• 沒有找到結果。

Elementary Number Theory Chapter Two Congruences Section 2.1 Congruences Definition 2.1.1

N/A
N/A
Protected

Academic year: 2022

Share "Elementary Number Theory Chapter Two Congruences Section 2.1 Congruences Definition 2.1.1"

Copied!
4
0
0

加載中.... (立即查看全文)

全文

(1)

Elementary Number Theory Chapter Two Congruences

Section 2.1 Congruences

Definition 2.1.1 If an integer m, not zero, divides a − b, we say that a is congruent to b modulo m and write a≡ b(mod m).

Lemma 2.1.2 Let a, b, c, d be integers. Then

(a) a ≡ b(mod m), b ≡ a(mod m) and a − b ≡ 0(mod m) are equivalent statements.

(b) If a≡ b(mod m) and b ≡ c(mod m), then a ≡ c(mod m).

(c) If a≡ b(mod m) and c ≡ d(mod m), then a + c ≡ b + d(mod m).

(d) If a≡ b(mod m) and c ≡ d(mod m), then ac ≡ bd(mod m).

(e) If a≡ b(mod m) and d|m, d > 0, then a ≡ b(mod d).

(f) If a ≡ b(mod m) then ac ≡ bc(mod mc) for c > 0.

(g) Let f ∈ Z[x]. If a ≡ b(mod m) then f(a) ≡ f(b)(mod m).

(h) ax≡ ay(mod m) if and only if x ≡ y(mod (a,m)m ).

(i) If ax ≡ ay(mod m) and (a, m) = 1, then x ≡ y(mod m).

(j) a ≡ b(mod mi) for I = 1, 2, . . . , r if and only if a ≡ b(mod [m1, m2, . . . , mr]).

(k) If a ≡ b(mod m), then (a, m) = (b, m).

Example 2.1.3 (a) Show that 41|220− 1.

(b) Find the remainder of 1! + 2! + 3! +· · · + 99! + 100! dividing by 12.

(c) Find 999179(mod 1763)

Definition 2.1.4 (a) If x ≡ y(mod m) then y is called a residue of x modulo m.

(b) A set x1, x2, . . . , xm is called a complete residue system modulo m if for every integer y there is one and only one xj such that y ≡ xj(mod m) .

(c) A reduced residue system modulo m is a set of integers ri such that (ri, m) = 1, ri 6≡ rj(mod m) if i 6= j, and such that every x prime to m is congruent modulo m to some member ri of the set.

Definition 2.1.5 The number of positive integers less than m that are relatively prime to m is denoted by φ(m). This function is called Euler’s φ-function, sometimes the totient.

Lemma 2.1.6 All reduced residue systems modulo m will contain the same number φ(m), of members.

1

(2)

2

Lemma 2.1.7 Let (a, m) = 1. Let r1, r2, . . . , rn be a complete, or a reduced, residue system modulo m. Then ar1, ar2, . . . , arn is a complete, or a reduced, residue system, respectively, modulo m.

Lemma 2.1.8 Let p be a prime. Then (p

k

)≡ 0(mod p) for 1 ≤ k ≤ p − 1, Theorem 2.1.9 (Fermat) Let p denote a prime.. If p6 |a then ap−1 ≡ 1(mod p).

For any integer a, ap ≡ a(mod p). [Two proofs]

Example 2.1.10 (a) Find the residue of 538 modulo 11.

(b) (Testing the primality of n.) If an 6≡ a(mod n) for some a, then n must be composite. Since 2117 ≡ 44(mod 117), so 117 is a composite.

(c) Show that 1763 is a composite.

Lemma 2.1.11 If p, q are distinct primes with ap ≡ a(mod q) and aq a(mod p), then apq ≡ a(mod pq).

Example 2.1.12 Let p = 11, q = 31. We have 2341 ≡ 2(mod 341). Hence the converse of Fermat Theorem is false.

Definition 2.1.13 (a) A composite n for which an ≡ a(mod n) is called a pseudoprime to the base a.

(b) n is called an absolute pseudoprime (or Carmicheal number) if it is a pseudoprime to any base a.

