the kernel of
prime.
2.6
: Z/n[x] ! F (Z/n, Z/n) when n is not
More Problems
? Problem 45
Let n 2 N and let G be a proper subgroup of the multiplicative group (Z/nZ)⇥ . Show that there are infinitely many prime numbers p
such that [p] 62 G.
3
Arithmetic Functions
Some Stu↵ That Should Be in Section 1
Problem 46 Show every integer n has a unique expression in the form
n = a2 b such that if d2 |n then d2 |a2 .
If the a in this factorization is 1, then we say that n is square free (or
squarefree).
Problem 47 Show that every integer n has a unique largest squarefree divisor, which we’ll denote sqfr(n).2
This can be generalized.
Problem 48 Show that every n 2 N can be factored in the form
n = n1 · n22 · n33 · · · nrr
where (ni , nj ) = 1 for all i 6= j.
Earlier we introduced the notation !(n) for the number of distinct prime
factors of n. It is sometimes useful to get more detailed information about
the prime factors of n. We say that the prime p divides n with exponent
k if pk |n but pk+1 does not divide n; another notation for this is p||n. Let’s
agree to write
!k (n) = {p | p is prime and pk ||n}.
(I just made this notation up. It seems like good notation, so it might be
standard, but if so it’s an accident.)
Problem 49
(a) Show that n is squarefree if and only if !(n) = !1 (n)
P
(b) Show that !(n) = 1
k=1 !k (n).
(c) Referring to the factorization of Problem 48, show that
!k (n) = !(nk ).
2
Not standard notation.
3.1
Another Ring Structure for Functions into Z
We write F for the set of all functions f : N ! R (most of our functions will
actually land in Z, but there’s no harm in imagining the target is the larger
set R), and give it the structure of a ring in the following non-obvious way:
• addition actually is obvious: the sum f + g is the function defined by
the rule (f + g)(n) = f (n) + g(n);
• the product of two functions f and g is denoted f ⇤ g, and it is defined
by the formula
X
f ⇤ g (n) =
f (d) · g(n/d)
d|n
where the dot simply denotes ordinary multiplication of numbers, and
the sum runs over the positive divisors of n.
Exercise 50 Satisfy yourself that these rules give F the structure of a commutative ring.
Problem 51
(a) Show that F has a multiplicative identity, and figure out exactly
what it is; we’ll call it I.
(b) In view of part (a), it makes sense to ask the question: How can
you tell if f 2 F is an invertible in the ring F? Answer the
question!
In number theory we say that a function f : N ! R is multplicative if
f (xy) = f (x) · f (y) whenever (x, y) = 1. If f (xy) = f (x) · f (y) for all x and
y, then we say that f is totally multiplicative.
Problem 52
(a) Show that if f and g are both multiplicative, then so is f ⇤ g.
(b) Is it possible for f to be multiplicative and not invertible?
(c) Show that if f is multiplicative and invertible, then so is f 1 (here
f 1 denotes the multiplicative inverse of f in the ring F).
Problem 53 Use the operation ⇤ to express the functions
d(n) = the number of positive divisors of n
(n) = the sum of the positive divisors of n
in terms of the constant function
(defined by (n) = 1 for all n) and
the identity function id (defined by id(n) = n for all n). Are d and
multiplicative?
3.2
The Möbius Function
It frequently happens that we want to sum a function over the divisors of
n; that is, we’re given the function f and we find ourselves interested in the
related function F defined by the formula
X
F (n) =
f (d).
d|n
? Problem 54
Suppose f and F are related to one another as above.
(a) Show that f divides F in the ring F. That is, show that there is
an element g 2 F such that F = f ⇤ g. What is g?
(b) Show that g is invertible. Its inverse is known as the Möbius
function and denoted µ. Deduce that
X
f (n) =
µ(d) · F (n/d).
d|n
This is known as the Möbius inversion formula.
(c) Show that µ is multiplicative.
(d) Show that f is multiplicative if and only if F is multiplicative.
(e) You’ve proved the big result on the Möbius function without even
knowing how to compute it! Now find a formula for µ.
Hint Use the fact that µ is multiplicative.
Problem 55 Show that if n 2 Z is squarefree and d|n, then
µ(d) = ( 1)!(n) µ(n/d)
where !(n) is the number of distinct prime factors of n.
P
Problem 56 What is the function f (n) = d2 |n µ(d)?
Hint Write n = a2 b as in Problem 46.
? Problem 57
(a) Show that
P
(d) = n for every integer n.
P
(b) Show that (n) = n d|n µ(d)
for every integer n.
d
d|n
Theorem 7 If f : N ! R is a multiplicative function, then
X
Y
µ(d)f (d) =
(1 f (p))
d|n
p|n
where the product is taken over all positive prime divisors of n. How should
you interpret the product on the right when n = 1?
Problem 58 Prove Theorem 7.
Problem 59
(a) Show that
(b) Show that
(c) Determine
of n.
P
P
d|n
d|n
P
d|n
|µ(d)| = 2!(n) for every n 2 N.3
µ(d)d(d) = 0 for every n 2 N.4
µ(d) (d) where (d) is the sum of all the divisors
(d) Show that if n is even, then
Problem 60
(a) Show that
X
d|n
P
d|n
µ(d)f (d) =
µ(d) (d) = 0.
