Re: 幫個忙否?

作者: myflame (原來隱起來了)   2004-11-03 01:46:23
※ 引述《chaogold (H.Y. Chao)》之銘言:
: 1.Find all integers n>=1 such that
: (n^3+3)/(n^2+7)is an intege
let x=(n^3+3)/(n^2+7)
n=1 x=4/8 (x)
n=2 x=11/11=1 (o)
n=3 x=30/16 (x)
n=4 x=67/23 (x)
n=5 x=128/32=4 (o)
n>5 n>x>(n-1) (x)
[n>x,trivial
x>(n-1)=>(n^3+3)/(n^2+7)>(n-1)
=>n^3+3>n^3-n^2+7n-7
=>n^2-7n+10>0
=>n>5 or n<2 ]
so n=2 or 5
:

Links booklink

Contact Us: admin [ a t ] ucptt.com