I was wondering a problem: Is it possible to construct a $$$n \times n$$$ matrix (call it A) such that each number from $$$1$$$ to $$$n$$$ appears exactly $$$n$$$ times and the result of $$$A \times A$$$ (matrix multiplication) is a zero matrix (after modulo $$$n \times n$$$ terms in the matrix by $$$n$$$). I know it is obvious for odd $$$n$$$ but how about even $$$n$$$?
Upd: After roughly 2 weeks the solution is done by Phd Ben Grossmann and Litho on math stackexchange. You can find the solution and proof here. Thanks div4only for providing me with great support. I will state the construction here only.
Satisfied matrix construction
I do not know if this contributes much to competitive programming, but it is fun doing these types of math problems.
$$$A_{i, j} \neq 0 \rightarrow \forall 0 \leq k \leq n-1, A_{j, k} = 0$$$. However, there are only $$$n$$$ zeros and $$$n^2-n$$$ non-zeros. So for $$$n \geq 3$$$ there exists two distinct $$$j_1,\,j_2$$$ such that there are non-zeros on the $$$j_1$$$-th column and $$$j_2$$$-th column, respectively. So there have to contain $$$2$$$ zero rows of $$$A$$$, which is impossible.
Seems easy? Do I miss something? Is $$$A*A$$$ matrix multiplication or point-wise multiplication? You should make it clear.
I forgot to mention that the n*n terms in the final matrix should be modulo n. Sorry for missing information.
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