Analyzing Integer Properties: Consecutive and Odd Integers

The question asks us to evaluate the correctness of two statements related to the properties of integers. Let's examine each statement individually.

Statement 1: Of the two consecutive integers, one is even.

Consecutive integers are integers that follow each other in order, such as 5 and 6, or -2 and -1. We can represent any integer as either even or odd.

• An even integer is an integer that is divisible by 2, meaning it can be written in the form \(2k\) for some integer \(k\).
• An odd integer is an integer that is not divisible by 2, meaning it can be written in the form \(2k+1\) for some integer \(k\).

Let's consider two consecutive integers. We can represent the first integer as \(n\). The next consecutive integer would then be \(n+1\). There are two possibilities for \(n\):

1. If \(n\) is an even integer, then \(n = 2k\) for some integer \(k\). In this case, the next integer is \(n+1 = 2k+1\), which is an odd integer. Example: If \(n=4\) (even), then \(n+1=5\) (odd).
2. If \(n\) is an odd integer, then \(n = 2k+1\) for some integer \(k\). In this case, the next integer is \(n+1 = (2k+1)+1 = 2k+2 = 2(k+1)\). Since \(k+1\) is an integer, \(2(k+1)\) is an even integer. Example: If \(n=3\) (odd), then \(n+1=4\) (even).

In both cases, when we have two consecutive integers, one is even and the other is odd. This confirms that one of the two consecutive integers must be even.

Therefore, Statement 1 is correct.

Statement 2: Square of an odd integer is of the form 8n + 1

Let's consider an arbitrary odd integer. An odd integer can be represented in the form \(2k+1\) for some integer \(k\). Now let's find the square of this odd integer:

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

We can factor out 4k from the first two terms:

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

Now, consider the term \(k(k+1)\). This represents the product of two consecutive integers, \(k\) and \(k+1\). As we established in the analysis of Statement 1, one of any two consecutive integers must be even. This means either \(k\) is even or \(k+1\) is even.

• If \(k\) is even, then \(k = 2m\) for some integer \(m\). Then \(k(k+1) = 2m(k+1)\), which is an even number.
• If \(k+1\) is even, then \(k+1 = 2m\) for some integer \(m\). Then \(k(k+1) = k(2m)\), which is an even number.

So, the product of two consecutive integers \(k(k+1)\) is always an even integer. This means \(k(k+1)\) can be written in the form \(2m\) for some integer \(m\).

Substituting this back into the expression for the square of the odd integer:

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

\((2k+1)^2 = 8m + 1\)

Here, \(m\) is an integer. This shows that the square of any odd integer can be expressed in the form \(8m+1\). Using \(n\) instead of \(m\), we can say it is of the form \(8n+1\).

Let's test with some examples:

• Odd integer 3: \(3^2 = 9\). \(9 = 8 \times 1 + 1\). Here, \(n=1\).
• Odd integer 5: \(5^2 = 25\). \(25 = 8 \times 3 + 1\). Here, \(n=3\).
• Odd integer -7: \((-7)^2 = 49\). \(49 = 8 \times 6 + 1\). Here, \(n=6\).

The examples support the algebraic proof.

Therefore, Statement 2 is correct.

Conclusion

Based on the analysis of both statements, we find that Statement 1 ("Of the two consecutive integers, one is even") is correct, and Statement 2 ("Square of an odd integer is of the form 8n + 1") is also correct.

Thus, both statements are correct.

Revision Table: Properties of Integers

Additional Information: Integer Properties and Number Theory Basics

Understanding basic properties of integers is fundamental in number theory. Concepts like parity (even or odd) and divisibility rules are essential.

• Parity: Integers are classified by their remainder when divided by 2. Even numbers have a remainder of 0, and odd numbers have a remainder of 1.
• Consecutive Integers: Consecutive integers always alternate in parity (odd, even, odd, even...). This is why one of any two consecutive integers must be even.
• Product of Consecutive Integers: The product of any two consecutive integers is always even because one of them must be even, and the product of an even number with any integer is even.
• Forms of Odd Integers: While \(2k+1\) is the standard form for an odd integer, other forms like \(2k-1\) also represent odd integers. Using \(2k+1\) or \(2k-1\) leads to the same result for the square. Let's check with \(2k-1\): \((2k-1)^2 = 4k^2 - 4k + 1 = 4k(k-1) + 1\). Since \(k(k-1)\) is the product of two consecutive integers, it is even (\(2m\)). So, \(4k(k-1) + 1 = 4(2m) + 1 = 8m + 1\).
• Modular Arithmetic: The statement "Square of an odd integer is of the form \(8n+1\)" can also be expressed using modular arithmetic as \(( \text{odd integer} )^2 \equiv 1 \pmod{8}\). Odd integers can be written as \(4k+1\) or \(4k+3\).
  • If odd integer is \(4k+1\): \((4k+1)^2 = 16k^2 + 8k + 1 = 8(2k^2+k) + 1\), which is of the form \(8n+1\).
  • If odd integer is \(4k+3\): \((4k+3)^2 = 16k^2 + 24k + 9 = 16k^2 + 24k + 8 + 1 = 8(2k^2+3k+1) + 1\), which is also of the form \(8n+1\).
  This confirms the property from another perspective.