Carmichael function: Difference between revisions

Revision as of 23:10, 12 January 2022

Revision as of 23:15, 12 January 2022

Line 123:

:<math> a\,|\,b \Rightarrow \lambda(a)\,|\,\lambda(b) </math>

'''Proof.'''

'''Proof.''' The result follows from the Chinese remainder theorem.

By definition, for any integer <math>k</math>, we have that <math> b \,|\, (k^{\lambda(b)} \equiv- 1\mod{b} )</math> , and therefore <math> a \,|\, (k^{\lambda(b)} \equiv- 1)</math>. By the minimality property above, we have <math> \mod{lambda(a})\,|\,\lambda(b) </math>.

By the minimality property above, we have <math> \lambda(a)\,|\,\lambda(b) </math>.

===Composition===