For a fixed positive integer n, show that the determinant below is divisible by n For a fixed positive integer n, if   
$D = \left|\begin{array}{ccc} 
n! & (n + 1)! & (n + 2)! \\
(n + 1)! & (n + 2)! & (n + 3)! \\ 
(n + 2)! & (n + 3)! & (n + 4)!
\end{array} \right|$    
show that $\left(\dfrac{D}{(n!)^{3}} - 4 \right)$ is divisible by $n$.
Any ideas on how to go about solving this??
Thank You in advance.
 A: $$\begin{eqnarray*}
D &=&%
\begin{vmatrix}
n! & (n+1)! & (n+2)! \\ 
(n+1)! & (n+2)! & (n+3)! \\ 
(n+2)! & (n+3)! & (n+4)!%
\end{vmatrix}
\\
&=&%
\begin{vmatrix}
n! & (n+1)n! & (n+2)(n+1)n! \\ 
(n+1)n! & (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! \\ 
(n+2)(n+1)n! & (n+3)(n+2)(n+1)n! & (n+4)(n+3)(n+2)(n+1)n!%
\end{vmatrix}
\\
&=&n!^{3}(n+1)^{2}(n+2)\underset{\text{This determinant is 2 (see below)}}{\underbrace{%
\begin{vmatrix}
1 & (n+1) & (n+2)(n+1) \\ 
1 & (n+2) & (n+3)(n+2) \\ 
1 & (n+3) & (n+4)(n+3)%
\end{vmatrix}%
}} \\
&=&n!^{3}\left( 2n^{3}+8n^{2}+10n+4\right) 
\end{eqnarray*}$$
I have used the following property repeatedly: "If $B$ results from $A$ by multiplying one row or column with a number $c$, then $\det(B) = c \cdot \det(A)$". (Wikipedia). $n!$ is a factor of rows 1, 2 and 3. $n+1$ is a factor of rows 2 and 3. $n+2$ is a factor of row 3.
Thus
$$\frac{D}{n!^{3}}-4=2n^{3}+8n^{2}+10n=(2n^{2}+8n+10)n.$$

Evaluation of the last determinant expanding it by the minors of column 1.
$%
\begin{vmatrix}
1 & (n+1) & (n+2)(n+1) \\ 
1 & (n+2) & (n+3)(n+2) \\ 
1 & (n+3) & (n+4)(n+3)%
\end{vmatrix}%
$
$=%
\begin{vmatrix}
(n+2) & (n+3)(n+2) \\ 
(n+3) & (n+4)(n+3)%
\end{vmatrix}%
-%
\begin{vmatrix}
(n+1) & (n+2)(n+1) \\ 
(n+3) & (n+4)(n+3)%
\end{vmatrix}%
+%
\begin{vmatrix}
(n+1) & (n+2)(n+1) \\ 
(n+2) & (n+3)(n+2)%
\end{vmatrix}%
$
$=(n+2)(n+3)%
\begin{vmatrix}
1 & (n+3) \\ 
1 & (n+4)%
\end{vmatrix}%
-(n+1)(n+3)%
\begin{vmatrix}
1 & (n+2) \\ 
1 & (n+4)%
\end{vmatrix}%
$
$+(n+1)(n+2)%
\begin{vmatrix}
1 & (n+2) \\ 
1 & (n+3)%
\end{vmatrix}%
$
$=(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)$
$=\left( n^{2}+5n+6\right) -\left( 2n^{2}+8n+6\right) +\left(
n^{2}+3n+2\right) =2$
A: Define $\rm\ (n+k)_k\ :=\ (n+k)!/n!\: =\: (n+k)\ \cdots\ (n+1)\ \equiv\ k!\ \ (mod\ n)\:.\ $ Then
$$\rm \frac{D}{n!^3}\ \ =\ \ \begin{vmatrix}
1 & n+1 & (n+2)_2 \\ 
n+1 & (n+2)_2 & (n+3)_3 \\ 
(n+2)_2 & (n+3)_3 & (n+4)_4
\end{vmatrix}\ \ \equiv\ \ \begin{vmatrix}
1 & 1 & 2! \\ 
1 & 2! & 3! \\ 
2! & 3! & 4!%
\end{vmatrix}\ \ \equiv\ \ 4\ \ \ (mod\ n) $$
Note that $\rm\:det(a_{i\:j})\ \equiv\ det(a_{i\:j}\: mod\ n)\ \ (mod\ n)\ $ because $\rm\:det\:$ is a polynomial in the $\rm\:a_{i\:j}\:.$ Hence it's just a special case of $\rm\ f\:(g_1(n),\cdots,g_k(n))\ \equiv\ f\:(g_1(0),\cdots,g_k(0))\ \ (mod\ n)\ $ for polynomials $\rm\:f,\:g_i\:.$ Note how performing such modular reductions greatly simplifies the arithmetical calculations.
