I need to show that for $q$, a prime power not equal to 2 or 3, the polynomial $f(x,y) = \lambda x+(1-\lambda)y$ represents a self-orthogonal latin square of order $q$, where $\lambda \in F_{q}$ is any field element except $0, 1, 2^{-1}$.

(In case you're wondering :D, a self-orthogonal LS is an LS which is orthogonal to it's transpose if you consider them as matrices)

It's easy to show $f$ is an LS, so I'll cut to chase, I can't show it's orthogonality to it's transpose. If we denote the polynomial that represents the transpose by $f^{T} (x,y)$ then we must have $$f^{T} (x,y) = \lambda y+(1-\lambda)x$$ as a matter of fact I considered $f^{T}$ to be the same as $f$ with only the places of $x$ and $y$ interchanged.

So bearing this in mind, here's what I did to show the orthogonality:

$$ (f (x_1, y_1), f^T(x_1, y_1)) = (f(x_2, y_2), f^T (x_2, y_2))$$

$$\Rightarrow \lambda x_1 + (1-\lambda)y_1 = \lambda x_2 + (1-\lambda)y_2$$


$$\lambda y_1 + (1-\lambda)x_1 = \lambda y_2 + (1-\lambda)x_2$$

which imply that $$ x_1 + y_1 = x_2 + y_2$$

but this doesn't imply that the LS represented by $f$ is orthogonal to that represented by $f^T$. We need $x_1 = x_2, y_1 = y_2$ for that.

Something's wrong here. Any help will be appreciated.


Continuing from here: \begin{align*} \lambda x_1 + (1-\lambda)y_1 &= \lambda x_2 + (1-\lambda)y_2 \\ \lambda y_1 + (1-\lambda)x_1 &= \lambda y_2 + (1-\lambda)x_2. \end{align*} We rearrange these to obtain \begin{align*} x_1-x_2 &= \frac{-(1-\lambda)}{\lambda}(y_1-y_2) \tag{1}\\ y_1-y_2 &= \frac{-(1-\lambda)}{\lambda}(x_1-x_2) \tag{2} \end{align*} since $\lambda \neq 0$.

Substituting $(2)$ into $(1)$ gives $$x_1-x_2 = \frac{(1-\lambda)^2}{\lambda^2} (x_1-x_2).$$

From here we can use the field properties to show that if $x_1 \neq x_2$, then $\lambda=2^{-1}$. But since we assume $\lambda \neq 2^{-1}$, we must have $x_1=x_2$. Similarly, $y_1=y_2$.

  • $\begingroup$ That was brilliant, Dr. Stones. $\endgroup$ – Erfan Jul 6 '15 at 20:51

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