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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?

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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 $4\mid n$.

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  • $\begingroup$ [+1] Very interesting proof $\endgroup$ – Jean Marie Feb 21 at 8:39

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