CHAPTER 1 Prime factorisation Flashcards

1
Q

Motivation: Diophantine eq

A

x^n +y^n=z^n

has only trivial solutions for n>=3

how many integer solutions?

x^2+y^2=z^2 has integer solutions 3^2+4^2=5^2

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2
Q

N

A

N := {0, 1, 2, . . . }

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3
Q

integers

A

Z := {. . . , −2, −1, 0, 1, . . .}.

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4
Q

Definition 1.1.1 (Divisibility)

A

Let n, d ∈ Z. We say that d is a factor n, or d is a divisor of n, or that d divides n if there is some q ∈ Z such that n = qd. When this is the case, we write d | n, otherwise we write d ∤ n.

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5
Q

d divides n if

A

Take n,d in Z
d| n if there is k in Z s.t nk=d
d is a divisor/factor of n
12|24 8 doesnt divide 15
10|10a+20b

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6
Q

Example 1.1.2. 5 | 10, 5 ∤ 11, −5 ∤ 10. Moreover, note that n | 0 for all n ∈ Z. When is it
true that 1 | n? And what about 0 | n?

A

The number 1 divides n for any n ∈ Z because n = n · 1. On the other
hand, 0 | n only when n = 0, because n = q · 0 implies n = 0.

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7
Q

5|0?

A

yes 0 = 5 x 0

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8
Q

n|0?

A

Every n|0 including 0 itself

.

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9
Q

1|n?

A

for every n in N n=n x 1

The number 1 divides n for any n ∈ Z because n = n · 1.

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10
Q

0|n?

A

On the other
hand, 0 | n only when n = 0, because n = q · 0 implies n = 0

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11
Q

d|a and d|b then

A

d| (a+b)
a=dxk b =d x c
a+b = d(k+c) by def k+c in integers

more generally d|(ax+by) for every x,y in Z. if we write a = qd, b = rd, then ax + by = (qx + ry)d.

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12
Q

a | b, b | c imply that a | c

A

TRANSITIVE

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13
Q

a | b, b | c imply that a | c

A
  • It is reflexive: n | n because n = n · 1.
  • It is transitive: suppose that n | m and m | k. This means that m = nq, k = mr for some q, r ∈ Z. Then k = n(qr), so n | k.
  • It is not symmetric: for instance, 2 | 4 but 4 ∤ 2.

An additional detail: within N, divisibility is antisymmetric. Indeed, if a | b | a,
there are k, ℓ ∈ N such that b = ka, a = ℓb, so a = kℓa, hence k = ℓ = 1, therefore a = b. This makes | a partial order on N.

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14
Q

gcd definition 1.1.4

A

Let a, b ∈ Z not both zero. The highest common factor or greatest common divisor of a and b is the largest positive integer d such that d divides both a and b.

It is denoted gcd(a, b) := d, or sometimes (a, b) := d. We also set gcd(0, 0) := 0.

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15
Q

coprime

A

If gcd(a, b) = 1, then we say that a and b are coprime or relatively prime

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16
Q

gcd(8, 12) =
gcd(9, 0) =

A

gcd(8, 12) = 4, gcd(9, 0) = 0,

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17
Q

gcd(0,0)

A

=0

as common divisors are all naturals so can we take a max?

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18
Q

are 7 and 9 coprime?

A

gcd(7, 9) = 1 (hence 7 and 9 are coprime).

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19
Q

Exercise 1.1.6. What is gcd(n, n)?

A

1,n are definitely common divisors
gcd(n, n) = |n|

By def gcd(0,0)=0 so consider case
n ̸= 0:
|n| divides n, since |n|=n or |n|=-n so gcd(n,n) ≥ |n|.

Moreover, if d ≥ 0 divides n, meaning that n = qd for some q, then |n| = |q||d|, thus |d| ≤ |n| because |q| ≥ 1, hence gcd(n, n) ≤ |n|.

It follows that gcd(n, n) = |n|.

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20
Q

remark lcm

A

lcm(a, b) := ab/gcd(a,b) is
the ‘least upper bound’ of a, b (where lcm(a, b) = 0 when one of a, b is zero).

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21
Q

Lemma 1.1.7 DIVISION ALGORITHM

A

Let a, b ∈ Z, where b≠ 0. Then there exist unique q,r ∈ Z (respectively quotient, remainder) such that
a = qb + r and 0 ≤ r < |b|.

