I have this problem in my book:

Solve the congruences:

$\displaystyle x^3+2x-3\equiv0(mod 9)$;

$\displaystyle x^3+2x-3\equiv0(mod 5)$;

$\displaystyle x^3+2x-3\equiv0(mod 45)$ .

Does this question ask me to solve this as a system of congruences? I'm unsure on this.If not, here is my way to solve each individually, can someone check if what I'm doing is correct. Thanks.

Since 9=3x3, by the CRS we can solve the system $\displaystyle x^3+2x-3\equiv0(mod 3) and x^3+2x-3\equiv0(mod 3)$. I check {0,1,2} and see that all of them are solutions, so the answer for the first congruence are {0,1,2}.

Do the same thing for the second congruence, and I get {1,3}

Since 45=3x3x5. We can solve the third congruence by solving a system of

$\displaystyle x^3+2x-3\equiv0(mod 3) , x^3+2x-3\equiv0(mod 3) and x^3+2x-3\equiv0(mod 5) $.

So,the solution for the third congruence is the common solution for the first and second, which is {1}.