Modular arithmetic
Divisibility Given p ositive numb ers a b if a we can write
b aq r
1
for appropriate integers q r such that r a The numb er r
is the remainder We say that a divides b or ajb if r and so
b aq i e b factors as a times q
Primes A numb er is prime if it can t b e factored as a pro duct
of two numb ers greater or equal to If a numb er factors i e it is
not prime then we say that it is comp osite
Exercise Find all prime numb ers smaller than
Here are several imp ortant questions
How do we determine whether a numb er is prime
How do we factor a numb er into primes
How can we nd big prime numb ers
How many prime numb ers are there
Exercise Factor
Exercise Show that every numb er is either prime or divisible by
a prime numb er
Theorem There are in nitely many prime numbers
Proof Supp ose that there are only nitely many prime numb ers P P
1 n
then N P P P is not divisible by any prime hence it do es
1 2 n
factor and hence it is a new prime not ˜
Divisibility tricks Recall that a numb er is even if its last digit
is divisible by it is divisible by if the sum of its digits is divisible
by it is divisible by if its last digit is divisible by etc
Why is this Can we nd other divisibility tricks eg for
One p ossible explanation Let abc b e a digit numb er so that abc
a b c Then
abc a b c and this is an integer only if c is an
integer
abc a b c and this is an integer only if c is an
integer
abc a b a b c and this is an integer only if
a b c is an integer
A b etter explanation is given by congruences
1
In Python notation r b a and q b a 1
2
Congruences We say that a is congruent to b mo dulo N i e
a b mod N i N divides a b or equivalently i a N b N So a
is congruent mo dulo N to any numb er in the arithmetic progression
f a N a N a a N a N g
Here are some examples
mod
mod
mod
mod
Exercise Is divisible by How ab out
100
Exercise What is the last digit of
111
Of
What are the last two digits of the numb ers ab ove
Solution Every integer mo dulo is congruent to one of
f g
we call this a system of residues mo dulo We now compute the
100
residue of We have
2 3 4
Then
100 4 25 25
Arithmetic of residues mo dular arithmetic Residues mo d
ulo f g
Z
Exercise Write down the addition and multiplication tables
We can compute etc We have
Check Now a b makes little sense eg
1 1
what do es mean We should write a b eg From the
1
multiplication table we see that so that and then
1
Exercise Write down the addition and multiplication ta
bles for
Z Z
Find the additive and multiplicative inverses of mo d
mo d mo d and mo d
Find the order of mo d
Find the orders of mo d Do you see a pattern
2 2
Can you solve x mo d and x mo d
Show that if a b c d then a b b d and ac bd
3
The Euclidean Algorithm
The greatest common divisor By de nition the greatest com
mon divisor of two numb ers a b is the largest numb er that divides
b oth a and b We denote it by GC D a b or just by a b Eg
To see the last one write
and so that the sets of divisors are
f g and f g
We can compute the GCD in the following ways
By brute force Find all the divisors and compare
Find all divisors of a and throw out the ones that do not divide
b
Compare prime factors and multiply the highest p owers that
divide b oth numb ers
The Euclidean algorithm is by far the fastest metho d for big numb ers
The Euclidean algorithm Start with a b and
let r a b b e the remainder Replace a by b and b by r and rep eat
The last nonzero remainder is the GCD
a b r q
The GCD is
Exercise Find the GCD of and of
Exercise Factor the ab ove numb ers
Python Co de def gcd a b
r a b
print a b r
while r
a b r b r b r
print a b r
return b
It works as follows
gcd
Well ordering principle Each nonempty set of positive integers
has a least element
4
Note that this is not true for the rationals Eg has no smallest 0
Q
