Determinant of the matrix with $a_{i,j} = (i+j)^2$ Determinant of the matrix with $ a_{i,j} = (i+j)^2 $
I was trying to solve, but it is impossible
 A: When your matrix is big enough it gets easier.
\begin{bmatrix} 2^2&3^2\cdots &(n+1)^2\\
\vdots & \ddots &\vdots\\
(n+1)^2&&(2n)^2\end{bmatrix}
Now lets do some row opperations.
Subtract the row above from every row.
Leave the top row as it is
\begin{bmatrix} 2^2&3^2\cdots &(n+1)^2\\
5&7&\cdots 2n+3\\
\vdots & \ddots &\vdots\\
2n+1&&4n-1\end{bmatrix}
and do it again
\begin{bmatrix} 2^2&3^2\cdots &(n+1)^2\\
1&-2\cdots &2-n^2\\
2&2\cdots& 2\\
\vdots & \ddots &\vdots\\
2&&2\end{bmatrix}
A singular matrix when $n\ge 4$
A: i will look at the case of $3 \times 3$ determinant
$f_3(x) = det \pmatrix{x^2&(x+1)^2&(x+2)^2\cr
                    3^2 & 4^2 & 5^2\cr
                    4^2 & 5^2 & 6^2}$  
clearly $f$ is a quadrtaic in $x$ and vanishes at $x = 3$ and at $x = 4.$ therefore $f_3(x) = k(x-3)(x-4).$  you can now evaluate $k$ by setting a convenient value for $x,$  for example $x = -1.$ 
using the same idea we can show that for $n > 3, f_n = 0.$  this follows from the fact that $f_n(x)$ is quadratic in $x$ and that $f_n(3)=0, f_n(4)=0, f_n(5)= 0. \ldots.$
A: Another way of showing that the matrix is singular for $n\ge 4$:
$$ A {\bf b} = 0 \iff \sum_{j} b_j (i+j)^2=0 \, (\forall i) \tag{1}$$
Equivalently
$$ \sum b_j j^2 + 2 i \sum b_j j+  n i^2 \sum  b_j=0  \tag{2}$$
We can always find a non trivial solution by setting $ \sum b_j j^2 =\sum b_j j=  \sum  b_j=0$, this corresponds to solving
$$ C{\bf b} =0\tag{3}$$  with $$C=\begin{pmatrix} 
1 & 1 &  1 &\cdots &1 \\
1 & 2 & 3 &\cdots &n \\
1 & 4 & 9 &\cdots &n^2 \\
\end{pmatrix} $$
As long as $n\ge  4$, $C$ is not full rank, hence we have a non-trivial solution for $(3)$, which is also a non-trivial solution for $(1)$. Hence $A$ is singular and $det(A)=0$.
Generalizations are straightforward. For example, if we had,say, $a_{i,j}=(5 i-4j)^3$ we can deduce that the determinant is zero for $n\ge 5$
