There exists exactly $p-1$ mututally latin squares for $p=q^d$ where $q$ is a prime number. A Latin square is an $n × n$ array filled with $n$ different symbols, each occurring exactly once in each row and exactly once in each column.
A pair of latin squares $A=(a_{ij})$ and $B=(b_{ij})$ are orthogonal iff the ordered pairs $(a_{ij},b_{ij})$ are distinct for all $i$ and $j$.
A set $\{L_1,\dots,L_t\}$ of $t \ge 2$ latin squares of order $n$ is orthogonal if any two distinct latin squares are orthogonal.  We call this a set of mututally orthogonal latin squares (MOLS).
Now Prove that :
There exists exactly $p-1$ mututally latin squares for $p=q^d$ where $q$ is a prime number.
Note: I found the proof on these links but i wan't a clean proof so that i can understand.
First proof is on page 45 of this link
Second proof is on page 11 of this link(similar to the first proof)
My first problem is that i don't know much about fields. In the first proof, it uses some of notations like $GF(q)$ or $PG(2,p)$ which i have no idea what they mean !!!
Second proof seems to be more readable but i think it's just a hint and is not complete.
Thank you in advance.
 A: Short answer: for the simple case of prime latin squares, you don't need fields, you only need to know that unique inverses exist which only happens for prime moduli.

This is not a rigorous answer but it should help some.
Lets look at $5$ first. We claim that there are $4$ MOLS. The first one is formed by making a $5\times 5$ addition table. Suppose two elements in the same row or column are the same. Then $a+b=a+c$, so $b=c$. This shows we have a latin square.
We can construct the other $3$ with a simple algorithm: add $0$ to the first row, $1$ to the second, $2$ to the third, $3$ to the fourth, and $4$ to the fifth (all mod $5$). We need to check that these are orthogonal. Thus we need to show that each pairing (x,y) shows up only once. Look at $(1,1)$ for example. One is in the top left corner of the first and second square. One is also in the second position of the second row of the first square, but in the second we have shifted it over by $1$. The same happens with every other row.
Why does this only work for $q^d$?: Because you will create something that is not a latin square if $q$ is not prime. You need to know a little bit about modular arithmetic to show this. In the $i$th row and $j$th column of our new latin square, we have the number $2i+j$. A problem will occur if we have the same number appearing twice in a column. If this happens, we have $2i+j=2i'+j$, or $2i=2i'$. However, since this is modular arithmetic, we can only divide by $2$ is $2$ is relatively prime to the modulus. In general, this algorithm only works if each number has a unique inverse mod $q$, which of course only happens with $q$ is prime.
Check out this website.