numb er
Prop osition Given positive integers a b let S fax by g
where x y are any integers Let g be the least element in S then
g GC D a b
Proof It is enough to show that g divides GC D a b and that GC D a b
divides g
Clearly GC D a b divides a and b so it divides any element in S and
so it divides g
To see that g divides GC D a b it is enough to show that g divides
a and that g divides b Consider a g q r then r a g q
a ax by q q ax bq y Now r g and so r
else r S contradicting the fact that g is the smallest element of S
Therefore g divides a and by the same argument it divides b and so
divides GC D a b it ˜
So we can always write
GC D a b ax by
This allows us to nd the inverse of any numb er a p mo dulo
p when p is prime In fact GC D a p so we have ax py so
1
mo dulo p we have ax i e x a mo dulo p
Finding x and y Use back substitution Eg for a
b then GC D a b Recall that
a b r q
So and
Then
Exercise Solve
x y
x y
Exercise Find the multiplicative inverse of
mo d
mo d
5
Python co de for the Extended Euclidean Algorithm
def egcd a b
q r divmo d a b
x y u v q
while r
qq rr divmo d b r
xx yy u qq x v qq y
a b r q x y u v b r rr qq xx yy x y
print a b r q x y u v
return u v b
Exercise What are the last digits of the multiplicative inverse
100
of mo d
Exercise I have enco ded my SSN using the algorithm f x x
10
mo d The result is What is my SSN You will need
a calculator that can compute with digit numb ers eg python
Modular Powers
We b egin by writing a table for the p owers mo dulo
Mo dulo
2 3 4 5 6 7
a a a a a a a
Note that after we get a column with all s and a at the top
the pattern must rep eat Therefore as so on as and if this happ ens
the table just keeps rep eating itself and so we do not b other to keep
writing the rep eated entries
Mo dulo
2 3 4 5 6 7
a a a a a a a
Here the th column has all s and a at the top
Mo dulo
2 3 4 5 6 7
a a a a a a a
6
Do you notice a pattern The rst pattern seems to b e that for all
2 4 6
a we have a mo d a mo d a mo d It is easy
to guess what the general pattern might b e
Theorem Fermat s Little Theorem Let p be any prime and
a p Then
p 1
a mo d p
22 22
Eg mo d Check Similarly
100 100
mo d Note that has more than digits so
checking explicitly might not b e so easy
Proof We claim that the set fa a p ag is the same as the set
f p g the elements are not in the same order Granting
the claim then we have
p 1
p a a p a p a
Since p do es not divide p it follows that GC D p p
so there is a numb er x p such that x p
mo d p But then
p 1 p 1
x p x p a a
as required
To see the claim we must check that
The map x ax mo d p induces a map
f p g fa a p ag
I e that for all x p we have that p do es not divide ax This
is clear as p is a prime numb er that do es not divide either of a and x
If x y p then ax ay mo d p i e this function is
to Supp ose in fact that ax ay mo d p As GC D a p there
is a numb er b such that ba mo d p Therefore bax bay so that
y x ˜
12345
Exercise Compute mo d
12345
Exercise Compute mo d
Exercise Compute the order of mo d and
Can you make a guess ab out the general pattern
Euler s function How do es Fermat s Little Theorem general
ize to numb ers m that are not prime The key thing in the pro of is that
f p g are coprime with p so that they have multiplicative
inverses in It is then natural to de ne
Zp
m fa a m and GC D a m g
This is Euler s function We now compute its value for some m s
p p for all prime numb ers p Eg
7
f g
f
g
2 2
p p p for any prime numb er p To see this note that
2
in the list f p g we must erase p p p p so we
2 2
have p p p p entries
n n n 1
p p p for any prime numb er p Explain this in
detail
f g
f g
f g
f g
p q p q for any prime numb ers p q To see this
note that in the list f pq g we must erase p p q
p and q q p q so we have pq p q
p q entries
m n m n whenever GC D m n
