Prove that there are no idempotent, commutative quasigroups of even order. I'm trying to prove this using a counting argument - Meaning of counting argument? 
I understand one can proved this for Latin Squares, as done in this post -Idempotent and commutative Latin squares of even order.
And I know a quasigroup is essentially just an unbounded Latin square, however I'm still stumped.
 A: EDIT: Actually Jens' answer in the link you provided on Latin squares shows a simpler proof. So all you need to do is realize that the multiplication table of an idempotent commutative quasigroup is a Latin square.

I'm not familiar with quasigroups, but using the definitions, here's my (perhaps cumbersome) approach:
Thinking about the multiplication table of a commutative idempotent quasigroup of order $n$, it is uniquely determined by the elements under the diagonal (or above, if you prefer). If $n$ is even, then $n-1$ can't be divided into two equal whole numbers, and in the $n-1$ columns under the diagonal there is at least one element appearing in over half of those columns (the number of elements below the diagonal is $n(n-1)/2$). By commutativity the same number of elements appearing below the diagonal also appears above, in the same number of columns.[Each time an element appears it must be in a different column, because of uniqueness of solutions]. But then there's an element appearing in at least $2 \times [(n-2)/2 +1] = n$ columns, in addition to the column containing the element on its diagonal - and we know that every column and every row should have a certain element only once.
Hopefully I'll think of a neater way to express this.
