Putting the chessboard on the plane coordinates, where the one of the corner is $(1,1)$. Main idea is to prove that the set $S = \{(1,k), (2,2k), \ldots, (p-1, (p-1)k)\}$ is tactical for $k \in\{2,3,\ldots, p-2\}$. (Here everything is modulo $p$) We need to prove that for each pair of squares $(a,b), (c,d) \in S$ satisfy: $a \not\equiv c$, $b \not\equiv d$ and $|a-c| \not\equiv |b-d| \pmod p$. It easy to see that $S$ satisfy the conditions Considering the family of these sets for $k=2,3,\ldots, p -2$ we get want we wanted. We need only to prove that in the family there are no repeats and no main diagonals squares. Which is pretty easy to prove, $(a,ak)$ and $(b,bl)$ are equal if $a = b$ and $k = l$ and if $(a,ak)$ is equal to $(x,x)$ or $(x,-x)$ then we get that $k = 1$ or $k= p -1$, which also contradicts the definition of $k$.