I know there are already several posts on how to prove the orthogonality of Latin squares that may be using a better proof, however there is a particular part in a proof I read that I could use some help to better understand. I will outline the proof below, and show what it is that I do not understand.
Let $p$ is a prime and $t$ and $u$ non-zero elements of $\mathbb{Z}_p$. We want to show that if $L_t(i,j) = ti+j$ and $L_u(i,j) = ui+j$, where $i,j\in\mathbb{Z}_p$, and $t\neq u$ then the squares $L_t$ and $L_u$ are orthogonal.
So we assume that there are two different positions $(i,j)$ and $(i',j')$ such that $L_t(i,j) = L_t(i',j')$ and $L_u(i,j) = L_u(i',j')$. Using the definitions of $L_t$ and $L_u$ we find that $t(i-i')=(j'-j)$ and $u(i-i')=(j'-j)$. If $i-i'=0$ then it follows that $j-j'=0$ which contradicts the assumption that $(i,j)$ and $(i',j')$ are different positions. Hence $i-i'\neq 0$, and $i-i'$ has an inverse in $\mathbb{Z}_p$. Solving the two equations for $t$ and $u$ we thus obtain $t=u=(i-i')^{-1}(j'-j)$. (The next step is what I do not fully understand.) So if we insist that $t\neq u$, then the symbols $L_t(i,j)$ and $L_u(i,j)$ can occur together in only one position.
What I am thinking is that if $t\neq u$ then we have again a contradiction. I am not sure how to formulate the contradiction, but something along the lines of "because something is not equal to itself"? And from this we can say that neither if $i-'i=0$ or $i-'i\neq 0$ is the hypothesis true, therefore it must be false (and it follows that every pair of symbols can occur in only one position).