Find an $n\times n$ integer matrix with determinant 1 and $n$ distinct eigenvalues Pretty much what the title suggests: for any positive integer $n$, I'm looking for an $n$-by-$n$ matrix with integer entries, determinant $1$ and $n$ eigenvalues.
In case it is absolutely useless to come up with such a matrix, I'm looking for a proof that such a matrix exists.

Take any  representation  of degree $n$ of  any  symmetric group $S_m$. All the matrix entries will be integers but determinant could be $\pm 1$. As suggested by others, we can change all the signs in the first row, if needed, and get integer matrices of determinant $+1$. As all matrices are of finite order, they will be diagonalizable and hence have $n$ eigenvalues.
