(n-14)可被(n^3-2)整除，求所有符合的n值

提示：整除：| n - 14 | = k．| n^3 - 2 |，k εN U { 0 }

1.在 k = 0 時
| n - 14 | = 0．| n^3 - 2 | = 0
n=14 ........................................................ n =14

2. 在 k > 0 時
(1). 當 n > 14 時
n – 14 = k．( n^3 - 2 )
Let f(x) = k ( x^3 – 2 ) - x + ( 14 – 2 k) ， xε( 14 , ∞ )
f ‘ (x) = 3k．x^2 – 1 > 0；因為 k > 0 => f 遞增
f(x) > f(14) = 0 ; 此區無 0 函數

(2). 14 > n > 3√2 時
14 - n = k ( n^3 - 2 )
k． n^3 + n - ( 14 + 2 k ) = 0

Let f(x) = k x^3 + x - ( 14 + 2 k )，xε(3√2 , 14 ) ; note: 3√2 是 2 的 1/3 次方

f ' ( x ) = 3k．x^2 + 1 > 0 ；因為 k > 0 => f 遞增
當 n=2 時 ; 2 = min { n > 3√2，nεN }
f ( 2 ) = 8 k + 2 - (14+2k) = 6 k - 12 = 6 ( k - 2 ) ; 可因數分解
令 f(2) = 0 , 得 k = 2 ......................................... n = 2

因為f 遞增，所以
當 14 > x > 2：f(x) > f(2) =0 ; 此區無 0 函數

(3) 3√2 > n > 0 時
此區只有一整數：1
| 1 - 14 | = 13
| 1^3 - 2 | = 1
=> k =13 / 1 = 13 ..................................................... n = 1

(4) n <= 0 時

14 – n = k．( 2 - n^3 )

Let f(x) = k ( 2 – x^3 ) + x -14
f ‘ (x) = -3 k．x^2 + 1
(a) 0 >= x > - 1 / √(3k) 時 , f ‘ > 0 ; 遞增
f(0) = 2k – 14 = 2 ( k – 7 ) ; 可因數分解
令 f(0) = 0 , 得 k = 7 ......................................... n = 0 ( 不符，因為 n 為 正整數 > = 1 )
所以
當0 > x > - 1 / √(3k)：f(x) > f(0) = 0 ; 此區無 0 函數

(b) x < - 1 / √(3k) 時 , f ‘( x) < 0 ; 遞減
max { n < - 1 / √(3k)，nεZ } = - 1
f ( - 1 ) = k ( 2 – (-1) ) – 1 – 14 = 3 k – 15 = 3 ( k – 5 ) ; 可因數分解
令 f(-1) = 0 , 得 k = 5 ......................................... n = -1 ( 不符，因為 n 為 正整數 >= 1 )
所以
當 x < -1： f(x) < f(0) = 0 ; 此區無 0 函數

由上述知
在 n = 1、2、14 時，n-14 可被(n^3-2)整除 #

note :
n= 1： n – 14 = - 13 ，n^3-2 = -1 => -13 / -1 = 13
n= 2： n – 14 = - 12 ，n^3-2 = -6 => -12 / -6 = 2
n= 14： n – 14 =0 ，n^3-2 = 2742 => 0 / 2742= 0

(n^3-2) > | n-14 | when n >= 3

Pf:
When n<14,
n^3-2>|n-14| <==> n^3+n-16>0 <==> (n^3-16)+n>0 <== n>=3.

When n>=14,
n^3-2>|n-14| <==> n^3-n+12>0 <== n^3>n <== n>=14.

Done.

not yet!!
1110
By the way, how do you prove that (n^3-2) > | n-14 | when n >= 3?

2008-01-24 02:37:56 補充：
Proving that (n^3-2) > | n-14 | when n >= 3 is not my method.
In fact, we can get (n^3-2) | k, in which k is an integer, having no unknown number n.

2008-01-25 21:16:05 補充：
No!!
(k+1)^3-2 = k^3+3k^2+3k-1= (k^3-2)+3k^2+3k+1> (k-14)+3k^2+3k+1 = 3k^2+4k-13 > (k+1)-14,
because 3k^2+3k>0 when k>0 or k<-1.
So, (k+1)^3-2 > |(k+1)-14| when k>=15.

2008-01-25 22:07:53 補充：
There's no need to prove that (n^3-2) > | n-14 | when n >= 3.

In fact, (n^3-2) | 2742=14^3-2= 2*3*457,

according to the polynomial division or the polynomial modular arithmetic.


So the answers are 2, 14.


If the divisor could be negative, 0,1 and -1 would be the answers, too.

1 . 2 ?

n >= 3

( n^3 - 2 ) > | n - 14 |

done ?

2008-01-24 08:05:15 補充：
when n = 3 ... 14 .

n^3 - 2 > | n - 14 |

when n>= 15

15^3 - 2 > 1

when n = k + 1

( k + 1 )^ 3 - 2 > k^3 + 1 - 2 > k + 1 + 1 = k + 2

by Mathematical ...

am i right ?


p is divided by q provided that

p = qk , where q is postive integer ?