Published on

Number Theroy & Mathematical Reasoning

Authors
  • avatar
    Name
    Daniel Jeong
    Twitter

Instructor: Dr. Daniel Saracino
Date: From 2024-08-29 to 2024-12-20

For a printable pdf file, please click here

Theorem 1

For all integers aa, bb, cc, dd:

  1. If a∣ba \mid b and b∣cb \mid c, then a∣ca \mid c.
  2. If a∣ba \mid b and a∣ca \mid c, then for all x,y∈Zx, y \in \mathbb{Z}, a∣(xb+yc)a \mid (xb + yc).
  3. If a∣ba \mid b and c∣dc \mid d, then ac∣bdac \mid bd.

Theorem 2

Suppose aa and bb are integers that are not both 0, so that (a,b)(a, b) exists.
Then for every n∈Zn \in \mathbb{Z},

(a,b)=(a+nb,b)=(a,b+na)(a, b) = (a + nb, b) = (a, b + na)

Theorem 3

Every integer n≥2n \ge 2 is the product of one or more primes.

Theorem 4

There exist infinitely many primes.

Theorem 5

For every integer n≥2n \ge 2, the factorization of nn into primes is unique, except for the order in which the factors are written.

Theorem 6

Suppose m∈Z+m \in \mathbb{Z}^+ and a,b∈Za, b \in \mathbb{Z}.
Then the following are equivalent:

  1. a≡b (mod m)a \equiv b \ (\text{mod}\ m), i.e., m∣(a−b)m \mid (a - b).
  2. a=b+mℓa = b + m\ell for some ℓ∈Z\ell \in \mathbb{Z}.
  3. a‾m=b‾m\overline{a}^m = \overline{b}^m. 1

Thus, the remainders are equal, and any one of (i), (ii), or (iii) proves the other two.

Theorem 7

Basic Properties of Congruence Modulo mm:

Suppose m∈Z+m \in \mathbb{Z}^+, then the following all hold:

  1. Reflexivity: For every a∈Za \in \mathbb{Z},
    a≡a (mod m)a \equiv a \ (\text{mod}\ m).

  2. Symmetry: For all a,b∈Za, b \in \mathbb{Z},
    if a≡b (mod m)a \equiv b \ (\text{mod}\ m), then b≡a (mod m)b \equiv a \ (\text{mod}\ m).

  3. Transitivity: For all a,b,c∈Za, b, c \in \mathbb{Z},
    if a≡b (mod m)a \equiv b \ (\text{mod}\ m) and b≡c (mod m)b \equiv c \ (\text{mod}\ m), then a≡c (mod m)a \equiv c \ (\text{mod}\ m).

When a relationship between elements of a set SS is reflexive, symmetric, and transitive, the relationship is called an equivalence relation on SS.

Thus, Theorem 7 says that "congruence modulo mm" is an equivalence relation on the set Z\mathbb{Z}.

Theorem 8

Suppose m∈Z+m \in \mathbb{Z}^+, and let a,b,c,d∈Za, b, c, d \in \mathbb{Z}.
Then, if a≡b (mod m)a \equiv b \ (\text{mod}\ m) and c≡d (mod m)c \equiv d \ (\text{mod}\ m), we have:

  1. a+c≡b+d (mod m)a + c \equiv b + d \ (\text{mod}\ m)
  2. ac≡bd (mod m)ac \equiv bd \ (\text{mod}\ m)
  3. Let m∈Zm \in \mathbb{Z} and a,b,c∈Za, b, c \in \mathbb{Z}.

Then

ca≡cb(modm)ca \equiv cb \pmod{m}

if and only if

a≡b(modm(c,m))a \equiv b \pmod{\frac{m}{(c,m)}}

Theorem 9

The congruence ax≡b(modm)ax \equiv b \pmod{m} has solutions if and only if (a,m)∣b(a,m) \mid b.
If (a,m)∣b(a,m)\mid b then there is a unique solutions x0x_0 such that 0≤x0<m(a,m)0 \le x_0 < \frac{m}{(a,m)}, and a complete set of solutions (modm)\pmod{m} is

{x0+m(a,m)t∣0≤t<(a,m)}\{x_0 + \frac{m}{(a,m)}t \mid 0 \le t < (a,m) \}

Chinese Remainder Theorem/CRT

Suppose b1,⋯ ,bn∈Zb_1, \cdots, b_n \in \mathbb{Z} and m1,⋯ ,mn∈Z+m_1, \cdots, m_n \in \mathbb{Z}^+.

