Technique to calculate rank of this matrix. 1) How can we calculate the rank of this matrix for $n > 3$ ?
\begin{matrix}
 1^2 & 2^2 & 3^2  & \cdots  & n^2  \\ 
 2^2& 3^2  & 4^2  &\cdots  & (n+1)^2 \\ 
 3^2& 4^2 & 5^2  &\cdots  & (n+2)^2 \\ 
 \vdots&\vdots  &\vdots  &\ddots   & \vdots\\ 
 n^2& (n+1)^2  & (n+2)^2  &\cdots  &(2n-1)^2 
\end{matrix}

2) What are some of the tricks one must learn to calculate rank of a matrix ? Can anyone summarize the important ones ?
 A: We claim that the determinant of matrix $M$ given by  $M_{i,j}=(a_i+b_j)^m$ where $i,j\in\mathbb{N}_n$ is $0$ if $n>m+1$. We have
$$
\begin{aligned}
\det(M)
&=\begin{vmatrix}
(a_1+b_1)^m & (a_1+b_2)^m & \ldots      & (a_1+b_n)^m\\
(a_2+b_1)^m & (a_2+b_2)^m & \ldots      & (a_2+b_n)^m\\
\ldots      & \ldots      & \ldots      & \ldots     \\
(a_n+b_1)^m & (a_n+b_2)^m & \ldots      & (a_n+b_n)^m\\
\end{vmatrix}\\
&=\begin{vmatrix}
\sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_1^{m-k_1} & \sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_2^{m-k_1} & \ldots & \sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_n^{m-k_1}\\
\sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_1^{m-k_2} & \sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_2^{m-k_2} & \ldots & \sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_n^{m-k_2}\\
\ldots & \ldots & \ldots & \ldots\\
\sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_1^{m-k_n} & \sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_2^{m-k_n} & \ldots & \sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_n^{m-k_n}\\
\end{vmatrix}\\
&=\sum\limits_{k_1=0}^m\sum\limits_{k_2=0}^m\ldots\sum\limits_{k_n=0}^m\binom{m}{k_1}\binom{m}{k_2}\ldots\binom{m}{k_n}
\begin{vmatrix}
a_1^{k_1} b_1^{m-k_1} & a_1^{k_1} b_2^{m-k_1} & \ldots & a_1^{k_1} b_n^{m-k_1} \\
a_2^{k_2} b_1^{m-k_2} & a_2^{k_2} b_2^{m-k_2} & \ldots & a_2^{k_2} b_n^{m-k_2} \\
\ldots                & \ldots                & \ldots & \ldots                \\
a_n^{k_n} b_1^{m-k_n} & a_n^{k_n} b_2^{m-k_n} & \ldots & a_n^{k_n} b_n^{m-k_n} \\
\end{vmatrix}\\
&=\sum\limits_{k_1=0}^m\sum\limits_{k_2=0}^m\ldots\sum\limits_{k_n=0}^m\binom{m}{k_1}\binom{m}{k_2}\ldots\binom{m}{k_n}a_1^{k_1}a_2^{k_2}\ldots a_n^{k_n} b_1^m b_2^m\ldots b_n^m
\begin{vmatrix}
b_1^{-k_1} & b_2^{-k_1} & \ldots & b_n^{-k_1} \\
b_1^{-k_2} & b_2^{-k_2} & \ldots & b_n^{-k_2} \\
\ldots                & \ldots                & \ldots & \ldots                \\
b_1^{-k_n} & b_2^{-k_n} & \ldots & b_n^{-k_n} \\
\end{vmatrix}\\
\end{aligned}
$$
Denote
$$
M_{k_1,k_2,\ldots,k_n}=
\begin{vmatrix}
b_1^{-k_1} & b_2^{-k_1} & \ldots & b_n^{-k_1} \\
b_1^{-k_2} & b_2^{-k_2} & \ldots & b_n^{-k_2} \\
\ldots                & \ldots                & \ldots & \ldots                \\
b_1^{-k_n} & b_2^{-k_n} & \ldots & b_n^{-k_n} \\
\end{vmatrix}
$$
Note that if $k_i=k_j$ for some $i,j\in\mathbb{N}_n$, then the determinant $M_{k_1,k_2,\ldots,k_n}$ have to equal columns and therefore equals zero. Now note that if $n>m+1$ there alway exist $i,j\in\mathbb{N}_n$ such that $k_i=k_j$, so $M_{k_1,k_2,\ldots,k_n}=0$ for all tuples $(k_1,k_2,\ldots,k_n)\in \{0,\ldots,m\}^n$ and as the consequence $\det(M)=0$. 
Now we return to the original problem. For $n=1,2,3$ we have the following matrices
$$
A=(1)\qquad 
A=\begin{pmatrix}1&4\\4&1\end{pmatrix}\qquad 
A=\begin{pmatrix}1&4&9\\4&9&16\\6&16&25 \end{pmatrix}
$$
respectively. Their ranks are $1$, $2$ and $3$ respectively. If $n>3$ by previous claim all minors of $A$ of size greater than $3$ are zero, hence $\operatorname{rank}(A)\leq 3$. Since $n>3$, then $A$ contains minor 
$$
\begin{pmatrix}1&4&9\\4&9&16\\6&16&25 \end{pmatrix}
$$
whose rank equal to $3$, so $\operatorname{rank}(A)=3$ for $n>3$. Finally
$$
\operatorname{rank}(A)=
\begin{cases}
1 & \quad\mbox{if}\quad n=1\\
2 & \quad\mbox{if}\quad n=2\\
3 & \quad\mbox{if}\quad n>2\\
\end{cases}
$$
