For a fixed positive integer n, show that the determinant below is divisible by n

Question

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.

Answer 1

$$\begin{eqnarray*} D &=&% \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)!% \end{vmatrix} \\ &=&% \begin{vmatrix} n! & (n+1)n! & (n+2)(n+1)n! \\ (n+1)n! & (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! \\ (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! & (n+4)(n+3)(n+2)(n+1)n!% \end{vmatrix} \\ &=&n!^{3}(n+1)^{2}(n+2)\underset{\text{This determinant is 2 (see below)}}{\underbrace{% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \\ 1 & (n+2) & (n+3)(n+2) \\ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% }} \\ &=&n!^{3}\left( 2n^{3}+8n^{2}+10n+4\right) \end{eqnarray*}$$

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

$$\frac{D}{n!^{3}}-4=2n^{3}+8n^{2}+10n=(2n^{2}+8n+10)n.$$

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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}% $

$=(n+2)(n+3)-2(n+1)(n+3)+(n+1)(n+2)$

$=\left( n^{2}+5n+6\right) -\left( 2n^{2}+8n+6\right) +\left( n^{2}+3n+2\right) =2$

Answer 2

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.