I. \(p\) is always odd
II. \(p\) is a perfect square
Which of the statements given above is/are correct?
Analyzing the expression \(p = n(n + 1)(n+2)(n + 3) + 1\)
We are given an expression \(p = n(n + 1)(n+2)(n + 3) + 1\), where \(n\) is a natural number. We need to determine if the statements "\(p\) is always odd" and "\(p\) is a perfect square" are correct.
Proving Statement I: \(p\) is always odd
Let's examine the term \(n(n + 1)(n+2)(n + 3)\). This represents the product of four consecutive natural numbers.
- Among any four consecutive integers, there must be at least one (and in fact, two) even numbers.
- The product of any integer and an even number results in an even number.
- Therefore, the product \(n(n + 1)(n+2)(n + 3)\) is always an even number.
Now consider the full expression for \(p\): \( p = \underbrace{n(n + 1)(n+2)(n + 3)}_{\text{Even}} + 1 \)
Adding 1 to any even number always results in an odd number.
Thus, \(p\) is always an odd number. Statement I is correct.
Proving Statement II: \(p\) is a perfect square
Let's try to simplify the expression algebraically. We can rearrange the terms to group them strategically:
\( p = [n(n + 3)][(n+1)(n+2)] + 1 \)
Expand the grouped terms:
- \(n(n + 3) = n^2 + 3n\)
- \((n+1)(n+2) = n^2 + 2n + n + 2 = n^2 + 3n + 2\)
Substitute these back into the expression for \(p\):
\( p = (n^2 + 3n)(n^2 + 3n + 2) + 1 \)
To make this simpler, let's substitute \(x = n^2 + 3n\). The expression becomes:
\( p = x(x + 2) + 1 \)
Now, expand this:
\( p = x^2 + 2x + 1 \)
This is a perfect square trinomial, which factors as:
\( p = (x + 1)^2 \)
Now substitute \(x = n^2 + 3n\) back:
\( p = (n^2 + 3n + 1)^2 \)
Since \(n\) is a natural number, \(n^2\), \(3n\), and \(1\) are integers. Therefore, \(n^2 + 3n + 1\) is also an integer. Because \(n \ge 1\), \(n^2 + 3n + 1\) will always be a positive integer.
This shows that \(p\) is always the square of the integer \((n^2 + 3n + 1)\). Thus, \(p\) is always a perfect square. Statement II is correct.
Conclusion
Both statements, "\(p\) is always odd" and "\(p\) is a perfect square", have been proven to be correct.
Therefore, the correct option is the one stating that Both I and II are correct.
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