Question

For \(n \in N_{\geq 2}\) and \(a \in \mathbb{Z}\) let \(C_{n}(a)\) be the number of solutions \(g \in\\{0, \ldots, n-1\\}\) of the cubic congruence \(g^{3} \equiv a \bmod n\). (i) Show that the following hold for an odd prime \(p\) : \- \(C_{p}(a) \leq 3\). \- \(C_{p}(a)=1\) if \(p \mid a\) or \(p=3\). o \(C_{p}(a) \neq 2\), and for any value \(C \in\\{0,1,3\\}\) there is an odd prime \(p\) and an integer \(a\) such that \(3 \neq p \nmid a\) and \(C_{p}(a)=C\). (ii) Let \(p>3\) be a prime and \(e \in \mathbb{N}_{>0}\). Show that \(C_{r}(a)=C_{p}(a)\) if \(p \nmid a\), and give a counterexample when \(p \mid a\). (iii) Now let \(n \in \mathbb{N}\) such that \(\operatorname{gcd}(n, 6)=1\), and let \(n=p_{1}^{e}+\ldots p p_{r}^{e}\) be its prime factorization, with distinct primes \(p_{1}, \ldots, p_{r} \in \mathbb{N}\) and positive integers \(e_{1}, \ldots, e_{r}\). Find a formula expressing \(C_{n}(a)\) in terms of \(C_{p_{1}}(a)_{\ldots}, C_{p_{0}}(a)\) in the case where \(a\) and \(n\) are coprime. (iv) Compute all cube roots of 11 modulo \(225625 .\)

Short answer

Solutions involve counting distinct cubic roots, verifying conditions, applying theorems, and solving modulus systems.

Step-by-step solution

Introduction to Cubic Congruences

A cubic congruence like \(g^3 \equiv a \pmod{n}\) seeks to find all integers \(g\) such that when \(g^3\) is divided by \(n\), the remainder is \(a\). Here, \(C_n(a)\) counts the number of such solutions for a given \(n\) and \(a\).

Situation with Odd Prime p

For a prime \(p\), any number raised to the cubic power modulo \(p\) can have at most 3 distinct values. This is due to the nature of the polynomial \(g^3 - a\) which can have at most three roots over any field, here being the integers modulo \(p\). Hence, \(C_p(a) \leq 3\).

Behavior When p Divides a or p=3

When \(p \mid a\), it means \(a\equiv 0 \pmod{p}\). The only solution \(g\equiv 0 \pmod{p}\) makes \(C_p(a)=1\). If \(p=3\), a similar simplification occurs due to the symmetry in cubic roots modulo 3 which makes \(C_3(a)=1\).

Exploration of C Values

For \(C_p(a) eq 2\), it is due to non-existence of exactly two distinct roots in modular arithmetic for cubics over odd primes. For \(C_n(a)\) to be \{0,1,3\}, constructive examples must exist. Consider primes \(p eq 3\) and a carefully chosen \(a\) to validate each possibility.

Property for p>3 and e>0

If \(p mid a\), then \(C_p(a)=C_r(a)\). This follows from the fundamental theorem of arithmetic and Chinese Remainder Theorem under coprime conditions. A counterexample for when \(p \mid a\) shows dependence, displaying the failure of direct correspondence under this condition.

Derivation for Coprime a and n

For \(n\) with a prime factorization \(n=p_1^{e_1} \cdots p_r^{e_r}\) we use Chinese Remainder Theorem. The formula becomes \[C_n(a) = C_{p_1}(a) \times C_{p_2}(a) \times \ldots \times C_{p_r}(a).\] This holds under the condition that \(a\) and \(n\) are coprime, leveraging the structure of modular solutions.

Example: Cube Roots of 11 Modulo 225625

The number 225625 factors as \(5^2 \times 13^2\). We evaluate cube roots modulo each factor and use the Chinese Remainder Theorem to solve the system of congruences. Given \(x^3 \equiv 11 \pmod{25}\) and \(x^3 \equiv 11 \pmod{169}\), find all solutions for both, then combine them.

Key concepts

Modular Arithmetic

Modular arithmetic is an essential mathematical concept when dealing with congruences. It involves performing arithmetic operations, such as addition, subtraction, multiplication, and exponentiation, with a wrap-around at a certain number, called the modulus. This approach is similar to the way we view time on a clock, where after reaching 12, it starts again at 1.

In mathematical terms, if we have numbers, say, 7 and 20, we perform the operation 7 + 20 and get 27. In modular arithmetic, with modulus 12, 27 is equivalent to 3 because 27 minus 2 full cycles of 12 gives a remainder of 3. This is expressed as 27 ≡ 3 (mod 12).

When addressing cubic congruences like the ones in the original task, we apply modular arithmetic to look for solutions to the equation \( g^3 \equiv a \pmod{n} \), where we're interested in finding the values of \( g \) that satisfy this equation after exponentiation.

Chinese Remainder Theorem

The Chinese Remainder Theorem (CRT) is a powerful tool in solving problems where you're working with several congruences. Essentially, CRT allows us to solve systems of simultaneous linear congruences, provided the moduli are coprime, that is, their greatest common divisor is 1.

For example, if you have two congruences \( x \equiv a \pmod{m} \) and \( x \equiv b \pmod{n} \), and given that \( m \) and \( n \) are coprime, CRT guarantees a unique solution modulo \( mn \). This theorem is particularly useful in our scenario for solving cubic congruences across multiple moduli, like in the example from the original problem, where the moduli can be treated separately before combining the solutions.

CRT is analogous to piecing a puzzle together where each piece fits in perfectly due to the conditions of coprimeness, allowing us to form a complete picture with unique solutions up to the product of the moduli.

Prime Factorization

Prime factorization involves breaking down a composite number into a product of its prime numbers. This concept is fundamental in number theory and plays a vital role in modular arithmetic problems, especially when dealing with cubic congruences.

Consider a number \( n \) which can be expressed as \( n = p_1^{e_1} \times p_2^{e_2} \times \ldots \times p_r^{e_r} \), where \( p_1, p_2, \ldots, p_r \) are distinct primes and \( e_1, e_2, \ldots, e_r \) are their respective powers. This breakdown is crucial because, using the prime factorization, we can employ the Chinese Remainder Theorem to tackle large modulus problems by breaking them into manageable smaller problems.

The step-by-step solution of cube roots or any polynomial congruences often requires understanding the structure of the moduli through their prime factors, allowing for efficient and successful problem-solving.

Polynomial Roots

Finding the roots of a polynomial is at the core of solving congruences like \( g^3 \equiv a \pmod{n} \). For a given polynomial \( f(g) = g^3 - a \), the goal is to find all possible values of \( g \) for which the polynomial equals zero within a specific modulus.

When working over the integers, polynomial roots can generally be straightforward; however, things can become more intricate within modular arithmetic because a polynomial of degree \( n \) can have up to \( n \) roots. In the context of the problem, we consider the reductions modulo prime and composite moduli.

The exploration of polynomial roots in modular arithmetic applies techniques like evaluating residue classes or leveraging the symmetry properties of specific powers when the modulus has particular characteristics. Understanding these aspects aids in finding complete solutions and forms the foundation for determining the number of valid solutions to a modular cubic equation.