# Please teach me. I am completely blank with it. :(

Let alpha = (3+sqrt(-3))/2 belongs to Q[sqrt(3)].

Show that if x is congruent to 1 mod alpha, then x^3 is congruent to 1

(mod alpha)^3.

Similarly, show that if x is congruent to -1 mod alpha, then x^3 is congruent to

-1 (mod alpha)^3 , and that if x is congruent to 0 mod alpha, then x^3 is congruent to 0 (mod alpha)^3.

Hint: you can factor x^3 -1 in Q[sqrt(d)] completely into linear factors.

## To prove these congruence relationships, we need to show that if x is congruent to a certain value mod alpha, then x^3 is congruent to a certain value mod (alpha)^3.

First, let's understand what it means for two numbers to be congruent mod alpha. If x is congruent to a mod alpha, it means that their difference is a multiple of alpha. Mathematically, x ≡ a (mod alpha) implies x - a = m*alpha, where m is an integer.

Let's start with the first part: if x is congruent to 1 mod alpha, then we need to show that x^3 is congruent to 1 mod (alpha)^3.

We can start by expanding (x - 1)^3 using the binomial theorem: (x - 1)^3 = x^3 - 3x^2 + 3x - 1.

Now, we have to prove that (x - 1)^3 is congruent to 0 mod alpha. This means that (x - 1)^3 - 0 = m*alpha, where m is an integer.

To simplify the task, we can factorize (x - 1)^3 - 0 = m*alpha into linear factors. The hint provided suggests that we can factorize it completely in Q[sqrt(d)], where d = -3.

Let's factorize (x - 1)^3 - 0. We have (x - 1)^3 - 0 = x^3 - 3x^2 + 3x - 1 - 0 = m*alpha.

Now, substitute alpha as (3+sqrt(-3))/2 in (x - 1)^3 - 0:

(x - 1)^3 - 0 = (x^3 - 3x^2 + 3x - 1) - 0 = m*((3+sqrt(-3))/2).

Simplifying further, we get: x^3 - 3x^2 + 3x - 1 = m*((3+sqrt(-3))/2).

Now, to prove the congruence relationship, we need to show that the right-hand side of the equation is a multiple of (alpha)^3.

Let's evaluate (alpha)^3:

(alpha)^3 = ((3+sqrt(-3))/2)^3 = (27+9*sqrt(-3)+9*sqrt(-3)+(-3))/8 = (27-3+18*sqrt(-3))/8 = (24+18*sqrt(-3))/8 = 3+9*sqrt(-3) = 3alpha.

Therefore, we can rewrite the equation as: x^3 - 3x^2 + 3x - 1 = m*(3alpha).

This shows that x^3 is congruent to 1 mod (alpha)^3 when x is congruent to 1 mod alpha.

Similarly, you can follow a similar approach to prove the other two statements:

- If x is congruent to -1 mod alpha, you need to show that x^3 is congruent to -1 mod (alpha)^3.

- If x is congruent to 0 mod alpha, you need to show that x^3 is congruent to 0 mod (alpha)^3.

By following the same steps and substitutions, you can demonstrate these congruence relationships.

Remember to substitute alpha as (3+sqrt(-3))/2 in your calculations and use the fact that (alpha)^3 = 3alpha.

I hope this explanation helps you understand how to prove these congruence relationships. If you have any further questions, feel free to ask!