"Concrete Mathematics"--Number theory

Starting from this article, we began to explore the "forest" of number theory.

Divisible:

The problem of the division of numbers in number theory is nothing more than the approximate, multiple, approximate and multiple pairs of relative concepts, and if a is an approximate of B, then B is a multiple. We often use a|b to mean that B can divide a, that is, b/a is an integer, but "|" It is easy to confuse with absolute values, geometric definitions, conditional probabilities during use, so here we use a\b to denote that a can divide B.

Approximate: If b\a, it is said that B is an approximate of a.

Multiples: If b\a, then A is a multiple of B.

Greatest common divisor: gcd (A, b) = Max{k | K\a and k\b}.

Least common multiple: LCM (A, b) = min{k} k>0, a\k and B\k}

Let's explore the Euclidean algorithm for calculating gcd (A, B).

It is based on an important recursive--gcd (m,n) = gcd (n% m, m) and gcd (0,n) = N. (In fact, this recursive proof of the author in another article introduced.) )

Euclidean algorithm can also bring us more things, we generalize based on it, using it to solve the following equation of the solution (M ', n '):

M ' m + n ' n = gcd (m,n) ①

As can be seen, if M = 0, this equation obviously has countless groups of solutions.

If M! = 0, based on the Euclidean algorithm recursion, we take r = N m, you can convert ① equivalent, that is R ' r + M ' m = gcd (r,m) ②, you can see R = n (int) (n/m) m, we bring R into the ②, and do the following simplification operations.

R ' [n-(int) (n/m) m] + m ' m = gcd (r,m) = R ' n + [m '-R ' (int) (n/m)] = gcd (r,m) ③

It is easy to see that since gcd (r,m) = gcd (m,n), the equation ①③ is equivalent, so we will get the following equation:

n ' = R ' ④

M ' = M '-R ' (int) (n/m) ⑤

So for the ① solution we want to solve (m ', n '), we obviously need to know the solution of the ② (M ', R '), which forms a recursive solution pattern.

For the solution of the ① equation, many places in the field of number theory will be involved, here based on it there is also a simple inference: k\m, k\n is K\GCD (m,n) The sufficient and necessary conditions.

Prove that: adequacy, K is the male factor of M, N, obviously K is gcd (m,n) factor.

Necessity, K is the factor of gcd (M,n), obviously K is both the factor of M and the factor of N.

"Concrete Mathematics"--Number theory