Is there any proof that there doesn't exist a Hadamard matrix of size $6 \times 6$?

A matrix $$H \in {\pm 1}^n$$ is Hadamard matrix if $$HH^T=nI_n$$, where $$I$$ is $$n\times n$$ identity matrix. Hadamard's conjecture said that there exists Hadamard matrix of order 1,2 or $$4n$$, for every positive integer $$n$$.

My question is how to prove there does not exists a $$6 \times 6$$ Hadamard matrix?

Wlog (why?) the first row is $$(1\,1\,1\,1\,1\,1)$$ and all other rows have three $$+1$$ and three $$-1$$ entries. If the second and third row have $$r$$ entries $$+1$$ in common ($$0\le r\le 3$$), then they also have $$r$$ entries $$-1$$ in common and their scalar product is $$r\cdot 1^2+r\cdot (-1)^2+(6-2r)\cdot(-1)=4r-6\ne0$$, contradiction.
Upon closer look, this argument can be extended to show that $$H_n$$ can exist only when $$4\mid n$$ or $$n<3$$:
Assume $$n\ge 3$$. Again with the first row being wlog all-ones, every other row must have an equal amount of $$+1$$'s and $$-1$$'s, hence $$n$$ must be even. Then as above, if the second and third row have $$r$$ entries $$+1$$ in common, they also share $$r$$ entries $$-1$$, and their scalar product is $$4r-n$$. As this is $$0$$, it follows that $$4\mid n$$.