Question

Let \(p_{0}, p_{1} \in \mathbb{N}\) be distinct primes, \(m=p_{0} p_{1}, n \in \mathbb{N}\), and \(u_{0}, \ldots, u_{n-1}, v_{0}, \ldots, v_{n-1} \in \mathbb{Z}\). (i) Show that there exists an interpolating polynomial \(f \in \mathbb{Z}[x]\) such that \(f\) has coefficients in \(\\{0, \ldots, m-1\\}, \operatorname{deg} f

Short answer

(i) Existence: \( f \) exists if condition holds. (ii) Unique if \( u_i \not\equiv u_j \pmod{p_k} \). (iii) Use Lagrange interpolation.

Step-by-step solution

Understanding Conditions for Existence

The first task is to show that there exists a polynomial \( f \) such that \( f(u_i) \equiv v_i \pmod{m} \) if and only if certain conditions are met. The condition \( u_i \equiv u_j \pmod{p_k} \) implies \( v_i \equiv v_j \pmod{p_k} \) plays a crucial role. This ensures that for all intersecting equations mod \( p_0 \) and mod \( p_1 \), the values \( v_i \) are consistent. This condition is necessary to apply the Chinese Remainder Theorem, which assures the existence of such a polynomial.

Ensuring Uniqueness

Now, we must show that the solution is unique if \( u_i ot\equiv u_j \pmod{p_k} \) for any indices \( i, j \) and for both \( k=0,1 \). This condition prevents any overlap of the roots mod \( p_0 \) or mod \( p_1 \), ensuring that no two values \( u_i \) carry the same congruent information, which would potentially allow multiple solutions. Therefore, if every \( u_i \) is distinct mod \( p_0 \) and mod \( p_1 \), then the interpolating polynomial is unique.

Applying Facts to Compute Interpolating Polynomial

For part (iii), using the conditions, we find all interpolating polynomials for specific values of \( u_i \) and \( v_i \). The problem provides: \( f(1) \equiv 2 \pmod{15} \), \( f(2) \equiv 5 \pmod{15} \), \( f(4) \equiv -1 \pmod{15} \). Given that \( 15 = 3 \cdot 5 \), we need to meet the same conditions for \( p_0 = 3 \) and \( p_1 = 5 \). Make sure that all \( u_i \) are distinct mod \( 3 \) and mod \( 5 \). Construct \( f(x) \) as a polynomial of degree less than 3 using these equivalences.

Using Lagrange Interpolation

Use the Lagrange interpolation formula to find the polynomial that fits the provided conditions. Start with: \[ f(x) = a(x-2)(x-4) + b(x-1)(x-4) + c(x-1)(x-2) \] and solve for coefficients \( a, b, c \) such that each given condition is satisfied. Solve these systems of congruences for \( a, b, c \) modulo 15 to get all polynomials.

Key concepts

Chinese Remainder Theorem

The Chinese Remainder Theorem is a powerful tool in number theory. It helps solve systems of linear congruences with different moduli. When numbers or equations are expressed modulo several co-prime integers, the theorem guarantees a solution. This solution exists and is unique modulo the product of these integers.
For example, if you have systems modulo two co-prime numbers, say 3 and 5, as in this exercise, the theorem assures that there is a unique solution modulo 15, which is the product of 3 and 5.
The application of the Chinese Remainder Theorem in polynomial interpolation ensures that the polynomial with values determined by the conditions at hand can be constructed. Each congruence relation needs to be consistent across the moduli, ensuring the solution unambiguously meets the given conditions at different moduli.

Lagrange Interpolation

Lagrange Interpolation is a method used to find a polynomial that passes through a given set of points. This technique is particularly useful when the degree of the polynomial needs to be lower than a certain value. It constructs a polynomial by ensuring it fits the given points exactly.
In the context of this problem, Lagrange Interpolation helps find a polynomial of degree less than 3, which fits points derived from the congruence relations:

• \( f(1) \equiv 2 \pmod{15} \)
• \( f(2) \equiv 5 \pmod{15} \)
• \( f(4) \equiv -1 \pmod{15} \)

The Lagrange interpolation formula is applied to combine the known values by constructing individual polynomial components. These components ensure each given point is fulfilled without affecting others. This formula is particularly efficient for small sets of data points.

Congruence Relations

Congruence Relations are fundamental in modular arithmetic. They describe equivalences between integers under a given modulus. Understanding these relations makes it easier to solve equations where normal arithmetic doesn’t apply.
In this exercise, the congruence relations join points and values to create conditions for constructing a polynomial. The rule \( u_i \equiv u_j \pmod{p_k} \) implies \( v_i \equiv v_j \pmod{p_k} \) for certain indices. This condition ensures solutions are consistent across different moduli.
When multiple conditions need to be met simultaneously, congruence relations act as the framework ensuring each condition doesn’t contradict another under different moduli. In polynomial interpolation, they help ensure individual data points result in predictable and unique outputs.