Consider the system

x≡b1(modm1)x≡b2(modm2)⋮x≡bn(modmn)\begin{align*} x &\equiv b_1 \pmod{m_1} \\ x &\equiv b_2 \pmod{m_2} \\ \vdots \\ x &\equiv b_n \pmod{m_n} \end{align*}

If m1,⋯ ,mnm_1, \cdots, m_n are pairwise relatively prime, meaning (mi,mj)=1(m_i, m_j)=1 where i≠ji \neq j, then the system has a solution.

And all the solutions are ≡(modm1∗m2∗⋯∗mn)\equiv \pmod{m_1 * m_2 * \cdots * m_n}

Fermat's Little Theorem(1640)

Let pp be prime then:

  1. If a∈Za \in \mathbb{Z} and (a,p)=1(a, p)=1, then ap−1=1(modp)a^{p-1}=1 \pmod{p}

  2. For all integers aa, ap≡a(modp)a^p \equiv a \pmod{p}

Wilson's Theorem

For every prime pp, (p−1)!≡−1(modp)(p-1)! \equiv -1 \pmod{p}

Euler's Theorem

If a∈Za \in \mathbb{Z} and (a,n)=1(a,n)=1 then

aφn=1(modn)a^{\varphi{n}} = 1 \pmod{n}

Notice how aa can be any number coprime to nn

If n≥2n \ge 2 and the prime power decomposition of nn is n=p1e1×⋯×pkekn = p_1^{e_1} \times \cdots \times p_k^{e_k} then

φ(n)=p1e1−1(p1−1)×p2e2−1(p2−1)×⋯×pkek−1(pk−1)\varphi(n) = p_1^{e_1-1}(p_1 - 1) \times p_2^{e_2-1}(p_2 - 1) \times \cdots \times p_k^{e_k-1}(p_k - 1)

e.g,

(3,100)=13φ(100)=1(mod100)100=22⋅52φ(100)=22−1(2−1)×52−1(5−1)=40340=1(mod100)\begin{align*} (3,100) &= 1 \\ 3^{\varphi(100)} &= 1 \pmod{100} \\ 100 &= 2^2 \cdot 5^2 \\ \varphi(100) &= 2^{2-1}(2-1) \times 5^{2-1}(5-1) = 40 \\ 3^{40} &= 1 \pmod{100} \end{align*}

Theorem 11

If nn is composite, then nnth Mersenne number is composite.

Fermat's Mersenne Test

If pp is an odd prime, then every prime factor qq of MpM_p has the form q=2kp+1q=2kp+1 for some k∈Z+k \in \mathbb{Z}^+

Perfect Number

nn is perfect if and only if σ(n)=2\sigma(n) = 2

For every prime p,

σ(p)=p+1σ(pk)=1+p+p2+⋯+pk=pk+1−1p−1\begin{aligned} \sigma(p) &= p + 1 \\ \sigma(p^k) &= 1 + p + p^2 + \cdots + p^k = \frac{p^{k+1}-1}{p-1} \end{aligned}

Theorem 12

σ\sigma is multiplicative

Euclid's Theorem on Perfect Numbers

If nn is an even perfect number, then there exists a Mersenne prime 2p−12^p-1 such that

n=2p−1(2p−1)n=2^{p-1}(2^p-1)

i.e., all even perfect numbers comes from n=2p−1(2p−1)n=2^{p-1}(2^p-1)

Theorem 13

Suppose f:S→Tf: S \rightarrow T and g:T→Ug: T \rightarrow U Then:

