a-maths

如下:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/CrazyMI8.jpg

5^(2(1)-1) -3^(2(1)-1) -2^(2(1)-1) = 0 能被15整除
assume 5^(2k-1) -3^(2k-1) -2^(2k-1)能被15整除
5^(2k-1) -3^(2k-1) -2^(2k-1) = 15m, m 係整數
5^(2(k+1)-1) -3^(2(k+1)-1) -2^(2(k+1)-1)
= 5^(2k+1) - 3^(2k+1) - 2^(2k+1)
= (25) 5^(2k-1) - (9) 3^(2k-1) - (4) 2^(2k-1)
= (25) [5^(2k-1) - 3^(2k-1) - 2^(2k-1)]
+ (25) 3^(2k-1) + (25) 2^(2k-1)
- (9) 3^(2k-1) - (4) 2^(2k-1)
= (25)(15m) + (16) 3^(2k-1) + (21) 2^(2k-1)
= (25)(15m) + (15) 3^(2k-1) + (15) 2^(2k-1)
+ 3^(2k-1) + (6) 2^(2k-1)
如果 3^(2k-1) + (6) 2^(2k-1) 能被15整除
(25)(15m) + (15) 3^(2k-1) + (15) 2^(2k-1)
+ 3^(2k-1) + (6) 2^(2k-1) 就能被15整除, 即係證完
宜家變多一條問題,要證明
3^(2k-1) + (6) 2^(2k-1) 能被15整除
3^(2(1)-1) + (6) 2^(2(1)-1) = 15 能被15整除
assume 3^(2q-1) + (6) 2^(2q-1) 能被15整除
3^(2q-1) + (6) 2^(2q-1) = 15p, p 係整數
3^(2(q+1)-1) + (6) 2^(2(q+1)-1)
= 3^(2q+1) + (6) 2^(2q+1)
= (9) 3^(2q-1) + (24) 2^(2q-1)
= (9) [ 3^(2q-1) + (6) 2^(2q-1) ] - (9)(6) 2^(2q-1)
+ (24) 2^(2q-1)
= (9)(15p) - (30) 2^(2q-1)
= (15)[9p - (2) 2^(2q-1)]
能被15整除, 因 [9p - (2) 2^(2q-1)] 係整數
做完, 明未?