The theorem of the same modulus remainder

Statement: Learn from the master!

First, co-yu

For the problem of dividing an integer by a positive integer, the concept of congruence is generated if only the remainder is concerned.

definition 1 with the given positive integer m in addition to the integer a, b, if the remainder is equal, then a, b to modulo M congruence, as A≡b (mod m), such as 56≡0 (mod 8).

theorem 1 integer A, A/b to modulo m congruence is sufficient and necessary to be divisible by m (i.e. M|a-b).

A=MQ1+R1, 0<=r1<m;

B=MQ2+R2, 0<=r2<m.

If A≡b (mod m), by definition 1,r1=r2, so a-b=m (Q1+Q2), that is m|a-b.

Conversely, if the m|a-b, that is M|m (Q1-A2) +r1-r2, then m|r1-r2, but |r1-r2|<m, so r1=r2, that is, A≡b (mod m).

The necessary and sufficient condition for inference a≡b (mod m) is a=mt+b (T is an integer).

The expression of the modulo m congruence relation is called the congruence of modulo m, the same remainder, and the notation of the congruence is Gauss (Gauss) was first used in 1801.

theorem 2 The congruence relation has reflexivity, symmetry and transitivity, namely

1) a≡a (mod m);

2) if A≡b (mod m), then B≡a (mod m);

3) if a≡b (mod m), B≡c (mod m), then A≡c (mod m).

theorem 3 if a≡b (mod m), C≡d (mod m), then

1) a+c≡b+d (mod m);

2) a-c≡b-d (mod m);

3) ac≡bd (mod m).

More than two of the same-mode congruence can also be added and subtract multiplication. The following inference is also available for multiplication:

Inference if a≡b (mod m), n is the natural number, then an≡bn (mod m).

theorem 4 if CA≡CB (mod m), (C,m) =d, and A, B is an integer, then a≡b (mod m/d).

Inference if CA=CB (mod m), (c,m) = 1, and A/b is an integer, then a≡b (mod m).

theorem 5 if a≡b (mod m), a≡b (mod n), then a≡b (mod [m,n]).

Inference if a≡b (mod mi), i=1,2,..., N, then a≡b (mod [m1,m2,.., mn]).

Example 1 proves that the positive integer A is a multiple of 9 and only the sum of the numbers of the figures of a is a multiple of 9.

Certificate Set a=an.10n+an-1.10n-1+...+a0

by 10≡1 (mod 9) 10k≡1 (mod 9), k=0,1,2,..., N,

So Ak.10k≡ak (mod 9), k=0,1,2,..., N.

So A≡a0+a1+...+an (mod 9)

So the necessary and sufficient condition of 9|a is 9| A0+a1+...+an.

Example 2 set a=anan-1...a1a0, the necessary and sufficient conditions for 11|a.

Solution by 10≡-1 (mod 11), get 10k≡ ( -1) K (mod one), k=0,1,2,..., N

and a≡a0-a1+a2-... + ( -1) Nan (mod 11)

So the necessary and sufficient condition of 11|a is 11| A0-A1+A2-... + ( -1) Nan.

Example 3 the condition that a positive integer a can be divisible by 7.

Solution due to 1000≡-1 (mod 7), thereby 1000k≡ ( -1) K (mod 7), k=0,1,2,..., N,

So set a= anan-1...a1a0 (1000) This is A≡A0-A1+A2-... + ( -1) Nan (mod 7)

So the necessary and sufficient condition of 7|a is A0-A1+A2-... + ( -1) nan≡0 (mod 7) Here the AI is three digits (1000 binary).

such as when a=89101234579, because 579-234+101-89=357≡0 (mod 7), so 7|a.

definition 2 if M is a natural number, the set kr={x|x=mt+r,t is any integer},r=0,1,..., m

It is called K0,k1,..., Km-1 is the remainder class of modulo m.

For example, the remainder class of modulo 2 is an even class and an odd class, and the remainder class of modulo 3 is: k0={...,-6,-3,0,3,6,...},k1={...,-5,-2,1,4,7,...},k2={...,-4,-1,2,5,8 ...}.

The remaining classes have the obvious properties of the following columns:

1) The remaining class k0,k1,......,km-1 of modulo m are all non-empty-empty sets of integers;

2) Each integer must belong to and belong to only one remaining class;

3) The sufficient and necessary condition that two integers belong to the same remaining class is that they are congruent to modulo m.

Definition 3 any number of the remaining classes from modulo m, the resulting m number is called the complete residual system of modulo m.

