Question

For the prime \(p=7\), calculate for each integer \(a\) with \(1<\) \(a<7\) the smallest exponent \(m\) such that \(a^{m} \equiv 1(\bmod 7)\). Show that the theorem in Exercise 1 holds for all \(a\).

Short answer

For each a, the smallest m is as follows: - For a = 2, the smallest m is 3. - For a = 3, the smallest m is 6. - For a = 4, the smallest m is 3. - For a = 5, the smallest m is 6. - For a = 6, the smallest m is 2. These results are consistent with the theorem in Exercise 1, as all these m values divide 6 (p-1, where p is the prime number 7).

Step-by-step solution

List the integers between 1 and 7.

We have 5 integers between 1 and 7, which are: a = 2, 3, 4, 5, 6

Find the smallest exponent m for each a.

Calculate powers of a for each a (mod 7) until reaching 1: For a = 2: \(2^1 \equiv 2\), \(2^2 \equiv 4\), \(2^3 \equiv 1\) (mod 7), smallest m for a = 2 is 3. For a = 3: \(3^1 \equiv 3, 3^2 \equiv 2, 3^3 \equiv 6\), \(3^4 \equiv 4, 3^5 \equiv 5, 3^6 \equiv 1\) (mod 7), smallest m for a = 3 is 6. For a = 4: \(4^1 \equiv 4\) \(4^2 \equiv 2\) \(4^3 \equiv 1\) (mod 7), smallest m for a = 4 is 3. For a = 5: \(5^1 \equiv 5, 5^2 \equiv 4, 5^3 \equiv 6\), \(5^4 \equiv 2, 5^5 \equiv 3, 5^6 \equiv 1\) (mod 7), smallest m for a = 5 is 6. For a = 6: \(6^1 \equiv 6\), \(6^2 \equiv 1\) (mod 7), smallest m for a = 6 is 2.

Verify the theorem in Exercise 1.

According to the theorem in Exercise 1: For integers a coprime to the prime number p, there is a smallest positive integer exponent m such that \(a^m \equiv 1\) (mod p), and m divides p-1. From step 2, we obtained the following pairs (a, m): (2, 3), (3, 6), (4, 3), (5, 6), and (6, 2). We can see that all these m values divide (p-1) (p being 7): (7-1) = 6. Therefore, the theorem in Exercise 1 holds for all of the given integer values of a.

Key concepts

Modular Arithmetic

Modular arithmetic is like clock arithmetic, where numbers wrap around after reaching a certain value, called the modulus. It simplifies complex calculations by focusing only on the remainder after division.
For example, in the context of our original exercise, the modulus is 7. When we say that \( a^m \equiv 1 \ (\text{mod} \ 7) \), it means that when \( a^m \) is divided by 7, the remainder is 1.
Understanding modular arithmetic involves:

• Identifying the modulus: It's the number after 'mod', such as 7 in our case.
• Calculating remainders: Perform the calculations to see which number a power is congruent to.

This concept underpins many areas of mathematics, especially number theory, because it helps identify patterns and solve problems related to divisibility, congruence, and more.

Exponentiation

Exponentiation refers to the repeated multiplication of a number by itself. In mathematical terms, \( a^m \) represents the number \( a \) multiplied by itself \( m \) times.
In our original challenge, we needed to calculate the smallest \( m \) such that \( a^m \equiv 1 \ (\text{mod} \ 7) \) for various values of \( a \).
Here's how it works:

• Start with a base number: Choose the integer \( a \).
• Apply exponentiation: Multiply \( a \) by itself \( m \) times.
• Check modulus congruence: See if the result equals 1 modulus 7.

By understanding exponentiation in modular arithmetic, we can simplify complex numeric relationships and identify cycles in number patterns, essential in fields such as cryptography and computer science.

Prime Numbers

Prime numbers are unique because they have only two divisors: 1 and themselves. The number 7, used in our exercise, is a prime.
Prime numbers play a key role in our problem because they simplify finding congruence solutions.
Here's why prime numbers matter:

• Simplified calculations: Prime moduluses result in more straightforward modular arithmetic because fewer numbers share common factors.
• Key properties: Primes like 7 ensure properties like coprimeness, important for proving theorems like Fermat's Little Theorem.

Understanding these numbers enables mathematicians to prove important principles, such as Fermat's Little Theorem, which states any number \( a \) that is coprime to a prime \( p \) will satisfy \( a^{p-1} \equiv 1 \ (\text{mod} \ p) \). This theorem is a cornerstone of modern number theory.

Theorems in Number Theory

Number theory is rich with theorems that provide profound insights into integers and their relationships. In the context of our original exercise, one such theorem was fundamental: a variant of Fermat's Little Theorem.
This theorem tells us that for prime \( p \) and integer \( a \) where \( a \) is coprime to \( p \), there exists a smallest positive integer \( m \) such that \( a^m \equiv 1 \ (\text{mod} \ p) \).
Key takeaways from this theorem:

• The concept of coprimeness: \( a \) must have no common divisors with \( p \) except for 1.
• Divisibility: This \( m \) is a divisor of \( p-1 \), providing a constraint that simplifies finding \( m \).

Fermat's Little Theorem and related principles demonstrate the structured beauty of mathematics, explaining how properties like divisibility and congruence interconnect in modular systems.