推 FocusE :(h,k)=1 => (hk,h-k)=1 12/29 11:22
→ eqcolouring :可以再說明一下嗎?我還是不懂...謝謝! 12/29 12:03
推 hugogoss :反證,假如(hk,h-k)=e≠1,則會得到(h,k)為e的倍數 12/29 12:07
→ suker :d>0,設d|h ,(h,k)=1 故d不整除K d|hk 但d不整除h-k 12/29 13:21
→ suker :=> (hk,h-k)=1 12/29 13:21
→ suker :d>1才對打錯 12/29 13:21
→ eqcolouring :謝謝! 12/30 00:14