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22
Q

PROOF
Lemma 1.1.7 DIVISION ALGORITHM

A

Differs from notes

Proof. Suppose that b > 0. The set Q = {q′: q′b ≤ a} is bounded from above (for
instance, Q ≤ |a|), so it has a maximum q.

Note in particular that a < (q + 1)b. Let r = a − qb. By construction, 0 ≤ r < b.

When b < 0, we first use the conclusion with −b > 0 in place of b. We find that
a = q′(−b) + r where 0 ≤ r < −b = |b|. Now simply set q =−q′: we have
a =(−q′)b + r = qb + r, as desired

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23
Q

50 = 2 · 24 + 2

division 50 hours by 24

A

2 x 24 +2

2 days 2 hours

Example 1.1.9 (Clocks). Times are the prototypical example of integer division.

90 minutes are equal to 1 hour and 30 minutes, because 90 = 1 · 60 + 30.

50 hours are 2 days and 2 hours because 50 = 2 · 24 + 2

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24
Q

e.g consider sequence of digits

a=dₙ …d₃d₂d₁d₀

from 0-9
format

Example 1.1.10 (Digits).

A

a=10ⁿ dₙ +…+10¹d₁+d₀

The last digit of a number is its remainder after division by 10

d₀ remainder

a= 10q + d₀

where q=10ⁿ⁻¹ dₙ +…+d₁

25
Q

Example 1.1.8
divide 17 by 5

A

Divide 17 by 5: we find 17 = 3 · 5 + 2 (here q = 3, r = 2).
Divide −17 by −5: we get −17 = 4 · (−5) + 3 (so q = 4, r = 3).

26
Q

Theorem 1.1.11 (Bézout’s Lemma (Bachet de Méziriac, 1624))

A

Given a, b ∈ Z,
there are s, t ∈ Z s.t.
gcd(a, b) = sa + tb.

Moreover, if a, b > 0, then we may assume that s > 0 and t < 0.

27
Q

Theorem 1.1.11 (Bézout’s Lemma (Bachet de Méziriac, 1624))
PROOF

A

Specifically for integers.
Let’s take the set
I={ as+tb , s,t s,t∈ Z} ∩ N ⁺¹
subset of the naturals and greater than or equal to 1

Special case: if a=b=0 then I=empty set
gcd(0,0)=0, 0 ∈ I s=t=0 and conclusion holds

For a≠0 or b≠0: (We want to show that gcd(a, b) ∈ I).
The condition gives I is non empty: contains at least one non-zero natural. E.g. it contains |a| and |b| (s=+-1 and t=0 and s=0 t=+-1).
By the induction principle it contains minimum h>0.

CLAIM: h=gcd(a,b). ( we will show h|a and h|b)

Fix some s, t in Z s.t h=sa+tb

Apply the division algorithm and write a= qh+r where 0 ≤ r< h.
(h is minimum and in set I)
Rearrange
r=a-qh=a-q(sa+tb)
= (1-qs)a-qtb
implies r ∈ I.
(If r>0 then r is in I but r <h as is the remainder but h is the minimum of I: contradiction )
By minimality of h, we must have r = 0, remainder must be 0 and so that h | a. By the same argument with b in
place of a, we find that h | b.

It follows that h ≤ gcd(a, b).

CONVERSELY
gcd(a, b) divides a and b,
so it divides h = sa + tb,
gcd(a,b)|h for any s,t in Z
hence gcd(a, b) ≤ h.

It follows that h = gcd(a, b).

suppose a,b>0 Choose s>0 and t <0
Fix s’,t’ in Z s.t. gcd(a,b)=s’a+t’b (we know exists by prev part of proof)

Take a large s.t. s’ +kb>0
(add multiples of b s.t. it becomes positive e.g. take k=|s’|+1)
Take t=t’-ka
taking k large makes t negative
sa+tb= s’a+kab+t’b-kab
=s’a+t’b=gcd(a,b)

(these multiples arent unique; infinitely many ways of writing this GCD)

(differs notes)
Choose k ∈ N large enough so that s′ = s + kb > 0 and t′ = t − ka < 0.
But then s′a +t′b = sa + kab + tb − kab = sa + tb = gcd(a, b), as desired

28
Q

IMMEDIATE CONSEQUENCES OF BEZOUTS LEMMA

meaning behind GCD

Corollary 1.1.12.