For Die m, its complete residual system is a lot of, often using:

0,1,2,..., m-1;

,..., m;

-(M-1)/2,...,-1,0,1,..., M/2 (M is odd),

-m/2+1,...,-1,0,1,..., M/2 (M is even),

-M/2,...,-1,0,1,..., m/2-1 (M is even).

theorem 6 k integers A1,A2,..., ak the necessary and sufficient conditions for the complete residual system of the modulus M is k=m, and the m number is different than the modulus m 22.

theorem 7 if x1,x2,..., XM is the complete residual system of modulo m, (a,m) =1,b is an integer, then Ax1+b,ax2+b,...,axm+b is also the complete residual system of modulo m.

two Euler functions

Definition 1 in the complete residual system of modulo m, the number of all and M-interconnects is called the simplified residual system of modulo m. For example, 1,3,7,9 is a simplified residual system for modulo 10.

Definition 2 If the number of arbitrary natural number m, with the notation Ф (m) to denote the 0,1,2,..., m-1 and M-Mutual, then called Ф (M) is the Euler function.

For example Ф (10) =4,ф (7) =6,ф (1) = 1.

theorem 1 k integers a1,a2,..., ak constituent modulus m the necessary and sufficient conditions for simplifying the remainder system are k=ф (m), (ai,m) =1,i=1,2,..., ф (M), and this ф (m) is different than the modulus m 22.

Theorem 2 if (a,m) =1,x1,x2,..., Xф (M) is a simplified residual system of modulo m, then ax1,ax2,..., Axф (M) is also a simplified residual system of modulo m.

theorem 3 ( Euler theorem) if (a,m) = 1, then Aф (m) ≡1 (mod m)

The x1,x2,..., Xф (M) is a simplified residual system of modulo m, according to the theorem 2,ax1,ax2,..., Axф (M) is also a simplified residual system of modulus M.

It is x1,x2,..., Xф (m) in any number must be the same as the Ax1,ax2,..., Axф (m) in a certain number of modulo m, whereas Ax1,ax2,..., Axф (m) must be the same as the x1,x2,..., Xф (m) in a number of modulo m, which is:

Ax1ax2...axф (M) ≡x1x2...xф (m) (mod m), also (X1x2...xф (m), m) = 1, so Aф (m) ≡1 (mod m).

Example 1 known x=h is the smallest positive integer set ax≡1 (mod m), proving H|ф (m).

AH-1=MT (t is an integer) (a,m) = 1, so

Aф (m) ≡1 (mod m).

Make Ф (m) =hq+r,0<=r

Substituting the above congruence, can get ar≡1 (mod m), so r=0, so H|ф (m).

inference (Fermat theorem) if p is a prime number, then

1) when (a,p) =1, ap-1≡1 (mod p);

2) ap≡a (mod p)

Certificate of first 1). By P is the prime number, known 0,1,2,..., p-1 there are p-1 numbers and P-reciprocity, so Ф (p) =p-1. Also because (a,p) = 1, so according to Theorem 3 is 1).

Re-license 2). When (A,p) =1, by 1) known 2) was established, when (a,p) not equal to 1 o'clock, P|a, the remainder of the same 0, 2) is also established.

Euler proved theorem 3 in 1760, so it is called Euler's theorem. Fermat, in 1640, presented the above inference, which proved to be Euler's completion in 1736, which is often called the Fermat theorem.

Example 2 set A is an integer, proving a5≡a (mod 30).

Because of the 30=2.3.5, and according to the Fermat theorem, there are

A5≡a (mod 5) (1)

A3≡a (mod 3) (2)

A2≡a (mod 2) (3)

By (2) A5≡a3≡a (mod 3) (4)

By (3) A5≡a4≡a2≡a (mod 2) (5)

Thereupon by (1). (4), (5), and 2,3,5 22 biotin, so A5≡a (mod 30).

theorem 4 if p is a prime number, then Ф (PA) =pa-pa-1. (Ф (PA) Calculation formula)

Full residual system 0,1,2,..., p,..., 2p,..., pa-1 (1)

(1) The number of the PA is only a multiple of P 0,p,2p,..., (pa-1–1) p, which has a total of P A-1, so (1) and PA in the number of the Pa-p A-1, so Ф (PA) =pa-pa-1.

theorem 5 if (m,n) = 1, then Ф (MN) =ф (m) ф (n).

Inference if the positive integer m1,m2,... mk 22, then Ф (M1M2...MK) =ф (M1) Ф (m2) ... Ф (MK).

theorem 6 If the standard decomposition formula of M is M=p1a1p2a2...pkak,

Then Ф (m) =p1a1-1p2a2-1...pkak-1 (p1-1) (p2-1) ... (pk-1).

Example 3 is set (n,10) = 1, the verification n101 is the same as the last three digits of N.

Certificate: In order to prove n101-n≡0 just prove n100≡1 (mod 1000).

In fact by (n,125) =1,φ (125) =φ (5^3) =5^3-5^2=100, there is n100≡1 (mod 125);

Again by N is odd known 8|n^2-1, and then n^100≡1 (mod 8), and (125,8) = 1, the proof.

The theorem of the same modulus remainder