I understand what the author is doing in Theorem 1.3.1 intercalate proof, however I don't see how this proof relies on the fact that $n$ needs to be odd. It seems to me the same logic holds for any $n$ (though, I do know there are latin square defined by $L_{i,j} = i + j\mod{n}$ with intercalates when $n$ is even). Hope this is clear, thanks in advance.
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It seems to me that they desire to partition the $n$ rows into $2$ (approximately) equal-size sets. However, you can't do that when $n$ is odd; instead, you can divide $n-1$ rows into two equal sized sets. But I may be way off... – apnorton Jan 28 '15 at 21:42
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Their proof is terribly written, here is a better version which shows why $n$ needs to be odd.
With $L$ as they defined it, an intercalate would consist of $i\neq \ell$ and $j\neq k$ so that $L_{i,j}=L_{\ell,k}$ and $L_{i,k}=L_{\ell,j}$, so that $$ i+j=\ell +k $$ $$ i+k=\ell+j $$ Subtracting these, we get $$ j-k=k-j=-(j-k) $$ So, $j-k$ solves the equation $x=-x$. This equation has only one solution when $n$ is odd, namely, $x=0$ (can you see why? check it for $n=3$, $n=5$ if not). This is a contradiction, since $j-k=0$ contradicts $j\neq k$.
The reason this proof doesn't work when $n$ is even is because $x=-x$ has two solutions in that case. For example, , both $0=-0$ and $2=-2$ (mod $4$).
Mike Earnest
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