Exercise Prove the last fact
Exercise Compute
Theorem Euler s Formula If GC D a m then
(m)
a mo d m
5243
Exercise Compute mo d
90
Exercise Compute mo d Is prime
m
So if we pick a random numb er a m and we have a a mo d
m
m then m is not prime If a a mo d m then m mayb e prime
m
If a a mo d m for many a s then m is probably prime
m
Exercise There are some comp osite numb ers such that a a
mo d m for all a m Can you nd one
1 000 000 000
Powers mo d m How would you compute mo d
by hand Or mayb e with a p o cket calculator
so
9 1 000 000 000 2200
so mo d
2 4 2 8 2 16 2
32 64 128 256 512
1024 2048
Now so
2200
mo d
This was remarkably quick It turns out that we did not need to
p erform op erations Just ab out l og so ab out
2
op erations note that in base the numb er is
1 000 000 000
mo d mo d mo d Exercise Compute
8
Exercise Estimate the numb er of doublings needed to compute
100
10 2
mo d
9990
Exercise Compute mo d without factoring Is
prime Factor
k th ro ots mo d m We wish to solve the equation
k
x b mo dm
5
For example x mo d
We have that and so
Now GC D and so we can invert mo dulo We
must write u mo d We use the Euclidean Algorithm to
get Therefore
5 3917 97922+1 9792 2
x x x x x mo d
3917
So x mo d
k
The general strategy to solve x b mo d m where GC D b m
and GC D k m is exactly the same
We compute m This is easy if we can factor m
Write k u m v Use the Euclidean Algorithm
1+ (m)v k u u u
Then x x x b Compute b via the metho d of
successive squaring
5 3
Exercise Solve x mo d and x mo d
RSA
The rst step is to convert the alphab et to numb ers Eg A
B C D E F G Y
Z So
GO O D M O R N I N G
This co de is very easy to break by frequency analysis So we want to
jumble the numb ers up in a way that will lo ok completely random to
the co de breaker
Fix p q two very large primes eg to digits Let m pq so
that m p q Pick k with GC D k m
Everyone knows m k they are public
p q and hence m are secret
For example let p q so that m Pick
k
Break a message up in to numb ers m Eg
k
Let b a mo d m We compute this by rep eated squaring to
i
i
get the enco ded message b b This is made public sent
1 r
77
to the intended receiver Eg b a mo d gives
i
i
9
k
To deco de solve x b mo d m So nd j with j k mo d
i
j
m Then a b mo d m Eg j Check that k j
i
i
8213
mo d and that mo d etc
This system is very easy to explain and implement on a computer
In order to deco de it you must know m So you must factor m
100
This is very hard Factoring a digit numb er requires ab out
computations In contrast it is easy to pro duce big primes We will
do this later
Question Couldn t we nd m without factoring m and so easily
break the co de
Answer No Since m pq if we know m p q
m p q then we know p q m m and so we can nd the
2
ro ots of x p q x m But these are the ro ots of x p x q
and hence we have found p q
Exercise Enco de GOODMORNING using m digits at a
time Deco de
RSA signatures In practice each individual A B C will have
will have his her secret primes p q and will make m p q and
A A A A A
k public
A
Anyone can encrypt a message and send it to A or B C etc Just
do
k
k
A
A
a a a a b b
1 r 1 r
1 r
We assume that b b is public or could b e intercepted
1 r
Only A can deco de this message since only A knows m so only
A
j
A
A can solve j k mo d m and then compute a b mo d m
A A A i A
i
Supp ose that A is exp ecting a message from B How do es A know
that it really comes from the friendly B and not from the evil C
B can attach a signature as follows
j
j
B
B
B AR T S I M P S O N mo d m
1 s B
1 s
Only B can do this Only B can nd j
B
Now B enco des his message with the signature app ended
j j
k
j k k j k
B B
A
B A A B A
a a b b b b a a
1 r 1 r r +1 r +s
1 s 1 r 1 s
j
B j
B
Only A can deco de the message and see a a But
1 r
1 s
everyone knows k so A can compute
B
j
