Prove that $(a_1-b_1)^2(a_2-b_2)^2\cdots (a_n-b_n)^2$ is an even number. Let $a_1,a_2,\cdots ,a_n$ and $b_1,b_2,\cdots ,b_n$ be two permutations of $1,2,\cdots,n$ and $n$ is odd. Prove that 
$$(a_1-b_1)^2(a_2-b_2)^2\cdots (a_n-b_n)^2$$ 
is an even number.
I don't know where to start to solve this problem. Please help me. 

For the product to be even we require one pair $(a_i, b_i)$ to have the same parity. Since $n$ is odd, there are $\frac{n-1}{2}$ even numbers and $\frac{n+1}{2}$ odd numbers in $\{1, 2, \dots, n\}$, so it is impossible to pair each even number with an odd number, so one pair must have the same parity and we are done.
