Recent content by rjmack

Sorry, This is Not True in General Let k and b (both > 1) be such that b does not divide k-1 (with the exception that b may equal k-1) and k and b are relatively prime. If n is the smallest power so that k^n is equivalent to 1 mod b (so n is the length of the digit cycle of 1/b in base k)...

I am trying to solve this piece so I can show that For each prime p and k > 1 where k-1 = (a_1^b_1)*(a_2^b_2)*(a_3^b_3)*...*(a_n^b_n) in prime factorization, (k^p - 1) / c is squarefree where c = (a_1^(b_1 - 1))*(a_2^(b_2 - 1))*(a_3^(b_3 - 1))*...*(a_n^(b_n - 1))

Also... On a more general note we can say this: Let k and b (both > 1) be such that b does not divide k-1 (with the exception that b may equal k-1) and b has at least one factor that is relatively prime with k. If n is the smallest power so that k^n is equivalent to 1 mod c where c is the...

Sorry I forgot an important part... Let k and b (both > 1) be such that b does not divide k-1 (with the exception that b may equal k-1) and k and b are relatively prime. If n is the smallest power so that k^n is equivalent to 1 mod b (so n is the length of the digit cycle of 1/b in base k)...

Not sure if cycle is the best word here.. Let k and b (both > 1) be such that b does not divide k-1 (with the exception that b may equal k-1). If n is the smallest power so that k^n is equivalent to 1 mod b (so n is the length of the digit cycle of 1/b in base k), then bn is the...