Question

HELP! primitive roots #4
http://prntscr.com/4lfzxi

Answer 1

\[\large a^h \equiv 1 \pmod n \]
\[\large b^k \equiv 1 \pmod n \]

Answer 2

multiplying both gives
\[\large a^hb^k \equiv 1 \pmod n \]

Answer 3

\(\large \phi(n)=d_1h\)
\(\large \phi(n)=d_2k\)


then \(\large (ab)^g=1 \mod n\)
\(\large \phi(n)=d_3 g\)

hmmm

Answer 4

\(\large a^hb^k \equiv 1 \pmod n \) multiply both side by \(a^k b^h \)

\(\large (ab)^{hk} \equiv a^k b^h \pmod n \)
hmmming

Answer 5

http://math.stackexchange.com/questions/926117/assume-that-the-order-of-a-in-pmod-n-is-h-and-the-order-of-b-in-pmo/926130#926130

Answer 6

Or else :

\(\large a^hb^k \equiv 1 \pmod n \)

take "hk"th power both sides :

\(\large \left(a^hb^k\right)^{hk} \equiv 1^{hk} \pmod n \)

\(\large \left(ab\right)^{hk} \equiv 1 \pmod n \)

so \(\mathbb{ord(ab) ~| ~hk}\)

Answer 7

ohh
got it !
cool