Suppose gcd(a,b)=1\gcd(a, b) = 1gcd(a,b)=1. Which pairs of integers (x,y)(x, y)(x,y) solve the equation ax+by=1ax + by = 1ax+by=1?
No solution exists
There are infinitely many solutions of the form (x0+bdt,y0−adt)(x_0 + \frac{b}{d}t, y_0 - \frac{a}{d}t)(x0+dbt,y0−dat) for integer ttt where d=gcd(a,b)d = \gcd(a,b)d=gcd(a,b)
There is always exactly one solution
There are infinitely many solutions of the form (x0+bt,y0−at)(x_0 + bt, y_0 - at)(x0+bt,y0−at) for integer ttt