Let $A={a_1,a_2...,a_{2k+1}}$ where $k\geq 1$ ,be a set of $2k+1$ positive integers. \ show:that : for :any:premutation $a_{i1},a_{i2},a_{i2k+1}$ of A, the product $\prod_{j=1}^{2k+1}(a_i-a_j)$ is allways even:
My solution: i divide the problem to 2 boxes box1 is even numbers and box2 is odd numbers second,define the pigeon to be the numbers and the holes to be the boxes third,denote that ${a_{i1},a_{i2},...a_{i2k+1}}\bigcap {a_1,a_2...a_{2k+1}}\neq\emptyset$ beacuse if it was then we had 2k numbers in 2 boxes and this is not true. so we suppose $a_t=a_{ik}$ and therfore:$a_3-a_{ik}=a_3-a_t$ ,therefore,the factor $(a_t-a{ik})$ is even , and therefore we proved the existence of product of some 2 numbers is even.
is it valid?
I would try to prove by contradiction. Assume there exists some permutation such that the product is odd. In order for the product to be odd, every term of the multiplication must be odd. Terms of the product are all differences of terms of the original sequence. So, we can look at terms of the original sequence to try to determine its parity.
$a_i-a_j$ is even if $a_i$ and $a_j$ are either both odd or both even. The difference is odd if and only if the two have different parity. In order to construct a product that is odd, you must permute the numbers so that every even number is permuted to the position of an odd number in the original sequence and every odd number is permuted to the position of an even number in the original sequence.
Suppose you have $m$ even numbers and $n$ odd numbers such that $m+n=2k+1$. Then, the permutation yielding an odd product requires the $m$ even numbers to be permuted to the positions of the $n$ odd numbers and vise versa. This is only possible if $m=n$, which contradicts the fact that $m+n$ is odd.