X
µ(d)f (d)
d|sqfr(n)
for any f 2 F.
P
(b) Show that d|n dµ(d) = ( 1)!(n)
(n)sqfr(n)
.
n
Problem 61
(a) We know that if f is multiplicative, then ⇤ f is too, so it might
seem reasonable to guess that if f is totally multiplicative, then
⇤ f is also totally multiplicative (after all, is totally multiplicative). Give a complete list of all totally multiplicative functions f
such that g = ⇤ f is also totally multiplicative.
3
4
Remember, !(n) is the number of distinct primes that divide n.
Remember, d(n) is the number of divisors that divide n.
k
}|
{
(b) Identify
= ⇤ ⇤ · · · ⇤ . That is, give a number-theoretic
description of the meaning of ⇤k (n).
⇤k
z
Problem 62 There is a unique function ⇤ : N ! R such that
X
log(n) =
⇤(d)
d|n
for all n. Give an explicit formula for this function.
Project 63 Investigate the iterated products
k
µ
⇤k
z
}|
{
= µ ⇤ µ ⇤ ··· ⇤ µ.
Can you give a formula for µ⇤k ?
4
4.1
Continued Fractions
Finite Continued Fractions
Suppose we are given a real number x. Then we as number theorists might
like to think of x as approximately an integer:
x = n0 + y1 ,
where n0 2 N and 0  x1 < 1.
But now y1 is a pain: it is a real number, but its integer part is zero. So
here’s the trick: write y1 = x11 where x1 is a real number bigger than 1, so
we can write x1 = n1 + y2 where n1 2 N and 0  y2 < 1. This gives us
x = n0 +
1
n1 + y2
There’s no stopping us now, unless at some point yk+1 = 0.
Problem 64 Show that yk+1 = 0 for some k if and only if x 2 Q.
Hint Look back at how we found greatest common divisors by repeated
division.
If yk+1 = 0 for some k (and x 2 Q) then the expression we have is called
a finite continued fraction expression for x. For simplicity, it is written
x = hn0 , n1 , . . . , nk i
where each ni 2 N, except possibly n0 , which will be negative if x < 0.
But now that we have this notation, we can think about a function of real
numbers
fk (x0 , x1 , . . . , xk ) = hx0 , x1 , . . . , xk i.
Problem 65
(a) Show that
hx0 , x1 , . . . , xk i = x0 +
1
hx1 , . . . , xk i
(b) Show that
⌧
hx0 , x1 , . . . , xk i = x0 , x1 , . . . , xk
1
+
1
xk
Problem 66 Show that every rational number has at least two di↵erent
continued fraction expressions.
Problem 67 Think of hx0 , x1 , . . . , xk i as a function of xi (for a fixed i). Is
it increasing or decreasing?
Hint The answer will depend on i.
Project 68 Fix a sequence of real numbers a1 , a2 . . . , an , . . .
1 and define
fn (x) = han , an 1 , . . . , a1 , xi
Find a formula for the derivative of fn (x).
Hint Express fn in terms of fn 1 .
Theorem 8 Suppose we have two finite continued fraction expressions for
the same number:
hx0 , x1 , . . . , xr i = hy0 , y1 , . . . , ys i
in which xi , yj 2 N for 0 < i < r and 0 < j < s, and xr , ys > 1. Then r = s
and xi = yi for all i.
Problem 69 Prove Theorem 8 by induction on max{r, s}.
4.2
Infinite Continued Fractions
More interesting than the finite continued fractions are expressions like this
1+
1
1+
1
1+
1+
1
1
1
1+ ···
which extend forever. More generally, we could have infinite continued fractions with all kinds of denominators, which we could write like so:
hx0 , x1 , . . . , xn , xn+1 , . . .i = x0 +
1
x1 +
1
x2 +
x3 +
1
1
1
x4 + ···
with each xi 1 for i > 0. The first question to answer is: what do we mean
by such an expression?
We’ll answer this in the typical calculus approach to questions involving
infinity, by defining the infinite to be the limit of finite approximations, like
so
hx0 , x1 , . . . , xn , xn+1 , . . .i = lim hx0 , x1 , . . . , xn i
n!1
(provided, of course, that the limit exists).
Theorem 9 For any sequence x0 , x1 , . . . , xn , . . . of real numbers with xi
for all i > 0, the limit
lim hx0 , x1 , . . . , xn i
1
n!1
exists.
Problem 70 Work in the setup of Theorem 9; to simplify notation, let’s
write zn = [x0 , x1 , . . . , xn ].
(a) Show that z2n < z2n+1 for all n.
(b) Show that z2n < z2(n+1) for all n.
(c) Show that z2(n+1)+1 < z2n+1 for all n.
(d) Make a diagram showing the relative positions of the numbers
{zn | n 0}.
Hint Use Problem 67.
To complete the proof of Theorem 9 we need to estimate the di↵erence
between z2n and z2n+1 . Define two sequences of real numbers by the rules
h
h
k
k
2
1
2
1
= 0
= 1
and
hi = ai hi
1
+ hi
2
for i
0.
= 1
= 0
and
k i = ai k i
1
+ ki
2
for i
0.
