91Ó°ÊÓ

In mathematics, when we talk about odd integers, we're referring to any integers that cannot be divided evenly by 2. Odd integers follow the form of \(a = 2k + 1\), where \(k\) is an integer. This equation highlights that no matter what value \(k\) takes, the result will always be an odd number.

A key property of odd integers is that their square yields a consistent result when evaluated under certain modular conditions. Specifically, for any odd integer \(a\), when you compute \(a^2\), the remainder when divided by 8 is always 1. This is expressed as \(a^2 \equiv 1 \pmod{8}\).

Here's why: Consider \(a = 2k + 1\). Then,

• \((2k + 1)^2 = 4k^2 + 4k + 1\)
• This simplifies to \(4k(k + 1) + 1\), and since \(k(k + 1)\) is always even, \(4k(k + 1)\) is divisible by 8.
• Thus, \(a^2 \equiv 1 \pmod{8}\).

The Chinese Remainder Theorem (CRT) is a powerful tool in modular arithmetic. It provides a system to solve congruences that are relatively prime, which means that the numbers involved do not share any common divisors other than 1.

The theorem asserts that if you have several congruences:

• \(x \equiv a_1 \pmod{n_1}\)
• \(x \equiv a_2 \pmod{n_2}\)

For the pairs \((n_1, n_2, ..., n_k)\) that are pairwise coprime (i.e., any two \(n_i\)'s have GCD of 1), there is a unique solution modulo the product \(N = n_1 \cdot n_2 \cdot ... \cdot n_k\) that satisfies all the congruences.

In the context of our exercise, the CRT is used to prove that if an integer \(a\) is not divisible by 2 or 3, then \(a^2 \equiv 1 \pmod{24}\). Since:

• \(a^2 \equiv 1 \pmod{8}\)
• and \(a^2 \equiv 1 \pmod{3}\)

The CRT confirms \(a^2 \equiv 1 \pmod{24}\) since 8 and 3 are coprime.

Cubic residues refer to the set of integers that can be expressed as a cube of some integer, reduced modulo a prime number. These residues have a specific behavior dependent on the modulus involved.

In modular arithmetic, specifically with mod 7, any integer raised to the power of three yields one of a few possible residues. For any integer \(a\), the expression \(a^3 \pmod{7}\) can result in 0, 1, or 6. This means:

• Zero, indicating perfect divisibility by 7.
• One, showing equivalence and a recurring cycle.
• Or six, representing a specific pattern observed across these calculations.

To verify this, one can compute the cubes of integers from 0 to 6:

• \(0^3 \equiv 0\)
• \(1^3 \equiv 1\)
• \(2^3 = 8 \equiv 1\)
• \(3^3 = 27 \equiv 6\)
• \(4^3 = 64 \equiv 1\)
• \(5^3 = 125 \equiv 6\)
• \(6^3 = 216 \equiv 6\)

This set of results confirms the expected behavior of cubic residues.

Quartic residues are similar to cubic residues, but they involve the fourth power of integers under a given modulus. These can have interesting properties that are essential for solving various modular arithmetic problems.

In our exercise, any integer raised to the fourth power modulo 5 is considered. The result is always either 0 or 1, forming a distinct pattern:

• \(a^4 \equiv 0 \pmod{5}\) if \(a\) is a multiple of 5.
• Otherwise, \(a^4 \equiv 1 \pmod{5}\).

Calculating the fourth powers of small integers mod 5 yields:

• \(0^4 \equiv 0\)
• \(1^4 \equiv 1\)
• \(2^4 = 16 \equiv 1\)
• \(3^4 = 81 \equiv 1\)
• \(4^4 = 256 \equiv 1\)

These computations confirm the pattern that emerges with quartic residues.