Understanding Divisibility of Algebraic Expressions

The question asks us to evaluate the correctness of three statements regarding the divisibility of different algebraic expressions involving an integer 'n' by specific numbers.

We need to analyze each statement to determine if it holds true for all integers 'n'.

Analyzing Statement 1: \( n^3 - n \) is divisible by 6

Let's factor the expression \( n^3 - n \):

\( n^3 - n = n(n^2 - 1) \)

We can factor the term \( (n^2 - 1) \) further as a difference of squares:

\( n(n^2 - 1) = n(n-1)(n+1) \)

The expression \( (n-1)n(n+1) \) represents the product of three consecutive integers. Let's consider the properties of consecutive integers:

• Among any three consecutive integers, at least one must be even (divisible by 2).
• Among any three consecutive integers, exactly one must be divisible by 3.

Since the product \( (n-1)n(n+1) \) contains a factor that is a multiple of 2 and a factor that is a multiple of 3, the product must be divisible by both 2 and 3.

Since 2 and 3 are coprime (their greatest common divisor is 1), if a number is divisible by both 2 and 3, it is also divisible by their product, which is \( 2 \times 3 = 6 \).

Therefore, \( n^3 - n \) is always divisible by 6 for any integer 'n'. Statement 1 is correct.

Analyzing Statement 2: \( n^5 - n \) is divisible by 5

Let's factor the expression \( n^5 - n \):

\( n^5 - n = n(n^4 - 1) \)

We can factor \( (n^4 - 1) \) using the difference of squares formula repeatedly:

\( n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = n(n-1)(n+1)(n^2 + 1) \)

We can use Fermat's Little Theorem to analyze the divisibility by 5. Fermat's Little Theorem states that if 'p' is a prime number, then for any integer 'a', \( a^p \equiv a \pmod p \). This means \( a^p - a \) is divisible by 'p'.

In this statement, the prime number is \( p = 5 \) and the integer is 'n'. According to Fermat's Little Theorem, \( n^5 \equiv n \pmod 5 \). This implies that \( n^5 - n \) is divisible by 5.

Alternatively, consider cases for 'n' modulo 5:

• If \( n \equiv 0 \pmod 5 \), then \( n^5 - n \equiv 0^5 - 0 \equiv 0 \pmod 5 \).
• If \( n \equiv 1 \pmod 5 \), then \( n^5 - n \equiv 1^5 - 1 \equiv 1 - 1 \equiv 0 \pmod 5 \).
• If \( n \equiv 2 \pmod 5 \), then \( n^5 - n \equiv 2^5 - 2 \equiv 32 - 2 \equiv 30 \equiv 0 \pmod 5 \).
• If \( n \equiv 3 \pmod 5 \), then \( n^5 - n \equiv 3^5 - 3 \equiv 243 - 3 \equiv 240 \equiv 0 \pmod 5 \).
• If \( n \equiv 4 \pmod 5 \), then \( n^5 - n \equiv 4^5 - 4 \equiv (5-1)^5 - 4 \equiv (-1)^5 - 4 \equiv -1 - 4 \equiv -5 \equiv 0 \pmod 5 \).

In all cases, \( n^5 - n \) is divisible by 5. Statement 2 is correct.

Analyzing Statement 3: \( n^5 - 5n^3 + 4n \) is divisible by 120

Let's factor the expression \( n^5 - 5n^3 + 4n \):

\( n^5 - 5n^3 + 4n = n(n^4 - 5n^2 + 4) \)

The term \( (n^4 - 5n^2 + 4) \) is a quadratic in \( n^2 \). We can factor it by finding two numbers that multiply to 4 and add up to -5. These numbers are -1 and -4.

\( n(n^4 - 5n^2 + 4) = n(n^2 - 1)(n^2 - 4) \)

Now, we can factor the difference of squares terms \( (n^2 - 1) \) and \( (n^2 - 4) \):

\( n(n-1)(n+1)(n-2)(n+2) \)

Let's rearrange the terms in ascending order:

\( (n-2)(n-1)n(n+1)(n+2) \)

This expression represents the product of five consecutive integers: \( (n-2), (n-1), n, (n+1), (n+2) \). A fundamental property of consecutive integers is that the product of 'k' consecutive integers is always divisible by \( k! \).

In this case, we have the product of 5 consecutive integers, so it is divisible by \( 5! \). Let's calculate \( 5! \):

\( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Therefore, \( n^5 - 5n^3 + 4n \) is always divisible by 120 for any integer 'n'. Statement 3 is correct.

Conclusion on Divisibility Statements

Based on our analysis:

• Statement 1: \( n^3 - n \) is divisible by 6 (Correct, as it's the product of 3 consecutive integers).
• Statement 2: \( n^5 - n \) is divisible by 5 (Correct, by Fermat's Little Theorem or modular arithmetic).
• Statement 3: \( n^5 - 5n^3 + 4n \) is divisible by 120 (Correct, as it's the product of 5 consecutive integers).

All three statements are correct.

Revision Table: Divisibility Properties

Additional Information: Number Theory Concepts

Understanding concepts like divisibility, factorization, products of consecutive integers, and Fermat's Little Theorem is crucial for solving such problems involving algebraic expressions and number theory.

• Divisibility: An integer 'a' is divisible by an integer 'b' if there exists an integer 'k' such that \( a = bk \).
• Consecutive Integers: Integers that follow each other in order, like \( n, n+1, n+2, ... \). The product of 'k' consecutive integers is always divisible by \( k! \).
• Fermat's Little Theorem: If 'p' is a prime number, then for any integer 'a', \( a^p - a \) is divisible by 'p'. This can be written as \( a^p \equiv a \pmod p \).
• Factorials: The product of an integer and all the integers below it down to 1. Denoted by the symbol '!', e.g., \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).

These concepts help explain why the given expressions exhibit the specified divisibility properties.