Problem 71 Show that ki+1
ki + 1 for all i
1.
Problem 72 Continue to work in the setup of Theorem 9.
(a) Show that
ha0 , a1 , . . . , an 1 , xi =
xhn
xkn
+ hn
1 + kn
1
(b) Show that zn =
hn
kn
for n
(c) Show that hi ki
1
hi 1 ki = ( 1)i
(d) Show that zi
zi
1
=
2
for all x > 0.
2
0.
1)i 1
(
ki k i
1
for i
1
for i
1.
1.
(e) Explain why your work proves Theorem 9.
For finite continued fractions, we had a small amount of ambiguity: every
rational number had precisely two di↵erent expressions as a continued fraction. We discover that there is no ambiguity for infinite continued fractions.
Theorem 10 If ha0 , a1 , . . .i = hb0 , b1 , . . .i with an , bn 2 N for all n > 0, then
an = bn for all n.
Problem 73 Prove Theorem 10.
Hint Look at your proof of Theorem 8.
4.3
Numbers Expressible as Infinite Continued Fractions
In view of Theorem 10 we might think that we should just dispense with
the finite continued fractions and work with the infinite ones. Unfortunately,
we cannot do that because rational numbers do not have infinite continued
fractions (with integer entries).
Theorem 11 If a0 , a1 , a2 , . . . , an , . . . is a sequence of integers, all of which
are 1, except possibly a0 , then ha0 , a1 , . . .i is an irrational number.
Problem 74 Write z = ha0 , a1 , . . .i, and suppose z =
use the sequences {hn } and {kn } defined above.
(a) Show that 0 < |kn a
hn b| <
b
kn+1
for n
a
b
with a 2 Z, b 2 N;
1.
(b) Show that the result of (a) is impossible for n very large.
Simply knowing that every infinite continued fraction is irrational leaves
open the possibility that some irrational numbers cannot be written as continued fractions. But this is not true.
Problem 75 Show that every irrational number can be represented as a
continued fraction ha0 , a1 , . . . , an , . . .i with a0 2 Z and an 2 N for n 1.
5
Algebra Interlude
Here is a useful theorem that I want you to be able to use.
Theorem 12 If F is a field then every finite subgroup G of multiplicative
group F ⇥ is cyclic.
To prove it, we need a little information about orders of products of
elements in abelian groups.
Problem 76 Let x, y 2 G, where G is an abelian (multiplicative) group.
(a) Show that if (n, order(y)) = 1, then order(y n ) = order(y).
(b) Show that if (order(x), order(y)) = 1, then
order(xy) = order(x), order(y).
(c) Show by example that this is not true if (order(x), order(y)) 6= 1.
Can you give a formula for order(xy) in general?
Problem 77 Now let G be any finite subgroup of F ⇥ , say |G| = n, and let
m be the least common multiple of the orders of all of the elements of G.
(a) Show that m  n.
(b) Show that g m = 1 for all g 2 G.
(c) Show that m
n.
Hint This is where you use the fact that F is a field.
(d) Write n = pa11 pa22 · · · par r be the prime factorization of n. Show that
for each i there is an element gi 2 G with order(gi ) = pai i .
(e) Prove Theorem 12 by showing that there is an element g 2 G with
order(g) = n.
6
6.1
Solving Congruences
Linear Congruences
Let’s start with the simplest kind of linear congruence
? Problem 78
Find conditions on a, b and n that ensure the congruence
ax ⌘ b mod n
has a solution. How many solutions are there in Z/n?
Hint Try solving some examples! Here are some:
3x ⌘ 7 mod 19,
3x ⌘ 7 mod 18,
3x ⌘ 6 mod 18,
4x ⌘ 6 mod 18.
Now we move on to a more general problem, in which we want to find a
single number x simultaneously satisfying the all of the congruences
x ⌘ b1
x ⌘ b2
..
.
mod n1
mod n2
x ⌘ br
mod nr ,
where (ni , nj ) = 1 for all i 6= j.
? Problem 79
(a) Suppose x and y are two solutions to the system of congruences
above. Then x ⌘ y mod what? .
(b) Convert the problem above into a question involving rings of the
form Z/m for various values of m and homomorphisms between
them.
(c) Show that the problem always has solutions. How many solutions
does it have?
Hint Use Theorem 6.
You have proved the Chinese Remainder Theorem!
Knowing the existence of solutions is not the same thing as knowing how
to find them. So let’s get algorithmic.
? Problem 80
We continue to work with the problem above.
(a) Fix j between 1 and r; then explain how to find a solution xj in
the special case bi = 0 for all i except j.
(b) Explain how to combine your solutions x1 , x2 , . . . , xr from part (a)
to find solutions to the general problem.
Hint Look back at Section 2.5.
Problem 81 Solve the system of congruences
x ⌘ 1 mod 2
x ⌘ 2 mod 9
x ⌘ 3 mod 125,
Project 82 What can you say about solutions to the system
a1 x ⌘ b 1
a2 x ⌘ b 2
..
.
mod n1
mod n2
ar x ⌘ b r
mod nr ,
where (ni , nj ) = 1 for all i 6= j?