A

1) c | a and c | b IFF c | gcd(a, b).

** (gcd is a biggest common divisor, any other divisor divides it)**

2 If c | ab and gcd(a, c) = 1,
then c | b.

(3) If gcd(a, c) = 1 and gcd(b, c) = 1, then gcd(ab, c) = 1.

(4) If gcd(a, b) = 1 and a | c, b | c, then ab | c

29
Q

Corollary 1.1.12.
PROOF
1) c | a and c | b IFF c | gcd(a, b).

** (gcd is a biggest common divisor, any other divisor divides it)**

A

1) if c|a and c|b then
c|sa+tb for every integer s and t
so in particular c| gcd(a,b) by Bezouts lemma.

if c | gcd(a, b) then c|a and c|b

30
Q

Corollary 1.1.12.
PROOF
2 If c | ab and gcd(a, c) = 1,
then c | b.

A

c|ab now
Pick s, t ∈ Z such that
sa + tc = gcd(a, c) = 1.

Multiply by b, so that sab + tcb = b.

Since c | ab and obviously c | tcb, we find that c | b.

31
Q

Corollary 1.1.12.
PROOF

(3) If gcd(a, c) = 1 and gcd(b, c) = 1, then gcd(ab, c) = 1.

A

Let d = gcd(ab, c). Since d | c and gcd(a, c) = 1, any common divisor of d and a
cannot be more than 1, thus gcd(a, d) = 1. By the previous part, d divides b, hence
d ≤ gcd(b, c) = 1.

32
Q

Corollary 1.1.12.
PROOF

(4) If gcd(a, b) = 1 and a | c, b | c, then ab | c

A

Pick s, t ∈ Z such that sa + tb = 1. Write also c = qa = rb. Then
c = sac + tbc = sarb + tbqa = ab(sr + tq)
which shows that ab divides c

33
Q

Exercise 1.1.13 (⋆, in exercise sheet). Let a, b ∈ N be non-zero natural numbers and
let d = gcd(a, b).

Show that a/d, b/d are coprime.
gcd(a/d,b/d)=1

Show that if c is a common multiple of a and b (meaning that a | c and b | c), then ab/d | c.
Deduce that ab/d is the least common multiple of a and b.

A

**

34
Q

Finding s, t ∈ Z such that gcd(a, b) = sa + tb could be hard.

The proof of Bézout’s Lemma suggests searching among all s, t such that |s| ≤ |a| and |t| ≤ |b|, which is a lot of work: that’s (2|a| + 1)(2|b| + 1) possible pairs! We can do a lot better.

A

Euclid’s algorithm

algorithm takes at most |b| steps, because
r_k = 0 < · · · < r_{ℓ+1} < r_{ℓ} < . . .
forces k ≤ |b|.
This is quite an improvement over (2|a| + 1)(2|b| + 2).

35
Q

Theorem 1.2.1 (Euclid’s Algorithm, c. 300 BC)

A

Let a, b ∈ Z, with b ≠ 0.
(o/w gcd would be 0!)

Apply repeatedly the Division Algorithm, starting with
r₀ := a, r₁:= |b|, and obtaining remainders r₂,r₃, . . . :

a = r₀ = q₂r₁ + r₂,
where 0 ≤ r₂ < r₁ = |b|

r₁ = q₃r₂ + r₃, where 0 ≤ r₃ < r₂
.
.
.
r_ℓ = q_{ℓ+2}r_{ℓ+1} + r_{ℓ+2}
, where 0 ≤ r_{ℓ+2} < r_{ℓ+1}
.
.
.
Then some remainder r_k
is zero, and the last non-zero remainder r_{k−1}
is gcd(a, b).

strictly decreasing, must happen as natural numbers must have a minimum and it solved by induction principle

It works as GCD (a,b) =GCD(bxq+r, b) =GCD (b,r)
#binary digits of b steps

36
Q

e.g Euclidean Algorithm
Find the Greatest Common divisor of 42 and 112
the last non-zero remainder r_{k−1} is gcd(a, b).

A

GCD(42,112)

112=42x2 +28

42=28x1 +14

28=14 x 2

thus gcd is 14

37
Q

Exercise 1.2.2 (⋆, in exercise sheet). Prove that k is at most 2 log₂|b| + 3

(hint: show that r_{ℓ+2} ≤ rℓ/2).

