Analyzing Mathematical Statements on Number Properties

This problem asks us to evaluate the correctness of two statements involving mathematical expressions and specific types of numbers: natural numbers and odd integers. We need to determine if the results of these expressions always belong to the specified number sets for the given inputs.

Analyzing Statement 1: Natural Number Expression

Statement 1 claims that for any natural number \(n\), the expression \(\frac{n\left(n^2+2\right)}{3}\) is also a natural number.

A natural number is a positive integer (typically 1, 2, 3, ...). For the expression \(\frac{n(n^2+2)}{3}\) to be a natural number, it must be an integer greater than or equal to 1 for any natural number \(n\).

Let's examine the expression \(n(n^2+2) = n^3 + 2n\). We need to determine if \(n^3+2n\) is always divisible by 3 for any natural number \(n\).

We can check the divisibility by 3 by considering the possible values of \(n\) modulo 3:

1. Case 1: \(n \equiv 0 \pmod{3}\)
2. Case 2: \(n \equiv 1 \pmod{3}\)
3. Case 3: \(n \equiv 2 \pmod{3}\)

Case 1: \(n\) is a multiple of 3

If \(n \equiv 0 \pmod{3}\), then \(n\) is divisible by 3. The expression is \(\frac{n(n^2+2)}{3}\). Since \(n\) is divisible by 3, \(\frac{n}{3}\) is an integer. The expression can be written as \(\frac{n}{3} \cdot (n^2+2)\). This is the product of an integer (\(\frac{n}{3}\)) and an integer (\(n^2+2\)), which is always an integer.

Case 2: \(n \equiv 1 \pmod{3}\)

If \(n \equiv 1 \pmod{3}\), let's evaluate \(n^2+2\) modulo 3:

\(n^2+2 \equiv 1^2+2 \pmod{3}\)

\(n^2+2 \equiv 1+2 \pmod{3}\)

\(n^2+2 \equiv 3 \pmod{3}\)

\(n^2+2 \equiv 0 \pmod{3}\)

This shows that if \(n \equiv 1 \pmod{3}\), then \(n^2+2\) is divisible by 3. The expression \(\frac{n(n^2+2)}{3} = n \cdot \frac{n^2+2}{3}\) is the product of a natural number \(n\) and an integer \(\frac{n^2+2}{3}\). The product of a natural number and an integer is always an integer.

Case 3: \(n \equiv 2 \pmod{3}\)

If \(n \equiv 2 \pmod{3}\), let's evaluate \(n^2+2\) modulo 3:

\(n^2+2 \equiv 2^2+2 \pmod{3}\)

\(n^2+2 \equiv 4+2 \pmod{3}\)

\(n^2+2 \equiv 6 \pmod{3}\)

\(n^2+2 \equiv 0 \pmod{3}\)

This shows that if \(n \equiv 2 \pmod{3}\), then \(n^2+2\) is divisible by 3. The expression \(\frac{n(n^2+2)}{3} = n \cdot \frac{n^2+2}{3}\) is the product of a natural number \(n\) and an integer \(\frac{n^2+2}{3}\), which is always an integer.

Conclusion for Statement 1

In all possible cases for a natural number \(n\), the expression \(n(n^2+2)\) is divisible by 3. Furthermore, since \(n\) is a natural number (\(n \ge 1\)), \(n(n^2+2)\) will be \(1(1^2+2) = 3\) or greater. Thus, \(\frac{n(n^2+2)}{3}\) will be an integer greater than or equal to 1, which fits the definition of a natural number.

Therefore, Statement 1 is correct.

Analyzing Statement 2: Integer Expression for Odd Integers

Statement 2 claims that for any odd integer \(m\), the expression \(\frac{m^4+4m^2+11}{16}\) is an integer.

An integer is a whole number (..., -2, -1, 0, 1, 2, ...). For the expression to be an integer, the numerator \(m^4+4m^2+11\) must be divisible by 16 for any odd integer \(m\).

An odd integer \(m\) can be written in the form \(m = 2k+1\) for some integer \(k\).

Let's consider the square of an odd integer, \(m^2\):

\(m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1\)

The term \(k(k+1)\) is always the product of two consecutive integers, so it must be an even number. Let \(k(k+1) = 2j\) for some integer \(j\).

Then \(m^2 = 4(2j) + 1 = 8j + 1\).

This shows that the square of any odd integer \(m\) is always congruent to 1 modulo 8 (\(m^2 \equiv 1 \pmod{8}\)).

Now substitute \(m^2 = 8j+1\) into the numerator of the expression:

\(m^4+4m^2+11 = (m^2)^2 + 4(m^2) + 11\)

\(= (8j+1)^2 + 4(8j+1) + 11\)

Expand the terms:

\(= (64j^2 + 16j + 1) + (32j + 4) + 11\)

\(= 64j^2 + 16j + 32j + 1 + 4 + 11\)

\(= 64j^2 + 48j + 16\)

We need to check if this resulting expression is always divisible by 16.

\(64j^2 + 48j + 16 = 16(4j^2 + 3j + 1)\)

Since \(j\) is an integer, the expression \(4j^2 + 3j + 1\) is also an integer. Therefore, the numerator \(m^4+4m^2+11\) is always a multiple of 16 for any odd integer \(m\).

Thus, \(\frac{m^4+4m^2+11}{16}\) is always an integer.

Therefore, Statement 2 is correct.

Conclusion on Statement Correctness

Our analysis shows that both Statement 1 and Statement 2 are correct.

• Statement 1: For any natural number \(n\), \(\frac{n(n^2+2)}{3}\) is a natural number.
• Statement 2: For any odd integer \(m\), \(\frac{m^4+4m^2+11}{16}\) is an integer.

Revision Table: Key Mathematical Concepts

Additional Information: Further Exploration of Number Properties

Divisibility by 3

An integer is divisible by 3 if and only if the sum of its digits is divisible by 3. However, for algebraic expressions like \(n^3+2n\), checking modular properties (\(n \pmod{3}\)) is a more general and rigorous approach.

Squares of Odd Integers

As shown in Statement 2 analysis, the square of any odd integer \(m\) always leaves a remainder of 1 when divided by 8. This property (\(m^2 \equiv 1 \pmod{8}\)) is useful in number theory problems.

Generalizing Divisibility Proofs

To prove that an expression involving an integer variable is always divisible by a specific number, you can use methods such as:

• Testing cases based on modulo the divisor.
• Algebraically factoring the expression to show the divisor is a factor.
• Using proof by induction.