~~數論習題(Ⅴ)~~

2011-01-25 7:55 pm
證明或否定 :

x² + y² + z² = 999 沒有整數解。









回答 (1)

2011-01-23 10:14 pm
✔ 最佳答案
http://img600.imageshack.us/img600/7421/70735072.png

圖片參考:http://img600.imageshack.us/img600/7421/70735072.png


2011-01-23 14:31:11 補充:
Sometime wrong to be revised

2011-01-23 15:17:23 補充:
奇奇奇: 在不失一般性下,設x=2n+1,y=2m+1,z=2p+1
(2n+1)^2+(2m+1)^2+(2p+1)^2=999
4n^2+4n+1+4m^2+4m+1+4p^2+4p+1=999
n^2+n+m^2+m+p^2+p=249
n(n+1)+m(m+1)+p(p+1)=249
其中n(n+1),m(m+1),p(p+1)必為偶=>(矛盾)

2011-01-23 15:19:11 補充:
http://img541.imageshack.us/img541/2572/89685347.png


收錄日期: 2021-04-11 18:30:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110123000051KK00409

檢視 Wayback Machine 備份