6.2
Plan for Solving Congruences
In this section I have a slightly more general version of the discussion we had
in class on Tuesday (June 2), showing that solving congruences mod n can be
reduced to solving congruences mod pa for various primes p and exponents
a. You
Suppose we have polynomials fi 2 Z[x1 , x2 , . . . , xr ] for i = 1, 2, . . . , s in
some number of variables and we want to find solutions of the system of
congruences
f1 (x1 , x2 , . . . , xr ) ⌘ 0 mod n
f2 (x1 , x2 , . . . , xr ) ⌘ 0 mod n
..
.
fs (x1 , x2 , . . . , xr ) ⌘ 0 mod n
Problem 83
(a) Show that if the system is solvable mod pa for every prime p and
every a such that pa |n, then it is solvable mod n.
(b) Show that if the system is solvable mod pa+1 , then it is solvable
mod pa too.
The upshot of all of this is that in trying to solve congruences, there is
no loss in generality if we restrict our attention to the special case in which
the modulus is a power of a prime.
Now suppose we modify our problem so that we’re trying to solve the
system
Project 84 Investigate the extent to which you can reduce systems of the
form
f1 (x1 , x2 , . . . , xr ) ⌘ 0 mod n1
f2 (x1 , x2 , . . . , xr ) ⌘ 0 mod n2
..
.
fs (x1 , x2 , . . . , xr ) ⌘ 0 mod ns .
(where the moduli are (possibly) di↵erent from equation to equation) to
congruences modulo prime powers.
Plan of Attack. These results suggest a way to solve systems of congruences:
1. Make a list pa11 , pa22 , . . . of the prime powers relevant to the problem.
2. To solve the system mod pa , first solve it mod p; if there is no solution,
then there is no hope for solutions with higher exponents.
3. Work by induction: given a solution mod pb , try to find one mod pb+1 .
The first step is very easy, at least conceptually, and it turns out that the
third step is not too bad, either. Unfortunately the middle step does not
have any good general methods.
6.3
Working from One Power of p to the Next
We’ll examine in detail the problem of solving f (x) ⌘ 0 mod pa+1 given a
solution to f (x) ⌘ 0 mod pa . We saw in class that this could be thought of
as a mechanical operation which proceeds likes so:
1. choose your favorite solution y for f (x) ⌘ 0 mod pa
2. write x = y+kpa with the hope of finding a k that solves the congruence
f (x) ⌘ 0 mod pa+1 ; in our experiments there was be lots of cancellation
and we tended to ended up with k needing to solve a congruence modulo
p.
In this section we’ll work out the general pattern in more detail.
To do this we’re going to pull in a little bit of calculus. Nice functions f
can be evaluated by their Taylor series
f (x+h) = f (x)+
f 0 (x) 1 f (2) (x) 2
f (n) (x) n f (n+1) (x) n+1
h +
h +· · ·+
h +
h +· · · .
1!
2!
n!
(n + 1)!
All the terms suggested by the · · · are actually zero if f happens to be a
polynomial of degree  n. Your first job is to show that all of the nonzero
summands live in the world of number theory.
Problem 85 Let f 2 Z[x] be a polynomial in one variable with integer
coefficients. Show that for every r
0, the polynomial r!1 f (r) (x) also has
integer coefficients.
Hint Explain why it suffices to check this for f (x) = xn .
Suppose we’re given a specific number y such that f (y) ⇠
= 0 mod pa , with
a 1. We’ll look for solutions to the congruence modulo pa+1 that reduce
to the given solution y modulo pa . That is, we’ll write x = y + kpa where v
is some integer, and try to show that an intelligent choice of v will make x a
solution to f (x) ⌘ 0 mod pa+1 .
Problem 86
(a) Show that f (y + vpa ) ⇠
= f (y) + f 0 (y)kpa
mod pa+1 .
(b) Explain why pa |f (y) and show that f (y + kpa ) ⌘ 0 mod pa+1 if
and only
f (y)
k · f 0 (y) ⌘
mod p.
pa
(c) Under what conditions is the congruence of part (b) solvable?
We say that a solution y for f (x) ⌘ 0 mod pb is a regular solution if
f 0 (y) 6⌘ 0 mod p. We say that a solution x for f (x) ⌘ 0 mod pa lies over a
solution y for f (x) ⌘ 0 mod pb with b < a if x ⌘ y mod pb .
Theorem 13 If y is a regular solution for f (x) ⌘ 0 mod pb , then for every
a b there is a unique solution to f (x) ⌘ 0 mod pa lying over y.
Problem 87 Prove Theorem 13.
Put this into action.
Problem 88 Solve the congruence x3 + 5x2 + 3x + 2 ⌘ 0 mod 113 . How
many solutions are there mod 115 ?
7
Quadratic Congruences
This is a big and important part of number theory so it merits its own section.
7.1
Solving Quadratics Boils Down to Finding Square
Roots
Problem 89 Let p be an odd prime. Show that the congruence
ax2 + bx + c ⌘ 0 mod p
(with a, b and c not divisible by p) has a solution if and only if [b]2
has a square root in the ring Z/p.
4[a][c]
This problem clearly shows that the problem of identifying elements with
square roots is an important one.
Problem 90 Let p be an odd prime.
(a) Show that precisely half of the elements of (Z/p)⇥ have square
roots.
