#A10 · Algebra & Number Theory · Lv.3View in 3D Mind-Map

Modular Arithmetic & Fermat's Little Theorem

Pierre de Fermat, Euler & Gauss · 1640 CE

a^{p-1} \equiv 1 \pmod{p} \quad (\gcd(a, p) = 1)

a \equiv b \pmod{n} \iff n \mid (a - b)

In clock arithmetic modulo a prime p, raising any coprime integer a to the (p-1)-th power yields a remainder of 1—the mathematical backbone of RSA public-key cryptography.

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Base a

Exponent e

Modulus p

3^{6} \equiv 1 \pmod{7}, \quad 3^{6} \equiv 1 \pmod{7}

3^6 mod 7

페르마 a^(p-1) mod p

모듈러 역원 a^(p-2)

Prerequisite Path (5 STEP)

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Click steps in order to follow the 3D neon path from basics to target.

STEP 1#A01Linear Equation & Balance Axiom

ax + b = 0 \implies x = -\frac{b}{a}

STEP 2#A02Polynomial Expansion & Factoring

(a+b)^2 = a^2 + 2ab + b^2

STEP 3#A03Laws of Exponents & Radicals

a^m \cdot a^n = a^{m+n}

STEP 4#A08Binomial Theorem & Pascal's Triangle

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

STEP 5#A10Modular Arithmetic & Fermat's Little Theorem

a^{p-1} \equiv 1 \pmod{p} \quad (\gcd(a

Geometric & Intuitive Insight

12시간제 시계에서 15시는 3시와 합동(15 ≡ 3 mod 12)입니다. 소수 p로 나눈 나머지 세계에서는 {1, 2, …, p-1} 전체에 a를 곱해도 순서만 뒤섞일 뿐 집합 전체의 곱이 보존되어 a^(p-1) ≡ 1 (mod p)가 성립합니다.

Step-by-Step Derivation

Step 1.1단계: 소수 p와 서로소인 a에 대해 {a, 2a, 3a, …, (p-1)a} 를 p로 나눈 나머지는 모두 다릅니다.
Step 2.2단계: 따라서 이 나머지들의 집합은 {1, 2, 3, …, p-1} 과 순서만 다른 동일한 집합입니다.
Step 3.3단계: 모든 원소를 곱하면 a^(p-1) · (p-1)! ≡ (p-1)! (mod p) 이고, (p-1)! 을 약분하면 a^(p-1) ≡ 1 (mod p)

AI & Coding Real-World Use

HTTPS/TLS RSA public-key encryption, blockchain digital signatures, and hash table indexing

#RSA Cryptography#Blockchain Security#Hash Table Modulo

const modPow = (a: bigint, e: bigint, m: bigint): bigint => /* binary exp */

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