Question

Suppose \(n|(a^s-1)\) and \(n|(a^t-1)\)

Can we say anything about the relationship between \(s\) and \(t\) ?

Answer 1

cant see anything :\ first i thought n =a-1
but then ok not this form...

\(\Large a^s\equiv 1 \mod n \\\Large a^t\equiv 1 \mod n \\ \rightarrow\Large a^{s+t}\equiv 1 \mod n \\ \text{however this seems leads to nothing :\ } \)

Answer 2

i only can say W.L.O.G
s>t or so lol

Answer 3

that also gives
\[\large a^{s-t} \equiv 1 \pmod{n}\]

Answer 4

nope

Answer 5

for that we need ind thingy

Answer 6

because \(\large a^s \equiv a^t \pmod{n}\)
divide \(\large a^t\) both sides assuming \(\gcd(a,n)=1\)

Answer 7

oh got what your saying, but this goes for strict conditions also we need to be careful
\(a^{s-t}>n\)
else its false

Answer 8

so i won't say that applies :|

Answer 9

ok

Answer 10

but if n=a-1 then we can say
gcd(s,t)= min(s,t)

Answer 11

eh nvm seems im not thinking good of it

Answer 12

can we say this
\[\large n|(a^s-1)~~\text{and}~~n|(a^t-1) ~~\implies n|(a^{\gcd(s,t)}-1)\]

?

Answer 13

i dont know if thats true, but it is working for all the examplies i tried..

Answer 14

its true for n=a-1

Answer 15

lets see if its not

Answer 16

im not getting ideas will try ltr :\
\(\Large a^s\equiv 1 \mod n \\\Large a^t\equiv 1 \mod n \\ \)
assume gcd(s,t)=d
s=df
t=dg
\(\Large a^{df}\equiv 1 \mod n \\\Large a^{dg}\equiv 1 \mod n \\ \)