Proof that Hadamard matrices of order $4k+2$ don't exist It's known that Hadamard matrices can only exist for orders $1$, $2$ and $4k$. It's easy to show that there are no Hadamard matrices of order $2k+1$. But what is the proof that there are no Hadamard matrices of order $4k+2$?
 A: Assume the Hadamard matrix has $\ge3$ rows.
Consider the top row. You may as well assume it is all ones. (otherwise change the signs
of various columns). Then row two and row three each
consist of $n/2$ ones and $n/2$ minus ones.
So $n$ is even. As row $2$ and row $3$ are orthogonal, then they agree in $n/2$
entries. So if row $2$ and row $3$ both have ones in $k$ columns, then in $n/2-k$
columns, row $2$ has a one and row $3$ a $-1$ and so in $k$ columns, row $2$
and row $3$ both have $-1$s. So they agree in $2k$ entries: $2k=n/2$ and $n$
is a multiple of $4$.
A: The following is taken from this wikipedia page:

Any two rows of an $n\times n$ Hadamard matrix are orthogonal. For a $\{1, −1\}$ matrix, it means any two rows differ in exactly half of the entries, which is impossible when $n$ is an odd number. When $n \equiv 2\pmod 4$, two rows that are both orthogonal to a third row cannot be orthogonal to each other. Together, these statements imply that an $n \times n$ Hadamard matrix can exist only if $n = 1, 2$, or a multiple of $4$.