k j k
B
B B B
B AR T S I M P S O N
1 s
1 s
So A can tell that B sent the message and no one else can
Counting large primes
We already know that there are in nitely many prime numb ers but
how common are they
Eg Half of the numb ers are even
10
To b e more precise if we let
ev x fev en number s n w ith n xg
then ev ev ev ev ev etc So
ev x x if x is even and ev x x if x is o dd Then
ev x x if x is even and ev x x x if x is o dd
Then ev x x and x Eg ev and
ev
We will now consider the function
x fpr imes p such that p xg
Exercise Find Use a computer to
nd
We have
x
x
x x
It would b e reasonable to guess that x x as x This
means that for any we have that for big values of x the inequality
x x
The Prime Numb er Theorem
Theorem We have that x x l n x as x i e
x
lim
x!1
x l n x
This means that for large x the numb er x is very close to x l n x
10 10 10
Eg l n l n So the
10
probability of a random numb er n b eing prime is
10
Exercise How many primes are there How many primes
10 10 5
with digits How many primes b etween and
Finding large primes
We would like to pro duce big prime numb ers eg with digits
We could pick a random numb er p with digits and try and divide
1 2 100 1 2 50
by all the numb ers p This would take way
to o long
Exercise How long would this take assuming we can p erform
divisions each second
We will happily settle for a numb er which is prime with probability
say
11
Primality testing How can we test if p is prime By Fermat s
little theorem if p is prime then
p
a a mo d p for all a p
p p
So if a a mo d p then p is de nitely not prime If a a mo d p
p p
a a a we have some evidence that p might b e prime If a
2 1
2 1
p
a a mo d p we are tempted to conclude that p is probably
100
100
prime But what is the actual probability Also do es this test detect
all non prime numb ers
10 10 10 10 10 10
If n then and
mo d so we have negatives AKA witnesses and false p ositives
the numb ers Therefore this test is e ective
Eg n has negatives witnesses i e numb ers such that
p
a a mo d p So this test is accurate
Eg n accuracy Problem
561
satis es a a mo d for all a The
test is accurate
561
Proof It is enough to show that a a
mo d
2 560 2 280 280
Mo dulo If GC D a then a so a a
10 560 10 5
Mo dulo If GC D a then a so a a
56
16 560 16 10
Mo dulo If GC D a then a so a a
10
˜
p
Numb ers with this prop erty a a mo d p for all a p are
called Carmichael numb ers
Eg are all the Carmichael
numb ers less than
So mayb e these numb ers are very rare and this is not a problem
Exercise Can you show that there are in nitely many Carmichael
numb ers
560
Notice that mo d so mayb e we should b e testing
m 1 m
for a mo d m not a a mo d m It would b e even b etter to
nd a test that always works
The Miller Rabin test
k
Theorem If p is prime then we write p q where q
is odd If a p then either
q
a mod p or
k 1
q 2q 4q 2 q
a or a or a or or a mod p
Proof We will use that fact that if p is prime then the only numb ers
k k 1
2 q p 1 2 q
a so a that square to mo d p are Now a
12
k 1 k 2
2 q 2 q
If a we are done otherwise a Rep eat this
q q
pro cedure until p ossibly a if its a we are in case
q
a and we are in case Otherwise ˜
Notice that for a comp osite numb er such as we may have many
more square ro ots of eg
The Rabin Miller test for prime numb ers works as follows Pick n
k q
any o dd integer and write n q with q o dd If a mo d n
i
2 q
and a mo d n for i k then n is comp osite
Theorem If n is composite and odd then at least of the
numbers between and n fail the Rabin Mil ler test So at least
of such numbers are Rabin Mil ler witnesses
So if n passes the test for random numb ers then n is not prime
100 200 60
with probability less than
Exercise Apply the Rabin Miller test with a to n
The Primitive Element Theorem
We now return to the study of p owers of a numb er a mo d p where p is