A

*

38
Q

Theorem 1.2.1 (Euclid’s Algorithm, c. 300 BC)
proof
part 1:
claim every r_ℓ is divisible by gcd(a, b)

A

Since |b| = r1 > r2 > · · · ≥ 0, (decreasing remainders)

by the Induction Principle there is k with r_k = 0.
Take k minimum with this property (again, by the Induction Principle!).

claims: (equivalent)
*r_{k-1} = gcd(a,b)
* r_{k-1)as+bt
* gcd(a,b) divides r_ℓ for every ℓ.
* Every r_ℓ is of the form a s_ℓ + b t_ℓ
for some s_ℓ,t_ℓ in z

base case:
r_0=a r_1 =|b|= ±b (so of the required form)

inductively (looking at remainder two steps before )
claim r_ℓ and r_{ℓ+1} are of the required form

r_{ℓ+2} = r_ℓ - q_{ℓ+2}r_{ℓ+1}
= a s_ℓ + b t_ℓ - q_{ℓ+2}[a s_{ℓ+1} + b t_{ℓ+1}]

= (s_ℓ + q_{ℓ+2}s_{ℓ+1})a + (t_ℓ + q_{ℓ+2}t_{ℓ+1})b.

Thus every r_ℓ is divisible by gcd(a, b).

39
Q

part 2:
we also claim that r_{k−1} divides r_{k−ℓ}
for all ℓ > 0

A

(the last nonzero remainder divides all of the previous ones)
by induction again:
since r_k = 0 (algorithm stops when r_k=0)
r_{k-1} clearly divides r_{k-1}
, we have
r_{k−2} = q_k r_{k−1} + r_k (by definition)
= q_k r_{k−1}
r_{k-1} divides r_{k-2} (BASE CASE shown)

INDUCTIVE STEP:
Suppose r_{k−1} divides r_{k−ℓ−1} and r_{k−ℓ}
for some ℓ > 0,

r_{k−ℓ−2} = q_k r_{k−ℓ−1} + r_{k−ℓ}

By induction principle:
We can see that r_{k−1} divides r_{k−ℓ−2} as well.

Particularly,
r_{k−1} divides r_1 = |b| and r_0 = a,
Thus, by Bezouts lemma, r_{k−1} divides gcd(a, b).

40
Q

remark Euclids

if a| b and b|a then

A

divisibility on positive natural numbers is
(antisymmetric?)

if a| b and b|a then a=b

41
Q

def 1.3.1 prime and composite

A

A natural number p ∈ N with p > 1 is prime if its only positive divisors are 1 and p.

Otherwise, we say that p is composite

1 ISN’T A PRIME

42
Q

Lemma 1.3.2. A number p > 1 is prime if and only if

A

A number p > 1 is prime IFF for every a, b ∈ Z,

if p | ab implies p | a or p | b.

43
Q

PROOF
LEMMA 1.3.2
A number p > 1 is prime IFF for every a, b ∈ Z,

if p | ab implies p | a or p | b.

A

Suppose that p is prime (only positive divisors are 1 and p. ). If p ∤ a, then gcd(a, p) = 1 (only common divisor as p is prime), so p | ab implies p | b by Corollary 1.1.12.

Suppose that p is not prime, there is a divisor b s.t. 1<b< p thus p = ab for some positive a, b both different from 1 (and so from p as well):
we must have a < p, b < p, thus p ∤ a and
p ∤ b. condition fails

44
Q

Proposition 1.3.3. products of primes, when can we write as?

A

Proposition 1.3.3. Every natural number n > 1 can be written as a product of primes.

If n = prime, ‘product’ of a single prime.

n = 1 ‘empty product’ of primes.

45
Q

Proof
Proposition 1.3.3. Every natural number n > 1 can be written as a product of primes.

A

Suppose by contradiction that there is some n > 1 that cannot be written as a product of primes.

By the Induction Principle, we can take n minimum with this property.

But then n is not prime, (otherwise it would be a prod of primes)

thus n = ab for some a, b both different from 1 and n,
and in particular 1 < a, b < n.

By minimality of n, a and b can be written as products
of primes, ( By induction, since a,b are smaller than N
, they must each be a product of primes.)

so n = ab is a product of primes too, a contradiction.