(b) Show that if [x] 2 (Z/p)⇥ has a square root, then [x](p
1)/2
= [1].
(c) Show that there can be no more than (p 1)/2 elements of (Z/p)⇥
such that [x](p 1)/2 = [1].
(d) Conclude that [x] 2 (Z/p)⇥ has a square root if and only if
[x](p 1)/2 = [1].
(e) Show that if [x](p
1)/2
6= [1], then [x](p
Problem 91 For which primes p does
1)/2
=
[1].
[1] have a square root in Z/p?
The results of Problem 90 make this a good time to introduce the Legendre symbol
✓ ◆ ⇢
a
+1 if [a] has a square root in Z/p
=
1 if [a] does not have a square root in Z/p.
p
This is defined only for p a prime and a relatively prime to p; it should not
be confused with a fraction or division.
⇣ ⌘
Problem 92 Show that ap ⇠
= a(p 1)/2 mod p. Is this true only for odd
primes?
Project 93 Here’s another interpretation of the Legendre symbol. This
requires some setup. First of all a permutation of a set S is a function
f : S ! S that is one-to-one and onto. Write Sym(S) for the set of all
permutations of S; this is a group under composition of functions.
Among the permutations of S are the functions that swap s1 and s2 and
leave all the other elements alone. If S is a finite set, then f can be written
as a product of some number of simple swaps, and the fact is that either you
need an odd number of swaps or an even number of swaps. In the first case
f is an odd permutation and in the second f is an even permutation. Then
it makes sense to write ( 1)f to denote +1 if f is even and 1 if f is odd.
(a) Show that the function sign : Sym(S) ! {±1} given by sign(f ) =
( 1)f is a homomorphism of groups.
(b) For each a 2 (Z/p)⇥ we can define a permutation fa : (Z/p)⇥ !
(Z/p)⇥ by the rule fa (x) = ax. Show that the function µ :
(Z/p)⇥ ! Sym((Z/p)⇥ ) is a homomorphism of groups. Conclude
that sign µ : (Z/p)⇥ ! {±1} is also a homomorphism of groups.
(c) Show that L : (Z/p)⇥ ! {±1} given by L(a) =
morphism of groups.
⇣ ⌘
a
p
is a homo-
(d) Show that if G is a cyclic group of even order, then there is a
unique nontrivial homomorphism G ! {±1}.
⇣ ⌘
(e) Show that sign(fa ) = ap for every a 2 (Z/p)⇥ .
7.2
Quadratic Reciprocity
Suppose p and q are odd primes. Then we can form two di↵erent Legendre
symbols, namely
✓ ◆
✓ ◆
p
q
and
q
p
How do these compare?
Problem 94 Do a bunch of examples and look for patterns.
The answer to the question was found (and proved) by Gauss.
Theorem 14 (Quadratic Reciprocity) Let p and q be odd primes. Then
✓ ◆✓ ◆
p 1 q 1
p
q
= ( 1) 2 2 .
q
p
We’ll prove this in class.
Here’s a useful rephrasing.
Problem 95 Let p and q be odd primes. Show that
p ⌘ q ⌘ 3 mod 4.
⇣ ⌘
p
q
=
⇣ ⌘
q
p
unless
The Quadratic Reciprocity Theorem is the main ingredient in an e↵ective
method for deciding which elements of Z/p have square roots. For example,
we have
✓ ◆ ✓ ◆ ✓ ◆
✓ ◆
✓ ◆
19
79
3
19
1
=
=
=
=
= 1
79
19
19
3
3
since 1 obviously has a square root mod 3; we conclude that 19 has no
square root mod 79. This technique works very well as long as at each step
the numbers involed are prime, but breaks down otherwise:
✓ ◆
✓ ◆
✓ ◆
19
67
10
=
=
.
67
19
19
Here, though, we are saved by the multiplicativity of the Legendre symbol:
✓ ◆ ✓ ◆ ✓ ◆
ab
a
b
=
·
.
p
p
p
This formula allows us to continue
✓ ◆
✓ ◆ ✓ ◆
✓ ◆✓ ◆
10
2
5
2
19
=
·
=
=
19
19
19
19
5
✓
2
19
◆✓ ◆
4
=
5
✓
2
19
◆
2
since 45 = 1 obviously. So we run aground on trying to evaluate 19
. The
only
for the process to come to an end is at a symbol of the kind
⇣ ⌘other way
p 1
1
2
= ( 1) .
p
Problem 96 In our proof of Quadratic Reciprocity, we came up with the
equations
0p 1
1
p 1
⌫
✓ ◆
2
2
X
X
ja
a
(a 1) ·
j = p@
nA + 2R
and
= ( 1)n ,
p
p
j=1
j=1
valid for p an odd prime and (a, p) = 1, and where R is some number
⇣ ⌘defined
along the way. Use these formulas to derive a simple formula for p2 .
7.3
The Jacobi Symbol
In evaluating Legendre symbols this way, we need to check at every stage
whether our ‘numerator’ is prime and, if not, factor it into its prime pieces
and continue. This suggests an extension of the notation to handle composite
numbers, and this extension is called the Jacobi symbol. The notation is
exactly the same, and it is defined by two simple rules:
⇣ ⌘
1. ap is the ordinary Legendre symbol when p is prime, and
2.