prime By Fermat s Little Theorem we know that for all a p
p 1
we have a mo d p We would now like to answer the question
e
Are there any smal ler powers such that a mod p Let e a b e
p
e (a)
p
the smallest p ositive numb er such that a mo d p The numb er
e a is the exp onent of a mo d p
p
Exercise Find e a for a p and p Do you
p
see a pattern
There are two main observations that are true for all primes
Theorem Let p be any prime and a p Then
e a divides p
p
There are exactly p distinct numbers a p such
that e a p
p
Proof For if e a do es not divide p then g GC D e a p
p p
e a and we may write g xe a y p so that
p p
g xe (a)+y (p 1) e (a) x p 1 y x y
p p
a a a a
and this contradicts the fact that e a is the smallest integer such that
p
e
a mo d p
For we pro ceed as follows since e a always divides p we
p
b egin by counting how many a s with exp onent e p are there
n
If p q where q is prime then if e a p we have that
p
n 1
n 1 q
e a divides q and hence a is a ro ot of the p olynomial x
p
n 1 n n 1
There are at most q such ro ots and so there are at least q q
n
n
1
r
with r then if e a p primitive ro ots If p q q
p
r 1
13
we have that e a divides p q or e a divides p q so
p 1 p 2
(p 1) q (p 1) q
2 1
There are at most x a is a ro ot of x
p q p q distinct ro ots of this equation the ro ot is
1 2
counted twice It follows that there are at least
p p q p q p
1 2
q q
1 2
1 1
primitive ro ots We have used the fact that q so that
i
q q
1 2
We have so far veri ed that there is at least one primitive ro ot say
g So we know that
2 p 1
f p g fg g g g
i
as unordered sets The order of g is p exactly when GC D i p
exercise for the reader and so we have exactly p primitive ro ots
˜
Exercise Find an element of order mo d i e a primitive
element What are the other elements of order Find elements of
order Is there an element of order
Squares modulo a prime
Exercise Is a square mo dulo I e can we nd a numb er a
2
such that a mo d
Exercise Compute all squares mo d Do you notice
a pattern
Theorem Let p be a prime then there are p non zero
squares mod p
2 2
Proof Clearly a a so there can b e at most p distinct
squares Supp ose that there are less than p squares then there
is a numb er say a which is the square of at least di erent numb ers
2
say b c d mo d p But then b c d are distinct ro ots of x a This is
2
imp ossible In fact then x b x c and x d all divide x a so
2
a is divisible by a degree equation which is imp ossible x ˜
2
Exercise Show that if p is a prime and a is a ro ot of x bx c
2
then a and b a are the only ro ots of x bx c
2
Exercise Find a numb er m and an equation x bx c that
2
has at least solution Conclude that x bx c has more than one
We will call the squares mo d p Quadratic Residues or just QR
The non zero numb ers that are not QR will b e called Non Residues
or just NR
14
The next observation is that if a and b are squares mo dulo p then
ab is a square mo d p In other words
QR QR QR
Exercise What happ ens when you multiply QR NR NR
NR or NR QR mo d Exp eriment mo d Can you draw any
conclusions
Theorem Let p be a prime then modulo p we have that QR
QR QR QR NR NR QR NR and NR NR QR
Proof The primitive ro ot Theorem states that there is an integer g
2 p 1
such that g g g gives all numb ers x p mo d p
The QR s are exactly the even p owers of g and the NR are the o dd
p owers So to verify that NR QR NR we take a non residue of
2k +1 2j
the form g and a residue of the form g and we multiply them to
2k +1 2j 2(k +j )+1
g g g which is not a residue as k j is o dd get ˜
So multiplying QR s and NR s b ehaves like multiplying and
It makes sense to de ne the Legendre symb ol of a mo d p as follows
a a
if a is a QR mo d p and if a is a NR mo d p So if
p p
a is a QR and b is a NR mo d p then ab is a NR mo d p which can b e