  1. If ff and gg are both one-to-one, then so is g∘fg \circ f
  2. If ff and gg are both onto, then so is g∘fg \circ f

Theorem 14: Intervals Theorem

Suppose a<ba < b where a,b∈Ra, b \in \mathbb{R} Then:

  1. (a,b)∼(c,d)(a,b) \sim (c,d) for all c<dc < d in R\mathbb{R}
  2. (a,b)∼(c,∞)(a,b) \sim (c, \infty) for all c∈Rc \in \mathbb{R}
  3. (a,b)∼R(a, b) \sim \mathbb{R}

Theorem 15

The set of real numbers in the interval (0,1)(0, 1) is uncountable(equivalently, R\mathbb{R} is uncountable)

Theorem 16

If BB is countable, and f:A→Bf: A \rightarrow B is an injection, then AA is countable.

Theorem 17

If A1,⋯ ,AkA_1, \cdots , A_k are finitely many countable sets, then A1×A2×⋯×AkA_1 \times A_2 \times \cdots \times A_k is countable.

Theorem 18

the union of countably many countable sets are countable.

i.e., the union of countable collection of countable sets is countable.

note: the ®sets are allowed to overlap.


Lemma: Bezout's Lemma

If mm and nn are integers that are not both zero, such that (m,n)(m, n) exists,
then there exist x,y∈Zx, y \in \mathbb{Z} such that xm+yn=(m,n)xm + yn = (m, n).

Lemma: Euclid's Lemma

Suppose aa, bb, cc are integers such that a∣bca \mid bc and (a,b)=1(a, b) = 1.
Then a∣ca \mid c.

Lemma: Division Lemma

If aa and mm are integers and mm is positive, then there exist integers qq and rr such that a=qm+ra = qm + r and 0≤r≤m0 \le r \le m

Lemma: How to check if a congruence is solvable

The congruence ax≡b(modm)ax \equiv b \pmod{m} is solvable if and only if

(a,m)∣b(a,m) \mid b

Then there is a unique solution x0x_0 such that

0≤x0<m(a,m)0 \le x_0 < \frac{m}{(a,m)}

Then the complete set of solutions is

{x0+m(a,m)t∣0≤t<(a,m)}\{x_0 + \frac{m}{(a,m)}t \mid 0 \le t < (a,m) \}

There are exactly (a,m)(a,m) incongruent solutions mod  m\mod{m}

Lemma:

If SS and TT are finite sets with the same number of elements, and f:S→Tf: S \rightarrow T, then ff is one-to-one if and only if ff is onto.

∣S∣=∣T∣&f:S→T⇒onto⇔one-to-one|S| = |T| \& f: S \rightarrow T \Rightarrow \text{onto} \Leftrightarrow \text{one-to-one}

Lemma:

If xx is a nonempty countable set, then there exists an injection f:x→Z+f: x \rightarrow \mathbb{Z}^+

Lemma:

If S⊆Z+S \subseteq \mathbb{Z}^+, then SS is countable.


Definitions

Coprime

We say integers aa and bb are relatively prime (or "coprime") if (a,b)=1(a, b) = 1.

Prime

An integer pp is prime if p>1p > 1 and the only positive integers that divide pp are 11 and pp.

Euler's Function

φ(n)=the number of elements in {0,1,⋯ ,n−1} that are coprime to n\varphi(n) = \text{the number of elements in } \{0, 1, \cdots , n-1 \} \text{ that are coprime to } n

e.g,

φ(1)=1Since {(1-1)}={0} and 0 is coprime to 1\begin{gathered} \varphi(1) = 1 \\ \text{Since \{(1-1)\}=\{0\} and 0 is coprime to 1} \end{gathered}

Mersenne Numbers

If n∈Z+n \in \mathbb{Z}^+, the nnth Mersenne number is 2n−12^n-1

Proper Factors

If n∈Z+n \in \mathbb{Z}^+, the proper factors of nn are the positive integers that divide nn but are not nn.

Perfect Numbers

A positive integer is called perfect if it equals the sum of its proper factors.

σ(n)\sigma(n)

If n∈Z+n \in \mathbb{Z}^+, then σ(n)\sigma(n) denotes the sum of all positive integers that divide nn.