⇣ ⌘
a
xy
=
a
x
⇣ ⌘
a
y
.
a
x
Problem 97 True or False:
Z/x.
= 1 if and only if a has a square root in
Let’s write down the basic rules for Jacobi symboles.
Problem 98 Let a, b, x, y 2 Z.
(a) Show that
ab
x
(b) Show that xa
free part of a.)
= xa xb for any a, b, x 2 Z.
⇣
⌘
= sqfr(a)
. (Remember that sqfr(a) is the squarex
(c) Show that if a ⌘ b mod x, then
a
x
(d) Show that if x is odd, then
= ( 1)
(e) Show that if x is odd, then
1
x
2
x
b
x
=
= ( 1)
.
x 1
2
x2
8
1
.
.
Most importantly for our purposes is that the Jacobi symbol inherits the
Quadratic Reciprocity law.
Theorem 15 (Quadratic Reciprocity for Jacobi Symbols) If x, y 2 Z are
odd and relatively prime, then
✓ ◆⇣ ⌘
x 1 y 1
x
y
= ( 1) 2 2 .
y
x
Problem 99
(a) Show that if x and y are odd, then
x
1
2
+
y
1
2
⌘
xy
1
2
mod 2.
(b) Prove Theorem 15.
Hint This should follow fairly easily from the Quadratic Reciprocity theorem for the Legendre symbol.
Problem 100
(a) Evaluate
207
307
.
(b) Does 501 have a square root in Z/1000? What about 503?
(c) Does 101 have a square root in Z/1001?
8
Functions Defined by Continued Fractions
Suppose we have a list of real numbers x0 , x1 , . . . , xn , . . ., all greater than or
equal to 1. Then we may form the infinite continued fraction
hx0 , x1 , . . . , xn , . . .i.
We talked about letting one of the entries vary, thereby obtaining a function
of just one variable. Let’s write
fn (t) = the result of replacing xn with t in hx0 , x1 , . . . , xn , . . .i .
Thus we are able to define a whole family of functions fn : [1, 1) ! R. The
question I want to ask is: What kind of functions are these? One way to
phrase this is to define
F = {all functions of this form for all possible lists (xn )}.
How can we describe F? A good place to start on questions like this is
(a) Find some basic examples of functions in F
(b) Find some closure properties: rules that tell us that if we perform some operation on the functions in F then the result is also
in F.
After we’ve done this kind of thing for a while we might be tempted to
conjecture something like:
F is the smallest collection of functions that (a) contains basic examples ,
and (b) is closed under simple operations .
Often such theorems are not too hard to prove once you’ve formulated them.
Problem 101 Make a list of some functions in F. How simple can you make
them? How complicated?
Problem 102 Suppose f, g 2 F, c 2 R.
(a) Is f + g 2 F?
(b) Is f
g 2 F?
(c) Is c · f 2 F?
(d) Is
1
f
2 F?
(e) Is f invertible? Is f
1
2 F?
In each case, you should consider the possibility that the answer is ‘maybe’
and explore it: under what conditions on f, g and c does it turn out to be
true?
Hint Sometimes it can be useful to write f = fn for some list (xn ) and
g = gm for some list (yn ) and prove things by induction on max{m, n}.
What about finite continued fractions? What about continued fractions
with integer coefficients?
8.1
Periodic Contiued Fractions
Now let’s look at continued fractions of the form
x = hx0 , x1 , . . . , xr , x0 , x1 , . . . , xr , . . .i
where the sequence x0 , x1 , . . . , xr repeats forever. Then we have
x = hx0 , x1 , . . . , xr , xi
and so there is
ax + b
,
cx + d
meaning that x is a root of a quadratic. What if instead we have
x=
x = hy0 , y1 , . . . , ys , x0 , x1 , . . . , xr , x0 , x1 , . . . , xr , . . .i
where the x-section repeats. In this case we have
x=
ay + b
cy + d
where y is a root of a quadratic.
Problem 103 Let Q = {roots of quadratic polynomials in Q[x]}.
(a) Show that Q is a field.
(b) Show that the value of a purely repeating continued fraction is in
Q.
(c) Show that Q is closed under rational functions.
(d) Show that every eventually repeating continued fraction is in Q.
9
Recognizing Sums of Two Squares
Now we ask if it is possible to write a positive n in the form n = a2 + b2 with
a, b 2 Z. Certainly some numbers, such as 1, 2, 4, 5, . . . are such sums, while
others, such as 3, 6, 14, . . . are not. How can we tell which is which? Is 1001
a sum of squares? To attack this question, we’ll introduce some notation:
S = {n | n = a2 + b2 for some a, b 2 Z}.
Problem 104 Show that S is closed under multiplication: that is, if x, y 2
S, then xy 2 S as well.
This is great, because every time we get a new number in S it generates
a huge collection of other numbers too. It also suggests that we focus our
attention on prime numbers in S. We know of course that 2 = 12 + 12 2 S,
so we can concentrate on odd primes.
Problem 105 Show that if p is prime and p ⌘ 3 mod 4, then p 62 S but
p2 2 S.
Hint Work in the ring Z/p.
So now we are left with the question: what if p ⌘ 1 mod 4? We can see
that 5 = 22 + 12 2 S, and 13 = 32 + 22 2 S. That is suggestive!
