2020-2ν•™κΈ°, λŒ€ν•™μ—μ„œ β€˜ν˜„λŒ€λŒ€μˆ˜1’ μˆ˜μ—…μ„ λ“£κ³  κ³΅λΆ€ν•œ λ°”λ₯Ό μ •λ¦¬ν•œ κΈ€μž…λ‹ˆλ‹€. 지적은 μ–Έμ œλ‚˜ ν™˜μ˜μž…λ‹ˆλ‹€ :)

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2020-2ν•™κΈ°, λŒ€ν•™μ—μ„œ β€˜ν˜„λŒ€λŒ€μˆ˜1’ μˆ˜μ—…μ„ λ“£κ³  κ³΅λΆ€ν•œ λ°”λ₯Ό μ •λ¦¬ν•œ κΈ€μž…λ‹ˆλ‹€. 지적은 μ–Έμ œλ‚˜ ν™˜μ˜μž…λ‹ˆλ‹€ :)


μš°λ¦¬λŠ” $\mathbb{Z}_n$이 Ringμž„μ„ 이미 μ•Œκ³  μžˆλ‹€.

그런데, 이 $\mathbb{Z}_n$은 사싀 μ•„λž˜μ˜ Factor RingμœΌλ‘œλΆ€ν„° μœ λ„λ˜λŠ” Ring이닀!

\[\mathbb{Z} / n \mathbb{Z} \cong \mathbb{Z}_n\]


그리고 μš°λ¦¬λŠ” $\mathbb{Z}_p$의 경우 Ring보닀 μƒμœ„ 단계인 Fieldκ°€ 됨을 ν™•μΈν•˜μ˜€λ‹€!!

$\mathbb{Z}_p$μ—μ„œ κ³±μ…ˆμ— λŒ€ν•œ ꡰ인 $\mathbb{Z}^{*}_p$λ₯Ό μƒκ°ν•΄λ³΄μž.

\[\mathbb{Z}^{*}_p = \{1, 2, ..., p-1 \}\]

즉, $\lvert \mathbb{Z}^{*}_p \rvert = p-1$이닀.

μ΄λ•Œ, $a \in \mathbb{Z}^{*}_p$에 λŒ€ν•΄ $a$의 μœ„μˆ˜λŠ” $\mathbb{Z}^{*}_p$의 μœ„μˆ˜μ˜ μ•½μˆ˜μ΄λ―€λ‘œ $\lvert a \rvert \mid (p-1)$κ°€ λœλ‹€.

λ”°λΌμ„œ $a^{p-1} = 1$이 λœλ‹€.

이것을 $\mathbb{Z}$μ—μ„œ λ‹€μ‹œ μ“°λ©΄ μ•„λž˜μ™€ κ°™λ‹€.

\[a^{p-1} \equiv 1 \quad (\textrm{mod} \; p)\]

μ΄λ•Œ, $a$λ₯Ό $\mathbb{Z}^{*}_p$μ—μ„œ λ½‘μ•˜μœΌλ―€λ‘œ $a \not\equiv 0$ (mod $p$)이닀. 즉, $p \not\mid a$이닀.

μœ„μ˜ 사싀을 잘 μ •λ¦¬ν•˜μ—¬ κΈ°μˆ ν•œ 것이 λ°”λ‘œ Fermat’ Little Theorem이닀.



Fermat’s Little Theorem

Theorem. Fermat’s Little Theorem

Let $p$ be a prime,

If $a \in \mathbb{Z}$, and $p \not\mid a$,

then $p \mid (a^{p-1} + 1)$ $\iff$ $a^{p-1} \equiv 1$ (mod $p$).


μ΄λ•Œ, λ§Œμ•½ $p \mid a$라면,
$a^{p-1} \equiv 0$ (mod $p$)κ°€ λœλ‹€!

Corollary.

If $a \in \mathbb{Z}$, then $a^p \equiv a$ (mod $p$).

proof.