Function

We say ff is a function(or mapping) from SS to TT and we write

f:S→Tf: S \rightarrow T

if ff associates to each s∈Ss \in S a uniquely determined element f(s)∈Tf(s) \in T

One-to-one: Injection

If f:S→Tf: S \rightarrow T we say ff is one-to-one(or "1-1")

  1. if whenever s1≠s2s_1 \neq s_2 in SS, then f(s1)≠f(s2)f(s_1) \neq f(s_2) in TT.
  2. if and only if when s1,s2∈Ss_1, s_2 \in S and f(s1)=f(s2),thens1=s2f(s_1) = f(s_2), then s_1=s_2.

Onto: Surjection

If f:S→Tf: S \rightarrow T we say ff is onto if for every t∈Tt \in T there exists some s∈Ss \in S such that f(s)=tf(s)=t.

Bijection

ff is a bijection if ff is both one-to-ton and onto. When a function is a bijection, there is an inverse function and the domain eqals codomain.

Inverse Function

Suppose f:S→Tf: S \rightarrow T is a bijection. Then for each t∈Tt \in T there is a unique s∈Ss \in S such that f(s)=tf(s) = t. So we can define a function f−1:T→Sf^{-1}: T \rightarrow S by letting.

f−1(t)=(the unique s∈S such that f(s)=t)f^{-1}(t) = (\text{the unique } s \in S \text{ such that } f(s) = t)

f−1f^{-1} is the inverse function of ff.

Powerset

If xx is a set, then the powerset xx is the set whose elements are the subsets of xx. We denote the powerset of xx by P(x)P(x)

So P(x)={A∣A≤x}P(x) = \{A | A \le x\}

e.g., If x={1,2}x = \{1, 2\}, P(x)={∅,{1},{2},{1,2}}P(x) = \{\emptyset, \{1\}, \{2\}, \{1, 2\} \}

Number of Elements in a Finite Set: ∣x∣|x|

Cartesian Product

If AA and BB are sets, their cartesian product is the set A×B={(a,b)∣a∈Aandb∈B}A \times B = \{(a,b) | a \in A \text{and} b \in B\}

Note that if AA and BB are finite sets, then

∣A×B∣=∣A∣×∣B∣|A \times B| = |A| \times |B|

Composite Functions

If f:S→Tf: S \rightarrow T and g:T→Ug: T \rightarrow U, we define the composite function g∘f:S→Ug \circ f: S \rightarrow U by letting

g∘f(s)=g(f(s))g \circ f(s) = g(f(s))

for all s∈Ss \in S


Corollary 1

Every integer n≥2n \ge 2 has a unique representation in the form:

n=p∗1e∗1×p∗2e∗2×p∗3e∗3×⋯×p∗ke∗k n = p*{1}^{e*{1}} \times p*{2}^{e*{2}} \times p*{3}^{e*{3}} \times \cdots \times p*{k}^{e*{k}}

where the pp's are distinct primes and each ei≥1e_i \ge 1. This is called the prime-power decomposition of nn.

Corollary 2

Suppose n≥2n \ge 2 and the prime-power decomposition (ppd) of nn is:

n=p∗1e∗1×p∗2e∗2×p∗3e∗3×⋯×p∗ke∗k n = p*{1}^{e*{1}} \times p*{2}^{e*{2}} \times p*{3}^{e*{3}} \times \cdots \times p*{k}^{e*{k}}

Then if mm is any positive integer that divides nn,

m=p∗1f∗1×p∗2f∗2×p∗3f∗3×⋯×p∗kf∗k m = p*{1}^{f*{1}} \times p*{2}^{f*{2}} \times p*{3}^{f*{3}} \times \cdots \times p*{k}^{f*{k}}

for some integers f1,⋯ ,fkf_1, \cdots, f_k such that 0≤fi≤ei0 \le f_i \le e_i.

