Infinite Latin Square?

Solution #1: Construct a number line by placing 1 somewhere arbitrary, then 2 to the rightmost empty position possible on the same horizontal line, then 3 to the leftmost empty position possible, then 4 to the rightmost position, then 5 to the leftmost position, and so forth. The "middle" will be ..., 7, 5, 3, 1, 2, 4, 6, ...

Now take this number line, and copy all terms shifting up and one position to the left. Repeat infinitely.

Going back to the starting line, do the same copying all terms and shifting down and one position to the right.

This will generate an infinite Latin Square in all directions.

Solution #2: Let's try a different method which specifies an upper-left corner, but still makes an infinite grid. To make it to our goal we'd like a method to turn a finite $n$ by $n$ Latin Square filled with integers from $1$ to $n$ into a finite $2n$ by $2n$ Latin square filled with integers from $1$ to $2n$ such that any integer placed will not be moved on the iteration ; then we can repeat this process infinitely to cover the entire grid.

(Note that a 1 by 1 array with only the digit 1 works as a base case.)

Suppose we have $A,$ a finite $n$ by $n$ Latin Square filled with digits from $1$ to $n.$

Generate $B,$ a finite $n$ by $n$ Latin Square with digits from $n+1$ to $2n;$ this can be done by taking all elements of $A$ and adding $n$ to them.

Now generate a new $2n$ by $2n$ grid with subgrids laid out as shown:

Note that 1) No integer is repeated on a row or column 2) All integers from $1$ to $2n$ are included 3) No previously-existing integer placements (in the upper left subgrid $A)$ changed position.

Therefore, this process is repeatable infinitely, and we can make an infinite Latin Square.

Solution #3: Left as an exercise: you can perform a mapping on Solution #1 to make it be on a grid with a fixed upper left corner.

Solution #4: You can also perform a mapping on Solution #2 to make it be on a grid that is infinite in all directions.

Relevant wiki: Cardinality

Let $f\,:\mathbb{N} \to \mathbb{Z}$ be a bijection, and let the number in the $i$ th row and $j$ th column be $x_{i,j} \; = \; f^{-1}(i-j) \hspace{2cm} i,j \in \mathbb{Z}$ This is Jason's Solution #1 for a general bijection $f$ . Jason has $f(n) \; = \; \left\{ \begin{array}{lll} 2n & \hspace{1cm} & n \ge 1 \\ 1-2n & & n \le 0 \end{array} \right.$

Index the rows and columns with the non-negative integers 0,1,2, 3, ... Let the (i,j) entry be 1 + (i xor j).

HA HA HA HA ....This solution doesn't require even a bit of combinatorics. HIPPOC Take a $n * n$ array and we have atmost n arguments to be filled up from 1 to n. Indeed, we have exactly n arguments. Write 1 to n sequentially in the 1st row. Push the first row by 1 unit cyclically such that it is now, $2,3,4,...,n,1$ . This is your second row. Indeed, while moving from rth row to (r+1)th row, follow the same rule. We get exactly n distinct arrangements(or what you call permutations ) of the first n natural numbers with no set replicating the other in any cells. It is a Symmetric Matrix. ALSO NOTE THAT WHEN WE FIX THE FIRST ROW, WE HAVE $n!$ PERMUTATIONS(ROW-WISE). BUT AS THE FIRST ROW IS FIXED, WE HAVE EXACTLY (n-1) CHOICES TO BE MADE IN EACH COLUMN AND THAT TOO IS MADE DISTINCT SO THAT THE NUMBERS WONT REPEAT. OVERALL, A NUMBER 'G' CAN OCCUR IN A PARTICULAR CELL OF A ROW IN $n!$ WAYS, BUT IS CANCELLED BY ITS COLUMN REPETANT IN $(n-1)!$ WAYS. SO THE REQUIRE NUMBER OF WAYS IS $\frac{n!}{(n-1)!}=n$ WAYS.

ALSO NOTE THAT THE NUMBER OF CHOICES FOR EACH CELL DECREASES WHILE MOVING FROM TOP TO BOTTOM OR ROW WISE FROM LEFT TO RIGHT. THIS LEADS TO THE NUMBER OF PERMUTATIONS

I got it right on the first attempt and within 3 sec i got it right and I did not ask anyone but myself and I think I'm a genius

f(i, j) = smallest number not equal to f(i-k, j) or f(i, j-k) for any positive k.

Or in English, working outwards, always choose the smallest number that isn't a duplicate in the row or column. You're guaranteed not to miss any numbers because for any row R and number N, N is guaranteed to be within the first R+N-1 values, because at most R-1 times it would have been in another row and at most N-1 times there would have been a smaller eligible number.