expressed by the following equality
a b a
p p p
Exercise Find a primitive element mo d p Find all
QR s mo d p
Exercise Find the sum of all QR s a p mo d p
Do the same for the NR s Do you notice a pattern
1
Exercise Compute for p
p
Do you notice a pattern
The pattern is
if p mo d
p
if p mo d
p
This follows immediately from
Euler s Criterion If p is a prime then
a
(p 1) 2
mo d p a
p
1
2k
Clearly if p mo d then p k so that
p
1
2k +1
It and if p mo d then p k so that
p
remains to prove Euler s Criterion
15
2 (p 1) 2 p 1
Proof If a is a QR then a b so that a b as required
If a is a NR then a is an o dd p ower of a primitive element i e
2k +1
a g so that
(p 1) 2 (2k +1)(p 1) 2 p 1 k (p 1) 2 (p 1) 2
a g g g g
p 1 (p 1) 2
Now g so that g but since e g p i e p
p
e (p 1) 2
the smallest p ower e such that g then g is ˜
Next we would like to determine When is a QR mod p I e we
2
would like to compute for all o dd primes
p
2
for all primes p Do you see Exercise Determine
p
a pattern
The pattern which is hard to sp ot is
if p or mo d
p
if p or mo d
p
Proof We multiply the even numb ers
p
(p 1) 2
p
We now wish to rewrite the LHS p in a di erent way
Consider the numb ers p which are p p p
or equivalently So when we multiply p
we are actually multiplying all the even numb ers p and all
2
the o dd numb ers p with a minus sign Therefore
p
t
p
where t is the numb er of o dd integers a p Comparing
the two equations ab ove and canceling a p we get
(p 1) 2 t
mod p
We now can easily conclude via a case by case analysis
If p mo d then p k so that p k and t k
(p 1) 2 2k
Then mo d p and by Euler s Criterion is a
square mo d p
If p mo d then p k k k so that
p k k k and hence t k k k
(p 1) 2
and hence and by Euler s Criterion is a NR
QR mo d p NR ˜
1 2
Exercise Compute and for p
p p
2
Exercise Do es x x have a solution mo d
2
If p a p then p a so that p a i e p a
16
a
for any Quadratic recipro city We would like to compute
p
n
n
1
r
integer a p If we write a q q then we have
1 r
n
1 n
r
q
n
q
a
i
r
1
Now if n is even then q is clearly a square so
i
i
p p p
n n
i i
q q
q
i
i i
If n is o dd then Therefore we just need that
i
p p p
to compute
q
for any primes q p
p
q
Exercise Make a table for for the primes
p
Can you see a pattern Lo ok at the rows and columns for p
Now rub these out Is there a pattern
The pattern is the following
3
Theorem If p q are primes then
p q
if either p mo d or q mo d
p q
q p
if p mo d and q mo d
p q
44 51 91
Exercise Compute
53 101 127
We also have the Generalized Law of Quadratic Recipro city
Theorem Let a b be odd numbers then
1 1
if b mod and if b mod
b b
2 2
if b mod and if b mod
b b
a b a
if either a mo d or b mo d and
b a b
b
if a mo d and a mo d
a
Pythagorean triples
By de nition a Pythagorean triple is an integer solution to the
2 2 2
equation a b c It yields a rectangle triangle with sides of integer
length a b c You are probably familiar with the Pythagorean triple
Exercise Can you nd any other Pythagorean triples
Of course is also a Pythagorean triple and so is a a a
for any p ositive integer a We will say that a b c is a primitive
2 2 2
Pythagorean triple if a b c and a b c have no common factors
is another such triple
1 3 1
if p mo d and if p Together with the rules
p p
2 2
mo d and if p mo d and if p mo d this is
p p
known as the Law of Quadratic Recipro city
17
Exercise Can you nd any other primitive Pythagorean triples
Can you show that there are in nitely many such triples Write a
program to nd all such triples with a b
We would like to nd all primitive Pythagorean triples We notice
that if a and b are even then c is even so that is a common factor of
a b c and then a b c is not a primitive triple If a and b are b oth o dd
then a x b y and
2 2 2 2 2
c a b x x y y mo d
