Rank/determinant of the $n\times n$ matrix $((a_{ij}))$ where $a_{ij}=(i+j-1)^2$ I am trying to find the rank and determinant of the following $n\times n$ matrix : 
$$A=\begin{bmatrix}1^2&2^2&3^2&\cdots&n^2
   \\ 2^2&3^2&4^2&\cdots&(n+1)^2
   \\\vdots&\vdots&\vdots&\ddots&\vdots\\n^2&(n+1)^2&(n+2)^2&\cdots&(2n-1)^2
   \end{bmatrix}$$
I verified that the determinant of $A$ vanishes for $n> 3$, implying that $A$ has full rank for $n\le 3$. And the rank seems to remain $3$ for $n>3$. But I could not provide a rigorous proof.
Is there a way to deduce the rank/determinant without reducing $A$ to echelon form or any other shortcut? Or how can I easily find the number of linearly independent rows/columns? Any hints would be great.
Related question: Determinant of the matrix with $a_{i,j} = (i+j)^2$.
 A: Note that $(k+1)^2 - k^2 = 2k+1$.  Thus, we have
$$
((m+1)^2,(m+2)^2,\dots,(m+n+1)^2) = \\
(m^2,(m+1)^2,\dots,(m+n)^2) + (2m+1,2m+3,\dots,2m+2n+1)
$$
Conclude (inductively, if you like) that the vectors
$$
(1,2,\dots,n^2),(1,3,\dots,2n+1),(1,\dots,1)
$$
span the row/column space of $A$.  Or, if you prefer, you could use the basis
$$
(1,\dots,1),(0,1,\dots,n-1), (0,1^2,\dots,(n-1)^2)
$$

More generally: (I think that) as a consequence of the Newton interpolation formula, the rank of the matrix
$$
\pmatrix{f(1)& f(2) & \cdots & f(n)\\
f(2) & f(3) & \cdots & f(n+1)\\
\vdots & \vdots & \ddots & \vdots\\
f(n) & f(n+1) & \cdots & f(2n+1)}
$$
will have rank equal to at most $1 + \deg(f)$ (exactly $1 + \deg(f)$, whenever $n \geq 1 + \deg(f)$) for any polynomial $f$.
A: The fact that the upper left $3 \times 3$ submatrix of $A$ has nonzero determinant implies that the first three columns of $A$ are linearly independent, implying that the rank of $A$ is at least 3.
On the other hand, the column space of $A$ is contained in the image of the linear map which takes a quadratic polynomial $p \in P_2(\mathbb{R})$ to $(p(1), p(2), \ldots, p(n))$.  Since $P_2(\mathbb{R})$ has dimension 3, this implies that the rank of $A$ is at most 3.
