Im looking for an explanation of the following: a standard way to prove that, if there exists Hadamard matrix of dimension $n > 2$, then $4|n$, is to suppose that without loss of generality every column of the matrix starts with +. (Otherwise, one can multiply the column by -1, which doesn't change the Hadamard property).
Then there are only 4 possibilities of how the first three entries of each column look like: +++, ++-, +-+ and +--. Let's say there's $a$ columns of the first type, $b$ of the second, $c$ and $d$ of third and fourth respectively.
Obviously, since the matrix is $n\times n$, we obtain $a+b+c+d=n$.
But here comes the point of confusion: the final step says that because of the orthogonality relations we also obtain 3 more relations:
$$ a + b - c - d = 0 $$
$$ a + c - b - d = 0 $$
$$ a - b - c + d = 0 $$
which all put together yields $4a=n$. Im not sure, how can "number of columns of some type" be mixed with the fact that every two columns are orthogonal? How do we obtain these 3 relations? This might be a stupid question, but I can't really see something that might be obvious.
(How come the orthogonal property of the columns can yield something like "number of type 1 columns + number of type 2 columns - number of type 3 columns - number of type 4 columns" = 0 ??)
Hint: A Hadamard matrix $H$ has the property $HH^T=nI_n$. This implies that $H^T$ is Hadamard whenever $H$ is. Therefore if the columns all are orthogonal to each other, then so are the rows. Guess what you get with the inner products of the first three rows.