Example 2.1.14 561 is a Carmichael number.

Remark 2.1.15 (a) The pseudoprime to the base 2 are 341, 561, 645, 1105, . . . . (b) The smallest pseudoprime to the base 3 is 91.

(c) The smallest pseudoprime to the base 5 are 217.

(d) 561, 1105, 15841, 161038, 16046641, . . . are Carmichael numbers. The largest known to date is the product of 1101518 odd primes.

Proposition 2.1.16 If n is an odd pseudoprime to base 2, then so is 2n − 1.

Thus there are infinitely many pseudoprimes to base 2.

Theorem 2.1.17 Any Carmichael number must be square free.

Theorem 2.1.18 Let n = p1p2· · · pr be square free composite. If pi− 1|n − 1 for all i, then n is a Carmichael number.

Theorem 2.1.19 (Euler’s generalization of Fermat’s Theorem) If (a, m) = 1, then aφ(m) ≡ 1(mod m).

(3)

3

Example 2.1.20 Let n be an odd integer which is not a multiple of 5. then n divides a repunit Rk for some k. In fact, k = φ(n). [By Euler’s Theorem; by pigeon hole principle.]

Theorem 2.1.21 If (a, m) = 1 then there is an x such that ax ≡ 1(mod m).

Any two such x are congruent (mod m). If (a, m) > 1 then there is no such x.

Lemma 2.1.22 Let p be a prime. Then x2 ≡ 1(mod p) if and only if x ≡

±1(mod p).

Example 2.1.23 Let x = 2693. Then x2 ≡ 1(mod 1387, x ≡ 512(mod 1387.

Thus 1387 is a composite.

Theorem 2.1.24 (Wilson) If p is a prime, then (p− 1)! ≡ −1(mod p).

Remark 2.1.25 The converse of Wilson’s Theorem is also true.

Proposition 2.1.26 Let p be a prime. Then x2 ≡ −1(mod p) has solutions if and only if p = 2 or p≡ 1(mod 4).

Lemma 2.1.27 (Thue) Let p be a prime and let (a, p) = 1. Then the congruence ax≡ y(mod p admits a solution x0, y0 with 0 <|x0| < √p, 0 < |y0| < √p

Proposition 2.1.28 (Stated by Albert Girard, 1632; proved by Fermat, 1654.) A prime p is expressible as a sum of two squares if and only if p = 2 or p 1(mod 4).

Corollary 2.1.29 Any prime p of the form 4k + 1 can be represented uniquely as a sum of two squares of positive integers.

Lemma 2.1.30 Let q be a prime factor of a2 + b2. If q ≡ 3(mod 4) then q|a and q|b.

Theorem 2.1.31 (Fermat) Write the canonical factorization of n in the form n = 2α

p≡1(mod 4)pβ

q≡3(mod 4)qγ. Then n can be expressed as a sum of two squares of integers if and only if all the exponents γ are even.

Proposition 2.1.32 A positive integer n can be represented as the difference of two squares if and only if n6≡ 2(mod 4)

Corollary 2.1.33 An odd prime is the difference of two consecutive squares.

Example 2.1.34 25 = 32 + 42 = 52+ 02 745 = 272+ 42 = 242 + 132

(4)

4

24 = 72 − 52 = 52− 12

2.1.35 The Pollard rho factorization method.

Lemma 2.1.36 Factorize n = 36287.

2.1.37 The Fermat-Kraitchik factorization method.

Lemma 2.1.38 (a) Factorize n = 119143.

(b) Factorize n = 12499.

參考文獻

相關文件

在1980年代,非晶矽是唯一商業化的薄膜型太 陽能電池材料。非晶矽的優點在於對於可見光

Theorem 1.2.9 (Euclid) The number of primes is infinite... contains infinitely

Elementary Number Theory Chapter 1 Divisibility Section 1.0 Introduction.. Euclid (350 B.C.): There are infinitely many

[r]

[r]

Quality kindergarten education should be aligned with primary and secondary education in laying a firm foundation for the sustainable learning and growth of

[r]

[r]