The first conclusion is that if A and B are coprime, the necessary and sufficient conditions are: there must be m, and n is an integer, so that am+bn= 1. The proof of this conclusion is:

Necessity:

By stage:

Let two numbers be A and B (B < A), and find their greatest common divisor D as follows: A is divided by B to get A = Bq 1+R 1 (0 ≤ R < B). If r 1=0, then (a, b) = b; If r 1≠0, divide b by r 1 to get b = rq2+R2 (0 ≤ R2 < r 1). If r2 = 0, (a, b) = r 1, if r2≠0, continue to divide r 1 by r2, and so on until it is divisible. Its last nonzero remainder is d.

According to the toss and division, we can get:

a = bq 1+r 1(0 & lt； r 1 & lt； b)

b = r 1q 2+R2(0 & lt； r2 & ltr 1)

r 1 = r2q 3+R3(0 & lt； r3 & ltr2)

……

rk-2 = rk- 1qk+rk(0 & lt； rk & ltrk- 1)

……

rn-2 = rn- 1qn+rn(0 & lt； rn & ltrn- 1)

rn- 1=rnqn+ 1

Then (a, b) = (a-bq 1, b) = (b, r 1) = (r 1, R2) = … = (rn-1,rn) = rn.

Step by step from the previous formula, we can find out M and N, which proves the existence of M and N!

Let your d= 1, which is a b coprime.

Adequacy:

Let the greatest common divisor of a and b be d, then a = xd and b = yd x y are integers, and then substitute into the formula:

Xdm+ydn= 1, so d is the divisor of 1, so d= 1 is a and b coprime.

The following proves the original problem:

A m coprime means integers P 1 and Q 1, so that a * p1+m * q1=1.

B m coprime means that there is an integer P2, and Q2 makes b*p2+m*q2= 1.

Multiply the above two formulas to get:

a * b * p 1 * p2+m(a * p 1 * Q2+b * p2 * q 1+m * q 1 * Q2)= 1

Since p 1 p2 q 1 q2 a b m are all integers, so are p 1*p2, a * p1* Q2+b * p2 * q1+m * q1* Q2.

Take it, it's hard to find! !