Problem 106 Let p be a prime with p ⌘ 1 mod 4, and fix an integer s such
that s2 ⌘ 1 mod p.
(a) Show that there are integers u1 , u2 and v1 , v2 such that
• either u1 6= u2 or v1 6= v2 (or both)
p
• 0  u1 , u2 , v1 , v2 < p, and
• u1 sv1 = u2 sv2 .
(b) Show that p = (u2
u1 )2 + (v2
v1 ) 2 .
Let’s sum up the situation: the set S contains: (a) 2; (b) all primes p of
the form p = 4n + 3; (c) the squares p2 for all primes of the form p = 4n + 1.
It follows that every number n satifying
if p|n and p ⌘ 3 mod 4, then the largest power of p dividing n is
even
Theorem 16 This condition completely characterizes the set S; that is,
S = {n | if p2k 1 |n but p2k - n then p ⌘ 1 mod 4}.
Problem 107 Suppose p is prime and p|x2 + y 2
(a) Show that if p - x, then
1 has a square root mod p.
(b) Show that if p ⌘ 3 mod 4, then p|x and p|y.
(c) Show that if p ⌘ 3 mod 4, then the largest power of p that divides
x2 + y 2 is even.
(d) Prove Theorem 16.
10
Geometry of Numbers
We write R for the set of all real numbrers and Rn for the set of all ordered
n-tuples (x1 , x2 , . . . , xn ) of real numbers. Inside of Rn is the set Zn of all
ordered n-tuples of integers, called the integer lattice. The geometry
of numbers is a term that refers to theorems that guarantee that certain
subsets S ✓ Rn must contain points in Zn .
10.1
Volume
We need to have a little talk about volume. What do you know about
volume? Well,
1. the volume of an n-dimensional ‘rectangular solid’ (or ‘parallelpiped’)
R = [a1 , b1 ] ⇥ [a2 , b2 ] ⇥ · · · ⇥ [an , bn ] = {(x1 , x2 , . . . , xn ) | ai  xi  bi }
is just the product of its lengths:
volume(R) =
n
Y
(bi
ai ).
i=1
2. If you take a set S and move it (by translating, rotating or flipping it
over) somewhere else to get a new set T , the volume does not change:
volume(S) = volume(T ).
3. If you cut a set into nonoverlapping pieces:
[
S=
Si
with Si \ Sj = ? for i 6= j,
i
then the volume of S is the sum of the volume of the pieces:
X
volume(S) =
volume(Si ).
i
We know that we can use these three rules only to find formulas for the areas
of triangles, polygons, circles, and ultimately evaluate integrals; and also find
the volumes of polyhedra, spheres, and so on.
What do those derivations actually tell us, though? They tell us precisely
this:
If there is a volume function satisfying these rules, then it must
be true that the area of a triangle is bla, and the volume of a
tetrahedron is bla bla bla, and so on.
These derivations presuppose the existence of a consistent way of assigning volumes to subsets of Rn and deduce what values must be assigned to
triangles, tetrahedra and so on. Formally, we’d like there to be a function
volume : {subsets of Rn } ! [0, 1]
satisfying the three properties we listed above; is there such a function? It
took a long time before anyone thought to ask this question, and it was a big
surprise when we learned that the answer is no! There is no way to assign
volumes to all subsets of Rn that is consistent with our three rules.
Here’s a great example of why. Write B(x, r) for the closed spherical ball
in Rn with center x and radius r.
Theorem 17 (Banach-Tarski Paradox) Let x, y 2 Rn such that distance(x, y) >
2. Then it is possible to cut B(z, 1) into 5 disjoint pieces
B(z, 1) = P1 t P2 t · · · t P5
that can be moved (by translation, reflection or rotation) to give sets Q1 , Q2 , . . . , Q5
such that
B(x, 1) [ B(y, 1) = Q1 t Q2 t · · · t Q5 .
It is not possible to do this with four sets. (The square union symbol t is
notation for a union of sets that are disjoint from one another.)
Problem 108 Explain why the Banach-Tarski Paradox implies there can
be no volume function as described above.
Nevertheless, the concept of volume is clearly a very useful one, and the
way out is to give up the illusion that volume makes sense for all subsets
of Rn . Certain sets are just so weird that there is no sensible way to assign
them a volume. Conceptually, what we do nowadays is specify a collection
M of subsets of Rn , called the collection of measurable sets, and define a
volume function
volume : M ! [0, 1]
satisfying our three requirements. And it turns out that those three requirements are enough to completely determine the volume function!
Just about every set you can think of without looking for trouble is measureable, and in particular:
• If S and T are measurable sets, then so is S \ T .
• If S ✓ T are both measurable sets, then so is the complement T
S.
From now on, when we write volume(S), we are implictly assuming that
S is a measurable set.
Problem 109
(a) Show that if S ✓ T , then volume(S)  volume(T ).
(b) Show that none of the sets Qi in the Banach-Tarski Paradox can
be measurable.
10.2
Minkowski’s Theorem
The main work of proving Minkowski’s theorem is hidden in Blichfeldt’s
Principle.
