Thought of by indefinite equation--the selection of number theory

The so-called two yuan an indefinite equation, refers to the $x,y$, the shape of the $ax+by=c$ equation

Here we agree that the $a,b,c$ are integers.

So the question we have is: how do we judge this equation to have a solution?

Bezout theorem: Set $a,b,d$ are integers, and $ (A, B) =d$, there is $u,v$ make $ua+vb=d$

Proof: by $ (A, B) =d$ know $ and d|a and d|b$, then there is $d|ua+vb$

? Remember that the minimum positive value for $ua+vb$ is $s$, then $d|s$

? Do band Rest Division for a: $, and both are integers a=qs+r (0\leq r<s, and Q and R are integers) $

? Then $r=a-qs=a-q (UA+VB) = (1-qu) A + (-QV) b$, so R is a B linear combination (you can interpret the linear combination as an integer z can be represented as ax+by (both integers))

? and S is the minimum positive value of $ua+vb$ and $0\leq r<s$, so $r=0$, that is $a=qs$

? So $s|a$

? The same can be proven $s|b$

? So $ is s| (A, B) is s|d$ and $d|s$ known $d=s$

After we have defined the Bezout theorem, we can conclude that there is a condition for the solution of an indefinite equation of two yuan: $GCD (A, b) |c$

Proof: Set $D=GCD (A, B), a=a_1d,b=b_1d,c=c_1d$, then the original equation can be converted to $A_1X+B_1Y=C_1,GCD (a_1,b_1) =1$

? Known by the bezout theorem, there is an integer $x_0,y_0$, which makes $a_1x_0+b_1y_0=1$

? So $a_1 (x_0c_1) +b_1 (y_0c_1) =c_1$

? Thus the original equation has a set of integer solutions $x=x_0c_1,y=y_0c_1$

We have constructed a set of solutions to the equation in just the proof, it is obvious that if the solution is not feasible according to the above method, we must think of a method to find out a set of solutions to the indefinite equation.

In the above proof process, in addition to known quantity, we repeatedly mentioned another number $gcd (A, b) $

Let's see if we can find some clues from this number.

First solve this equation $ax+by=gcd (A, b) $

(1) If $b=0$, then $GCD (A, B) =a$, the original equation is $a x+0 y=a$, obviously $x=1,y=0$

(2) If $b\neq 0$, then $GCD (b) =gcd (b,a\ mod\ B) $

? This allows you to know $ax+by=bx_0+ (a\ mod\ b) y_0=bx_0+ (a-a\ div\ b*b) y_0=ay_0+b (x_0-a\ div \ b*y_0) $

So $x=y_0,y=x_0-a\ div\ b*y_0$

And we already know the value of the last $ with x_0 and y_0$

So go up and down the floor and push up.

Since all processes are based on the $GCD (A, b) $, this method is called extended Euclidean

The code is as follows

1#include <iostream>2#include <string>3#include <string.h>4#include <stdio.h>5#include <algorithm>6#include <vector>7#include <queue>8#include <map>9 using namespacestd;Ten   One voidEXGCD (intAintBint&x,int&y) A { -     if(b==0) {x=1; y=0;return;} -EXGCD (b,a%b,x,y); the     inttmpx=x,tmpy=y; -X=tmpy;y=tmpx-(A/b) *Tmpy; - } -   + intMain () - { +     intb; ACin >> a >>b; at     intx, y; - EXGCD (a,b,x,y); -cout << x <<" "<<y; -     return 0; -}

If the indefinite equation is $AX+BY=K*GCD (A, b) $, then directly on the basis of the $x,y$ obtained $k$ can be

After we have found out a set of solutions to the original indefinite equation, can we use it to introduce all the solutions?

Suppose that the solution we currently find is $ (x_0,y_0) $, the next set of solutions is $ (x_0+d_1,y_0+d_2) (where d_1 is the smallest positive integer that satisfies the condition) $

Then $ A (x_0+d_1) +b (y_0+d_2) =c,ax_0+by_0=c$

Brought into $ad_1+bd_2=0$, i.e. $\frac{d_1}{d_2}=-\frac{b}{a}$

$\FRAC{D_1}{D_2}=-\FRAC{\FRAC{B}{GCD (A, B)}}{\frac{a}{gcd (A, b)}}$

Thus two yuan the general solution of the $ax+by=c$ of an indefinite equation can be expressed as

$ $x =X_0+K*\FRAC{B}{GCD (A, B)},y=y_0-k*\frac{a}{gcd (A, b)}$$, where $k\in z$

Thought of by indefinite equation--the selection of number theory