Corollary 3

If

n=p∗1c∗1×p∗2c∗2×p∗3c∗3×⋯×p∗kc∗k n = p*{1}^{c*{1}} \times p*{2}^{c*{2}} \times p*{3}^{c*{3}} \times \cdots \times p*{k}^{c*{k}}

and

m=p∗1d∗1×p∗2d∗2×p∗3d∗3×⋯×p∗kd∗k, m = p*{1}^{d*{1}} \times p*{2}^{d*{2}} \times p*{3}^{d*{3}} \times \cdots \times p*{k}^{d*{k}},

where p1,⋯ ,pkp_1, \cdots, p_k are distinct primes, and the cic_i's and did_i's are nonnegative integers, then

(m,n)=p1min⁡(c1,d1)×p2min⁡(c2,d2)×⋯×pkmin⁡(ck,dk) (m, n) = p_1^{\min(c_1, d_1)} \times p_2^{\min(c_2, d_2)} \times \cdots \times p_k^{\min(c_k, d_k)}

Corollary 4

If nn is an integer such that n≥2n \ge 2, then there exists a positive integer mm such that n=m2n = m^2 if and only if all the exponents in the ppd of nn are even.

Corollary 5: Theorem 8-1

Suppose a≡b (mod m)a \equiv b \ (\text{mod}\ m). Then:

  • a≡b (mod m)a \equiv b \ (\text{mod}\ m)
  • a2≡b2 (mod m)a^2 \equiv b^2 \ (\text{mod}\ m)
  • a3≡b3 (mod m)a^3 \equiv b^3 \ (\text{mod}\ m)
  • ...
  • ak≡bk (mod m)a^k \equiv b^k \ (\text{mod}\ m) for every k∈Z+k \in \mathbb{Z}^+.

Thus, using part (ii) of Theorem 8 repeatedly, we get:

If a≡b (mod m)a \equiv b \ (\text{mod}\ m), then ak≡bk (mod m)a^k \equiv b^k \ (\text{mod}\ m) for every k∈Z+k \in \mathbb{Z}^+.

Corollary 6: Theorem 8-2

If p(x)p(x) is a polynomial with integer coefficients and a≡b (mod m)a \equiv b \ (\text{mod}\ m), then p(a)≡p(b) (mod m)p(a) \equiv p(b) \ (\text{mod}\ m).

φ(n)\varphi(n) Corollary

If n≥3n \ge 3, then φ(n)\varphi(n) is even.

For n=1 or 2n = 1 \text{ or } 2, φ(n)=1\varphi(n) = 1

If p∣n p \mid n, then (p−1)∣φ(n)(p-1) | \varphi(n)

Corolllary of Theorem 12

If the prime power decomposition of nn is

n=p1e1×⋯×pkekn = p_1^{e_1} \times \cdots \times p_k^{e_k}

then

σ(n)=σ(p1e1)×⋯×σ(pkek)=(p1e1+1−1p1−1)×⋯×(pkek+1−1pk−1)\begin{align*} \sigma(n) &= \sigma(p_1^{e_1}) \times \cdots \times \sigma(p_k^{e_k}) &= \left(\frac{p_1^{e_1+1}-1}{p_1-1}\right) \times \cdots \times \left(\frac{p_k^{e_k+1}-1}{p_k-1}\right) \end{align*}

Corolllary of Theorem 16

Subsets of countable sets are countable.


Sidenote 1: (a−m)x≡b(modm)(a-m)x \equiv b \pmod{m}

Given ax≡b(modm)ax \equiv b \pmod{m}, since a−(a−m)=ma-(a-m) = m,

a≡a−m(modm)a \equiv a-m \pmod{m}

In modular arithmetic, any integer is equivalent to itself plus or minus any multiple of the modulus.

so

ax≡(a−m)x(modm).ax \equiv (a-m)x \pmod{m}.

Thus

(a−m)x≡b(modm)(a-m)x \equiv b \pmod{m}

Sidenote 2

Every integer is congruent(mod9)\pmod{9} to one of {0,1,2,⋯ ,8}\{0, 1, 2, \cdots, 8\}

Footnotes

  1. Remainder notation created by Dr. Daniel Saracino ↩