(Case 1) $p \mid a$

(Case 2) $p \not\mid a$


Examples

Example 1.

$8^{103} \equiv \; ?$ (mod 13)

Sol.

By Fermat-, $8^{12} \equiv 1$ (mod 13)

$\implies$ $(8^{12})^{8} \cdot 8^7 \equiv 1 \cdot 8^7$ (mod 13)

$\implies$ $8^7 \equiv (-5)^7 \equiv (25)^3 \cdot (-5)$

μ΄λ•Œ, $25 \equiv -1$ (mod 13) μ΄λ―€λ‘œ

$\implies$ $(-1)^3 (-5) \equiv 5$ (mod 13)


Example 2.

$2^{11213} \equiv \; ?$ (mod 11)

Sol.

By Fermat-, $2^{10} \equiv 1$ (mod 11)

$(2^{10})^{1121} \cdot 2^3 \equiv 1 \cdot 8$ (mod 11)



Example 3.

Show that, for $\forall \; n \in \mathbb{N}$, $15 \mid (n^{33} - n)$.

Sol.

$15 \mid (n^{33} - n)$. 즉, $n^{33} - n = n(n^{32} - 1) \equiv 0$ (mod 15)μž„μ„ 보이면 λœλ‹€.


μ΄λ•Œ, $15 = 3 \times 5$이고, Fermat의 정리에 λ”°λ₯΄λ©΄

  • for $3 \not\mid n$, $n^2 \equiv 1$ (mod 3)
  • for $5 \not\mid n$, $n^4 \equiv 1$ (mod 5)


이제 $n^{32}$λ₯Ό Fermat 정리λ₯Ό μ΄μš©ν•΄ ν‘œν˜„ν•˜λ©΄ μ•„λž˜μ™€ κ°™λ‹€.

  • if $3 \not\mid n$, $n^{32} \equiv (n^2)^{16} \equiv 1$ (mod 3)
  • if $5 \not\mid n$, $n^{32} \equiv (n^4)^{8} \equiv 1$ (mod 5)


μΌ€μ΄μŠ€λ₯Ό λ‚˜λˆ„μ–΄ $n^{32} - 1$을 λΆ„μ„ν•΄λ³΄μž.


1. $3 \not\mid n$ and $5 \not\mid n$

then, $3 \mid (n^{32} - 1)$ and $5 \mid (n^{32}-1)$을 μ˜λ―Έν•œλ‹€.

λ”°λΌμ„œ $15 \mid (n^{32} - 1)$을 μ–»λŠ”λ‹€.

λ”°λΌμ„œ $(n^{32} - 1) \equiv 0$ (mod 15)이닀.


2. $3 \mid n$ and $5 \mid n$

이것은 $15 \mid n$λ₯Ό μ˜λ―Έν•˜λ―€λ‘œ $15 \mid n(n^{32}-1)$이닀.


3. $3 \not\mid n$ and $5 \mid n$

$3 \mid (n^{32} - 1)$μ΄λ―€λ‘œ $3 \mid n(n^{32} - 1)$이닀. λ˜ν•œ, $5 \mid n(n^{32} - 1)$μ΄λ―€λ‘œ

$15 \mid n(n^{32}-1)$이닀.


4. $3 \mid n$ and $5 \not\mid n$

3.의 κ²½μš°μ™€ λ§ˆμ°¬κ°€μ§€μ΄λ‹€.



Fermat’s Little Theorem은 Ring의 Multiplicative groupμœΌλ‘œλΆ€ν„° μ‰½κ²Œ μœ λ„ν•  수 μžˆλŠ” μ„±μ§ˆμ΄μ—ˆλ‹€ γ…Žγ…Ž

ν•˜.지.만.!

Euler ν˜•λ‹˜μ΄ 우리λ₯Ό μœ„ν•΄ FLTλ₯Ό μΌλ°˜ν™” ν•΄μ£Όμ…¨λ‹€ γ… γ… 

Euler’s Theorem