Permutation of set

Question

Suppose I have a set which contain finite elements from $1,2,\dots,n$, where $n$ is odd. Let there are two permutations $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,b_3,\dots,b_n)$ defined on this set. Now I have to prove that $$(a_1-b_1)^2(a_2-b_2)^2(a_3-b_3)^2\cdots(a_n-b_n)^2$$ is a even number.

My findings/questions are as follows :-

1. Every permutation of a set is either a cycle or product of disjoint cycles, so these to permutations are either be cycle or product of disjoint cycles.
2. Can the product of square terms can be a cycle of the set?
3. Or the given product of square numbers can be converted onto any one of permutations, so that can claim to even or odd.

Please clear it will be helpful for me

Answer 1

Since $n$ is odd, one of the factors $a_i-b_i$ is even, where $a_i,b_i$ are odd. Then the whole product is even.

Answer 2

A product of numbers is even if and only if at least on of the numbers is even, so it suffices to show that $(a_i-b_i)^2$ is even for some $i$. The square of a number is even if and only if the number you start with is even, so in fact it suffices to show that $a_i-b_i$ is even for some $i$.

Now $a_i-b_i$ is even if and only if either both $a_i$ and $b_i$ are even, or they are both odd. Since $n$ is odd, there are more odd numbers than even numbers in $\{1,\dots,n\}$ so some pair of odd numbers must be matched up by the permutations by the pigeon hole principle. This proves the claim.