Number Theory Introduction to Chapter 8 congruence

Number Theory Introduction to Chapter 8 congruence

We discuss how to solve the congruence $ ax \ equiv c (mod m) $

Obviously this is determined whether or not solvable congruence of determining a linear equation equivalent to \ (ax-my = c \ ) whether solvable

We respect \ (ax + by = c \ ) type of linear equations if and only if \ (gcd (a, b) | c \) when the solvability

Can be solved by way of the following codes to solve (expand Euclid)

//ax+by = gcd(a,b)
long long extend_gcd(long long a,long long b,long long &x,long long &y)
{
    if(a==0&&b==0) return -1;
    if(b==0) {x = 1; y = 0; return a;}
    long long d = extend_gcd(b,a%b,y,x);
    y -= a/b*x;
    return d;
}

Set \ (g = gcd (a, m) \) then

• If not divisible by g c This equation has no solution
  
• If \ (g \ mid c \) is exactly the formula it modified with different solutions of g

NOTE: The most important case is the linear congruence equations \ (gcd (a, m) = 1 \) In this case, the same below exactly one solution of formula I

At this time, and c = 1 is referred to a x m of the inverse element

Seeking inverse follows:

long long extend_gcd(long long a,long long b,long long &x,long long &y)
{
    if(a==0&&b==0) return -1;
    if(b==0) {x = 1; y = 0; return a;}
    long long d = extend_gcd(b,a%b,y,x);
    y -= a/b*x;
    return d;
}
ll getInv(int a,int mod)
{
    ll x,y;
    ll d=extend_gcd(a,mod,x,y);
    return d==1?(x%mod+mod)%mod:-1;
}