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.
 A: When $n$ is odd use the $n \times n$ matrix for the cyclic permutation $(a_1,a_2,...,a_n) \rightarrow (a_n,a_1,...,a_{n-1})$..  Then $A^n=I$ and since $n$ is odd, $det(A)=1$ and the entries of $A$ are all $1$ or $0$.  The eigenvalues are the distinct $n$'th roots of unity (it has characteristic polynomial $x^n-1$).  If you allow determinant $-1$ this will work in the even case as well. 
By derpy's suggestion below we can do the odd and even case at once using the matrix for the map:
$(a_1,a_2,...,a_n) \rightarrow ((-1)^{n+1}a_n,a_1,...,a_{n-1})$.
Then when $n$ is odd we get the cyclic permutation, and the quasi-cyclic one for $n$ even. In each case $det(A)=1$ and the eigenvalues are distinct roots of $\pm 1$.
A: 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.
