- #1
cragar
- 2,552
- 3
I just want to make sure I understand these number theory proofs.
[itex] b^{\phi (n)}=1mod(n) [/itex]
[itex] \phi (n) [/itex] is the order of the group, so b to some power will equal the identity. so that's why it is equal to one. [itex] b^p=bmod(p) [/itex]
[itex] b^p=b^{p-1}b [/itex]
[itex] b^{p-1} [/itex] produces the identity since p-1 is the order of the group. so that's why it equals b.
[itex] b^{\phi (n)}=1mod(n) [/itex]
[itex] \phi (n) [/itex] is the order of the group, so b to some power will equal the identity. so that's why it is equal to one. [itex] b^p=bmod(p) [/itex]
[itex] b^p=b^{p-1}b [/itex]
[itex] b^{p-1} [/itex] produces the identity since p-1 is the order of the group. so that's why it equals b.