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Magazine Chapter 5: Problem 17
If \(p\) and \(q\) are distinct primes, prove that for any integer \(a\), $$ p q \mid a^{p q}-a^{p}-a^{q}+a $$
Short Answer
Expert verified
\(pq\) divides \(a^{pq} - a^p - a^q + a\) by Fermat's Little Theorem and the Chinese Remainder Theorem.
Step by step solution
01
Understanding the Problem
We need to prove that \(pq\) divides \(a^{pq} - a^p - a^q + a\) for any integer \(a\), given \(p\) and \(q\) are distinct prime numbers.
02
Workshop on Expression
We need to re-write the expression \(a^{pq} - a^p - a^q + a\) in a form that is easier to work with in terms of divisibility by \(pq\).
03
Divisibility by p
Using Fermat's Little Theorem, which states that \(a^p \equiv a \pmod{p}\) for a prime \(p\), we see that \(a^{pq} - a^p - a^q + a \equiv 0 \pmod{p}\). This is because \ a^p \equiv a \Rightarrow a^{pq} \equiv a^q \equiv a \pmod{p}.
04
Divisibility by q
Similarly, use Fermat's Little Theorem to conclude \(a^{pq} - a^p - a^q + a \equiv 0 \pmod{q}\) because \ a^q \equiv a \Rightarrow a^{pq} \equiv a^p \equiv a \pmod{q}.
05
Combine Results for pq
Since \(a^{pq} - a^p - a^q + a\) is congruent to 0 modulo \(p\) and modulo \(q\), and because \(p\) and \(q\) are distinct primes, it follows from the Chinese Remainder Theorem that \(pq\) divides the expression \(a^{pq} - a^p - a^q + a\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fermat's Little Theorem
Many students encounter Fermat's Little Theorem when studying modular arithmetic. It's a powerful tool for simplifying calculations involving powers of integers. Fermat's Little Theorem states that if you have a prime number \( p \) and an integer \( a \), then \( a^p \equiv a \pmod{p} \). In simpler terms, when you raise \( a \) to the power of \( p \), the remainder is \( a \) when divided by \( p \). This result becomes handy when determining divisibility by a prime.
- Consider \( a^p \equiv a \pmod{p} \). This means subtracting \( a \) will always leave a multiple of \( p \).
- Similarly, raising \( a \) to multiples of \( p \), like \( a^{2p} \), results in \( a^{2p} \equiv a^p \equiv a \pmod{p} \).
Chinese Remainder Theorem
The Chinese Remainder Theorem (CRT) is an essential principle in number theory. It deals with solving simultaneous congruences and finding common solutions to multiple conditions. If \( p \) and \( q \) are distinct primes, CRT says we can uniquely determine a number modulo \( pq \) if we know its values modulo \( p \) and \( q \).
- For example, if we have two congruences \( x \equiv a \pmod{p} \) and \( x \equiv b \pmod{q} \), then a unique solution exists modulo \( pq \).
- This theorem ensures that conditions met individually by \( p \) and \( q \) will align when considered together for \( pq \).
Prime Numbers
Prime numbers are the building blocks of whole numbers. A prime number is greater than 1 and only divisible by 1 and itself. This fundamental property makes primes crucial in number theory.
- Each number greater than 1 can be uniquely factored as a product of prime numbers, known as prime factorization.
- Due to their unique divisibility properties, primes play a significant role in theorems related to number divisibility.
Divisibility
Divisibility is a key concept in mathematics. It refers to one number being able to be divided by another without a remainder. In particular, divisibility by products of primes is a common theme in number theory problems.
- When a number is divisible by another, the division leaves no remainder, i.e., \( a \mid b \) means \( b = ka \) for some integer \( k \).
- Divisibility rules help simplify expressions to see their factors easily, especially when breaking down components that share common divisors.
