Eigenvector of magic square

If every row in a matrix $A$ sums to $k$ then $k$ is an eigenvalue with eigenvector $v=[1,1,\ldots,1]^T$. Indeed, all the entries of the vector $Av$ are equal to $k$ trivially, so $Av=kv$.

$\begin{bmatrix}1&1&1&\dots&1\end{bmatrix}^T$ is an eigenvector since the sum of all rows must be the same. The sum of all the elements of an $n\times n$ square must be $\frac{n^2(n^2+1)}{2}$. Dividing this among the $n$ rows yields that each row must sum to $\frac{n(n^2+1)}{2}$, Thus, $$ M\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix}=\frac{n(n^2+1)}{2}\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix} $$ This is essentially SL2's idea, but I just filled in a few points.