Theorem 18 (Blichfeldt’s Principle) Let S ✓ Rn be any measurable subset
with volume(S) > 1. Then there are two distinct points a, b 2 S with
a b 2 Zn .
Exercise 110 Try to find a counterexample in the plane.
Here’s some notation to help us organize our proof. For z 2 Zn write
Cz = {x 2 | zi  xi < zi + 1 for all i}.
This is the unit cube based at z. For any set S ✓ Rn and x 2 Rn , the
translation of S by x is simply
S + x = {s + x | s 2 S}.
Problem 111 Show that if Theorem 18 is equivalent to
If volume(S) > 1 then S \ (S + z) = ? for all z 2 Zn .
Problem 112 Suppose S is a counterexample to Theorem 18. For each
z 2 Zn , write
Sz = S \ C z .
S
(a) Show that the sets Sz are pairwise disjoint and that S = z2Zn Sz .
(b) Show that the sets Sz
z are pairwise disjoint.
(c) Show that
volume(S) =
X
volume(Sz
z).
z2Zn
(d) Finish the proof of Theorem 18.
Hint Find an upper bound for the sum in part (c).
We’re going to use Blichfeldt’s Principle as a lemma in the proof of
Minkowski’s Theorem. To state it we need to recall two definitions:
1. A set S ✓ Rn is convex if for every x, y 2 S and t 2 [0, 1], the element
tx + (1 t)y is also in S.
2. A set S ✓ Rn is centrosymmetric if for every x 2 S,
x is in S too.
Exercise 113 Show that every nonempty convex and centrosymmetric subset S ✓ Rn contains 0.
Theorem 19 (Minkowski) Let S ✓ Rn be convex and centrosymmetric with
volume(S) > 2n . Then S contains a point z 2 Zn {0}.
Problem 114 Let S be a set as in Theorem 19, and let
T = 12 S = { 12 x | x 2 S}.
(a) Show that if x, y 2 T then x ± y 2 S.
(b) Use Theorem 18 to prove Theorem 19.
10.3
Lattices
A lattice in Rn is a discete addtive subgroup ⇤ ✓ Rn . If you know the
topological meaning of the word discrete, I have nothing more to say to
you. But if you don’t, you can use the following simple replacement:
there is a number r > 0 such that for any distinct elements x, y 2
⇤, the distance from x to y is at least r.
For example, Zn is a lattice in Rn because any two elements have distance at
least 1 from one another. A lattice ⇤ is a full lattice in Rn if Rn = span(⇤);
Zn is obviously a full sublattice of Rn .
Here’s a basic theorem that we’ll have to take for granted.
If ⇤ is a full lattice of Rn , then there is a linear isomorphism
T : Rn ! Rn such that T (Zn ) = ⇤.
As we know from linear algebra, every linear transformation T : Rn ! Rn
changes all volumes by a uniform ‘volume scale factor’ that depends on T .
This scale factor is ‘easily’ computed: it is the determinant of the matrix A
that represents T . Thus we have
volume(T (S)) = det(A) · volume(S)
for all measurable S ✓ Rn . This volume scale factor is an important invariant
of the lattice ⇤, and we’ll refer to it as d(⇤).
Theorem 20 Let ⇤ be full lattice in Rn , and let S ✓ Rn be convex and
centrosymmetric. If volume(S) > 2n d(⇤), then S contains a point of ⇤ {0}.
Problem 115 Derive Theorem 20 from Theorem 19.
10.4
Recognizing Sums of Four Squares
As an application of Minkowski’s Theorem, we’ll prove Lagrange’s theorem.
Theorem 21 (Lagrange) Every positive integer can be written a sum of
four squares.
Problem 116
(a) Show that three squares is not enough for certain integers. Can
you find an infinite family of integers that are not sums of three
squares?
We’ll prove this using the—by now—time-tested technique of studying
the entire collection
F = {n | n is a sum of four squares}.
of numbers for which the equation x21 + x22 + x23 + x24 = n has a solution.
Lemma 22 The set F is closed under multiplication.
We saw in class that the set of positive integers that can be written as a
sum of two squares is closed under multiplication: this was just an algebraic
identity which we could discover with a certain amount of determination.
This lemma can also be proved by finding an appropriate algebraic formula,
and I invite you to search for it. But I will also understand if you choose to
take it for granted and move on.
As a result of our lemma, it suffices to simply show that F contains every
prime number.
Problem 117 Let p be a prime number. Show that there are r, s 2 Z such
that r2 + s2 + 1 ⌘ 0 mod p.
Hint Study the sets
S = {s2 + 1 | s = 0, 1, . . . , p 2 1 }
and
T = { s2 | r = 0, 1, . . . , p 2 1 }.
Problem 118 Let p be a prime and choose, once and for all, r and s such
that r2 + s2 + 1 ⌘ 0 mod p. Consider the lattice ⇤ defined by the matrix
2
3
p 0 r s
6 0 p s
r 7
7
A=6
4 0 0 1 0 5
0 0 0 1
(a) What is d(⇤)?
(b) Show that if x 2 ⇤, then p|x21 + x22 + x23 + x24 .
p
(c) Let S = {x | |x| < 2p}. What is the volume of S?
Hint I don’t expect you to know this formula o↵ the top of your
head. Do some research or some integration.
(d) Prove Lagrange’s Theorem.