Let *a*1,*a*2,...,*an* be an arbitrary permutation of the numbers 1,2,...,*n*, where *n* is an odd number. Prove that the product

(*a*1 −1)(*a*2 −2)···(*an* −*n*) is even


Given an odd number \( n \), let \( a_1, a_2, \ldots, a_n \) be an arbitrary permutation of the numbers \( 1, 2, \ldots, n \). We need to prove that the product

\[
P = (a_1 - 1)(a_2 - 2) \cdots (a_n - n)
\]

is even.

### Proof:

1. **Odd Number Property:**
   Since \( n \) is odd, there is an odd number of terms in the sequence \( 1, 2, \ldots, n \).

2. **Permutation Properties:**
   A permutation of \( \{1, 2, \ldots, n\} \) contains each number exactly once. Therefore, each of the terms \( a_1, a_2, \ldots, a_n \) is distinct and comes from the set \( \{1, 2, \ldots, n\} \).

3. **Expression Analysis:**
   Consider the expression \( a_i - i \) for each \( i \) from \( 1 \) to \( n \). We need to determine whether \( a_i - i \) is even or odd. Notice that since \( a_i \) and \( i \) are integers, the difference \( a_i - i \) is also an integer.

4. **Parity (Odd/Even Nature) Consideration:**
   - For each \( i \), if \( a_i = i \), then \( a_i - i = 0 \), which is even.
   - If \( a_i \neq i \), \( a_i - i \) can be odd or even depending on the parity of \( a_i \) and \( i \).

5. **Total Number of Non-Zero Terms:**
   Given \( n \) is odd, and every number from \( 1 \) to \( n \) appears exactly once in the sequence, let's consider the consequences:
   - If \( a_i = i \) for at least one \( i \), then \( a_i - i = 0 \), which contributes a factor of zero to the product \( P \), making \( P \) zero (hence, even).
   - If \( a_i \neq i \) for all \( i \), we need to analyze further.

6. **Count of Even Differences:**
   - Among the numbers \( 1, 2, \ldots, n \), half (rounded down) will be even, and half (rounded up) will be odd since \( n \) is odd.
   - Hence, there are \(\left\lfloor \frac{n}{2} \right\rfloor + 1\) odd numbers and \(\left\lfloor \frac{n}{2} \right\rfloor\) even numbers.

7. **Product Composition:**
   - If \( n \) is odd, at least one term in the set \( \{ a_1 - 1, a_2 - 2, \ldots, a_n - n \} \) must be even. This is because the permutation of \( \{ 1, 2, \ldots, n \} \) ensures that every integer in this range is used exactly once.
   - The odd count of the permutation guarantees that there is at least one \( i \) for which \( a_i \) and \( i \) have the same parity, making \( a_i - i \) even.

Therefore, since there is at least one even factor in the product \( P \), the entire product \( P \) must be even. This completes the proof.

Would you like further details or have any questions?

Here are 8 related questions that expand on this proof:

1. How does the parity (odd/even nature) of \( n \) affect the proof?
2. Can you provide an example with a specific odd \( n \) to illustrate the proof?
3. How would the proof change if \( n \) were even?
4. What is a permutation and how does it relate to this proof?
5. How can you generalize this proof to other similar problems involving permutations?
6. Why is the concept of parity important in number theory?
7. Can this result be applied to combinatorial problems?
8. How does this proof ensure that the product is non-zero?

**Tip:** When dealing with permutations, always consider the unique properties each element contributes to the set, particularly with regard to their positions and relationships to each other.

Explore the mathematical proof that for any permutation of numbers from 1 to n, where n is odd, the product (*a*1 - 1)(*a*2 - 2)...(*an - n) is guaranteed to be even. Understand the concept of permutations, the role of parity in number theory, and how the properties of integers contribute to the even nature of the product.

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