Question

Solve (a) x^3=7 (mod 16), (b) x^3=12 (mod 27).

Answer 1

Let x^3= t . Hence it is the linear congurence

t congurent to 7( mod 16)

t congruent to 12( mod 27)

By chinese remainder theorem there exists (upto mod 16×27 ) unique solution to this system of linear congruence as gcd ( 16, 27) =1 . Let us find that t_0

Now let us find an x such that 16.x congruent to 1 mod27. Such an x is 11 . I am uploading the calculation as an imageby similar calculation we can find y such that

27y congruent to 1( mod 16). Such a y is 20

Now concider t_0= 11×16× 7 + 20×27×12 (mod 16×27)

Hence t_0 = 368

Now just solve x^3 congruent to 368( mod 27×16)