1. ## Congruence problem

I have this problem in my book:
Solve the congruences:
$x^3+2x-3\equiv0(mod 9)$;
$x^3+2x-3\equiv0(mod 5)$;
$x^3+2x-3\equiv0(mod 45)$ .

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 $x^3+2x-3\equiv0(mod 3) and x^3+2x-3\equiv0(mod 3)$. 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
$x^3+2x-3\equiv0(mod 3) , x^3+2x-3\equiv0(mod 3) and x^3+2x-3\equiv0(mod 5)$.
So,the solution for the third congruence is the common solution for the first and second, which is {1}.

2. Originally Posted by namelessguy
Since 9=3x3, by the CRS we can solve the system
Be careful, in the Chinese Remainder Theorem you must are relatively prime numbers, which 3 and 3 are not. Otherwise, it would mean than being divisible by 3 is the same as being divisible by 9...

Then I can't see a quicker solution than looking at the 9 cases (it is pretty quick though, since the remainder modulo 9 is just the sum of the digits). I think 1, 2 and 6 are solutions.

For the third one, using the CRS is correct however, since 5 and 9 have no common divisor except 1. What you get is that the solutions are the integers congruent to 1, 2 or 6 modulo 9, and to 1 or 3 modulo 5. This makes 3x2=6 possibilities of congruence systems. The CRS gives you that the solutions of these systems are the integers congruent to any particular solution modulo 45. For these particular solutions, I get 1 (congruent to 1 mod 5 and 1 mod 9), 11 (1 and 2), 6 (1 and 6), 28 (3 and 1), 38 (3 and 2) and 33 (3 and 6).

3. Hello,

Also note that $x^3+2x-3=(x-1)(x^2+x+3)$

So 1 is a solution in each equation !