Now, we know the equation
ax+by=gcd(a,b)
There is always an integer solutionxAndyHow many solutions are there and how to express them?
We are mutually exclusiveaAndbLet gogcd(a,b)=1, assuming(x_1,y_1)It’s an equationax+by=1A solution of. adoptx_1subtractbSum of multiples ofy_1addaOther solutions can be obtained. In other words, for any integerkWe get a new solution(x_1+kb,y_1-ka)By calculating
a(x_1+kb)+b(y_1-ka)=ax_1+akb+by_1-bka=ax_1+by_1=1
Still observinggcd(a,b)=1It can be proved that this method gives all solutions and hypotheses(x_1,y_1)And(x_2,y_2)It’s an equationax+by=1Two solutions of, namely
Ax_ 1+by_ 1 = 1 and ax_ 2+by_ 2=1
We use ity_1Multiply the first equation with they_2Multiply the second equation and subtract itbAfter sorting outax_1y_2-ax_2y_1=y_2-y_1
Similar, if you usex_2Multiply the first equation with thex_1Multiply the second equation and subtract itbx_2y_1-bx_1y_2=x_2-x_1
Ifk=x_2y_1-x_1y_2, then we get
x_2=x_1+kb\\
y_2=y_1-ka
Therefore, from the initial understanding of the solution(x_1,y_1)By going to differentkValue can be obtainedax+by=1Every solution of(x_1+kb,y_1-ka)
Ifgcd(a,,b)>1What will happen?
We orderg=gcd(a,b)The equation is known by Euclidean algorithmax+by=gAt least one solution(x_1,y_1), andgto be divisible byaAndbSo(x_1,y_1)It’s a simple equation
\frac{a}{g}x+\frac{b}{g}y=1
The solution. Through thekThe value of
(x_1+k·\frac{b}{g},y_1-k \frac{a}{g})
###Theorem of linear equation
set upaAndbIs a nonzero integer,g=gcd(a,b), equationax+by=gThere is always an integer solution(x_1,y_1)Then each solution of the equation can be obtained by the Euclidean algorithm(x_1+k·\frac{b}{g},y_1-k \frac{a}{g})Get, wherekCan be any integer.
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