46
Q

Theorem 1.3.4 (Euclid, c. 300 BC)
how many primes

A

There are infinitely many prime numbers.

47
Q

proof

There are infinitely many prime numbers.

A

Suppose by contradiction that there are only finitely many primes, and list
them all as p_1, . . . , p_ₙ. Note that N > 0 since we know for instance that 2 is a prime.

Let M = p₁ · · · p_ₙ + 1. Note that M ≥ 2 + 1 > 1.
(by construction we know the primes won’t divide this)

By Proposition 1.3.3, M is a product of
primes.

Since we listed all primes, we must have
M = p_{k₁} · · · p_{k_ℓ}.
On the other hand, p_i ∤ M for all i = 1, . . . , N, a contradiction.

48
Q

how many primes below M?

A

approx

mlogm

49
Q

e.g Suppose we thought that {2, 3, 5, 13} was the complete set of primes.

A

Suppose we thought that {2, 3, 5, 13} was the complete set of primes.
Then M = 2 · 3 · 5 · 13 + 1 = 391 is not divisible by 2, 3, 5, 13. We can factorise it and
find M = 17 · 23, and so find two new primes.

50
Q

Theorem 1.3.6 (Fundamental Theorem of Arithmetic)

A

Every natural number n > 1 can be written in a unique way as a product of primes, up to reordering the factors

90
= 2 · 3 · 3 · 5
= 3 · 5 · 2 · 3:
the two factorisations are simply a reordering of the same list 2, 3, 3, 5.

51
Q

PROOF
Theorem 1.3.6 (Fundamental Theorem of Arithmetic)

Uniqueness part only

A

We show uniqueness.

Write n = p_1 · · · p_r = q_1 · · · q_s where each p_i, q_j is prime. same number of primes?

We prove the conclusion by induction on r.
Since p_1 | n, we find that p_1 | q_j for some j (in particular, s ≥ 1) (p_1 is prime thus prev shown p_1 must divide one of the factors) ), thus p1 = qj since qj
is prime.

We reorder the second product so that in fact
p_1 = q_1. (Base case)

Induction hypothesis
Now divide n by p_1.
If r = 1 we find that n = p_1 = q_1, and we are done. Otherwise,
we now have p_2 · · · p_r = q_2 · · · q_s > 1. By induction hypothesis, after reordering the products, we may assume that p_2 = q_2, . . . , p_r = q_r and r = s, and we are done

52
Q

Remark 1.3.7. In practice, one usually applies the Fundamental Theorem of Arithmetic
to write positive integers as products of prime powers, t

A

in practice, one usually applies the Fundamental Theorem of Arithmetic
to write positive integers as products of prime powers, that is
n = p_1^{α_1}· · · p_k ^{α_k}
where the primes pi are distinct and the
α_i’s are natural numbers.

For example, 350 =2 · 5^2· 7 = 2 · 3^0· 5^2· 7 · 11^0

Given the above expression, one can immediately read off the following numbers:
* the divisors of n are exactly the numbers p_1^{γ_1}· · · p_k ^{γ_k} with γ_i ≤ α_i ;
* if m = p_1^{β_1}· · · p_k^{β_k}, then
gcd(n, m) = p_1 ^{min{α1,β1}}· · · p_k ^{min{αk,βk};
* with m as above, the least common multiple of n and m is
lcm(n, m) = p_1^{max{α1,β1}}· · · p_k ^{max{αk,βk}

53
Q

Exercise 1.3.8. Verify that gcd(n, m) · lcm(n, m) = nm

A
54
Q

1.3.9 exercise
. Given n = pα11· · · pαkk
, count the number of positive divisors of n

A

M. The number of positive divisors is (α1 +1)···(αk +1).

55
Q

1) c | a and c | b IFF

A

1) c | a and c | b IFF c | gcd(a, b).

** (gcd is a biggest common divisor, any other divisor divides it)**

56
Q

2 If c | ab and gcd(a, c) = 1,
then

A

2 If c | ab and gcd(a, c) = 1,
then c | b.

57
Q

(3) If gcd(a, c) = 1 and gcd(b, c) = 1, then

A

(3) If gcd(a, c) = 1 and gcd(b, c) = 1, then gcd(ab, c) = 1.

58
Q

(4) If gcd(a, b) = 1 and a | c, b | c, then

A

(4) If gcd(a, b) = 1 and a | c, b | c, then ab | c