Now is not a square mo d so this is imp ossible Therefore we
conclude that the set a b are not b oth even or o dd So we may assume
that
a is o dd b is even and c is o dd
We now consider the factorization
2 2 2
a c b c b c b
We claim that Both c b and c b are squares
Proof First of all g GC D c b c b as otherwise g divides
c b and c b so that g divides c c b c b and g divides
b c b c b Since c b is o dd g is o dd so g divides c and
2 2 2 2
b and hence g divides a as g divides a c b so that as a b c
have no common factor we have g
So if p is a prime numb er such that p divides c b then p do es not
t t
divide c b Let p b e the highest p ower of p dividing c b then p is
2
the highest p ower of p dividing a c b c b Therefore t is even
n
n
1
r
and so if we write c b p p with p distinct primes we have
i
1 r
that all the n are even so that c b is a square The same reasoning
i
shows that c b is a square also ˜
So we write
2 2
c b s and c b t
where s t are o dd integers Therefore we have that
Theorem Al l primitive Pythagorean triples are of the form
2 2 2 2
s t s t
b and a st c
for some odd integers s t
Which prime numbers are sums of two squares
Exercise Find all prime numb ers that are sums of two
squares do you notice a pattern
The pattern is
Theorem A prime number p is the sum of two squares if
and only if p mod
18
Proof If p is the sum of two squares then p mo d To see
2 2
this write p a b As p is o dd then we may assume that a is
2
even and b is o dd So mo d we have a f g so that a mo d
2
Similarly b f g mo d so that b mo d so p mo d
If p mo d then p is the sum of two squares This implica
tion is much harder The idea is as follows By quadratic recipro city
2 2 2
we know that A mo d p so letting B we have A B
2 2
mo d p i e A B M p If M we are done Otherwise we need
2 2
a pro cedure to nd new integers a b m such that a b mp and
m M We then rep eat this pro cedure until m
This is known as Fermat s Descent Pro cedure This is how it works
Let p b e any prime with p mo d eg we may write
2 2 2 2
A B M p with M p eg
1
We may pick u A v B mo d M such that M u v
2
1
2 2 2 2
M eg u v then u v A B mo d M
2
2 2 2 2
We have u v M r and A B M p for some r M
2 2 2 2
eg and
2 2
Notice that r since otherwise u v so u v
2
so A u mo d M and B v mo d M But then M
2 2
divides A B M p so that M divides p But M p so
that this is imp ossible
2 2
2 2
(M 2) +(M 2)
u +v M
Notice that r M since r
M M 2
M
If we multiply these together we get
2 2 2 2 2 2 2
uA v B v A uB u v A B M r p
2 2
Notice that uA v B A B M p mo d M and
v A uB BA AB mo d M So uA v B and
v A uB are divisible by M Eg uA v B
and v A uB
We have therefore obtained the equation
2 2
uA v B v A uB
r p
M M
2 2 2 2
with r M eg Rep eating this
we will get r as required
˜
Exercise Use Fermat s descent pro cedure to write as the sum
of squares Do the same for p
Fermat s Last Theorem
In the th century Fermat stated that the equation
n n n
a b c
19
has no solutions for n and a b c Rememb er that for n
this equation has in nitely many solutions and we can describ e all of
these So this fact is very surprising It also turns out to b e a very
deep result which was eventually proven by Wiles at the end of the
th century We will show the following
n n n
Theorem The equation a b c has no solution when n is
divisible by and a b c
4 4 2
Proof It is enough to show that the equation x y z has no
solution for x y z In fact if a b c is a solution of the original
n 4 n 4 n 2
equation then x a y b and z c are solutions to our
new equation The idea is once again to show that if there is a solution
0 0 0 0
x y z then we can nd a new solution x y z with z z We
can t rep eat this in nitely many times so there was no solution to
b egin with
To b egin with notice that we may assume that x y z have no com
2 2
mon factor Now if we let a x b y and c z then a b c is a
