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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.

  • 2
    Direct calculation leads to $D/(n!)^3 - 4= 2 n [5+n(4+n)]$. So it is not only divisible by $n$ but also by $2$ and by $5+n(4+n)$. – Fabian Mar 04 '11 at 08:52
  • First you can get rid of $n!^3$ by taking it out of every row. Then you are left with a simple 3x3 matrix. Calculate the determinant subtract 4 and there you go. – Fabian Mar 04 '11 at 09:26
  • can u show some steps on how to do it. Thanks –  Mar 04 '11 at 10:11
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    I'm just curious, why does something like this get a downvote? – Myself Mar 04 '11 at 12:46

2 Answers2

7

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

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}% $

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

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

Américo Tavares
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  • @ Christopher: I just mean $% \begin{vmatrix} 1 & (n+1) & (n+2)(n+1) \ 1 & (n+2) & (n+3)(n+2) \ 1 & (n+3) & (n+4)(n+3)% \end{vmatrix}% =2$ – Américo Tavares Mar 04 '11 at 13:35
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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.

Bill Dubuque
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  • @Chris: If you tell me what is not clear to you I'll be happy to elaborate. – Bill Dubuque Mar 04 '11 at 22:00
  • @Chris: I've revised the answer (the prior 2nd step was unneeded - based on a misreading of the problem). – Bill Dubuque Mar 06 '11 at 15:38
  • +1 for your elegant method. Now I think I understand it. Just to confirm: $k!$ is the independant term of $(n+k)_k$. And you cannot simplify the calculations further, by reducing $\mod 4$ all entries. – Américo Tavares Mar 06 '11 at 16:08
  • @Americo: Yes, indeed the $\rm:det:$ entries $\rm\ g(n) = (n+k):\cdots:(n+1)$ are polynomials in $\rm:n:$ so $\rm:g(n)\equiv g(0)\equiv k!\ (mod\ n):,:$ or, equivalently $\rm\ n\ |\ g(n) - g(0)\ $ the so-called Factor Theorem. The extra $\rm:mod\ 4:$ step in my initial version was unneeded (applied only to a misreading of the problem). – Bill Dubuque Mar 06 '11 at 16:25
  • Thanks for the re-corrections. This is much clearer to follow through with. +1 also for the cleverness of the solving strategy. Thank You. –  Mar 10 '11 at 11:57