03/04/02 01:18
>>918
a*b MOD c = 1 ‥‥‥(1)
⇒
∃a' s.t. (a'*c+1) MOD b = 0
∵
(1)より
a*b - 1 = a'*c
⇔
a*b = a'*c + 1
⇔
a = (a'*c + 1) ÷ b
∴(a'*c+1) MOD b = 0
03/04/02 01:18
>>918
a*b MOD c = 1 ‥‥‥(1)
⇒
∃a' s.t. (a'*c+1) MOD b = 0
∵
(1)より
a*b - 1 = a'*c
⇔
a*b = a'*c + 1
⇔
a = (a'*c + 1) ÷ b
∴(a'*c+1) MOD b = 0