primitive Pythagorean triple and so we may write
2 2 2 2
s t s t
2 2
x a st y b z c
2
for some o dd integers s t with no common factor Now st x is an
o dd square so that st mo d It follows that either
s t mo d or s t mo d
Also we have
2 2 2
y s t s t s t
Since s t are o dd and relatively prime then the only common factor
of s t and s t is Now divides s t so s t r where r is an
2
o dd numb er Now y s t s t so we have
2 2
s t u and s t v
where GC D u v Solving for s t we get
2 2 2 2
s u v and t u v
2
and so from the formula x st it follows that
2 4 4 2 4 4
x u v or x v u
2 2
We now let A x B v and C u so that
2 2 2
A B C
and A B C have no common factor i e A B C is a primitive Pythagorean
triplet Therefore we know that there are o dd relatively prime integers
S T such that
2 2 2 2
T S T S
2 2
x A S T v B u C
20
So we get
2 2 2
v S T S T S T
Now GC D S T S T S T are o dd and relatively prime So
we must have
2 2
S T x and S T y so
2 2 2 2
S x y and T x y so
2 2 2 2 2 2 2 2
S T x y x y
2 4 4
u x y
4 4 2
This is the new solution x y u to the original equation x y z
We must still show that u z but this is clear from the formula
2 2 2 2 2 2 2 2
s t u v u v
4 4
z u v
˜
Answers to the exercises
Ex
2
Ex
2
Ex do es not divide but it do es divide
100
Ex To compute the last two digits of we just need
2
to compute the last two digits of various p owers of
4 8 16 32 64 100
64+32+4 111 64+32+16+2+1
Ex mo d mo d mo d
1
mo d do es not exist
2 3 4 8
Mo d we have so
Mo d the p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
The p owers of are so the order is
2
the squares mo d are so x has no solution
2 2
mo d and
Ex
so GC D
21
Ex
and has no divisors We would need to use
a computer we got lucky here
Ex referring to ex
Ex so
so mo d
100
Ex abcd X So d mo d
i e d
c mo d i e c mo d so
that c mo d so that c
b mo d i e b mo d i e
b mo d so b
a mo d i e a mo d
so a mo d so a
10
Ex
So
10
10 1
So
10 10
mo d so SSN mo d
2 4 2 2 2 8 4 2 2
Ex
Since we have
12345 8 1543 1543
2 4 8 16 2
Ex
32 2
So
12345 100 123 45 123 32+8+4+1
3 6
Ex Mo d
4 8
Mo d The order is
5 6
Mo d The order is
6 7 8 9
Mo d
10 11 12
The order is
Ex
Ex
90
Ex mo d is not prime
22
9
Ex Mo d we have so and
9
10 16 4 8 2 16 2
Now
9 7 7
10 100 10 10
Mo d so that
9 4 8
Mo d
16 32 2 58 32+16+8+2
2 6
Ex and l og so ab out
2
doublings
9990
Ex mo d
5 5
Ex Mo d mo d so x x
5
mo d
Mo d and mo d so x
3 67 67 2 4 8 16
x mo d Now
32 64 67
So
Ex WELLDONE note that
and mo d
10 10 9
Ex
10 5 10
41 37 33
Ex s hr s years
4 35 70 2
Ex n mo d
140 2 280 2
mo d mo d mo d and
560 2
mo d So failed the Rabin Miller test therefore
it can not b e prime is a Rabin Miller witness for the numb er
4 3
Ex and mo d so the order of is The
5 7 1 2 3
other elements of order are has order has order
4 6
has order and has order
2
Ex We square all the residues mo d and we get
2 2 2 2 2 2 2 2 2 2
and
mo d So the answer is yes
Ex Mo d
Mo d
Mo d
Mo d
2
Ex x a x b a x b x a b a so we must show
2
that c a b a but this is clear as a is a ro ot of x b x c
2 2
so a ab c If z is a third ro ot of x b x c mo d p then
z a z b a so that p divides z a or it divides z b a
but then z a mo d p or z b a mo d p So these are the only two
ro ots
2 2
Ex are ro ots of x mo d so x x
x x x
Ex QR NR so QR NR NR QR
NR and NR NR QR
0 2 4 6 8
Ex Mo d QR
0 2 4 6 8 10
Mo d QR
23
0 2 4 6 8 10
Mo d QR
12 14
Ex Mo d Mo d
Mo d They are all mo d p
1 1 1 1 1
Ex
3 5 7 11
1 1 1 1 1
17 19 29 31
2 2 2 2 2
Ex
3 5 7 11 13
2 2 2 2 2 2
17 19 23 29 31 37
2 2 2
41 43 47
1 2
Ex and
p p
2 2 2
Ex x x x so is a
square mo d p Equivalently is a square mo d p
2 1
and so no Mo d
97 97
Mo d so no
Mo d so no
44 11 53 9 51 101 49
Ex
53 53 11 11 101 51 51
127 36 91
127 91 91
2 2 2 2 2 2 2 2
Ex
2 2 2 2 2 2
2 2
Ex A B so so M u
2 2
v so uA v B M and v A uB M so