Two versions of Eulers theorem and how they relate.

Question

I am familiar with the following Euler Theorem: (Note: In the following ø(n) is the Euler function)

version i)

If gcd(x,n)=1, then $x^{ø(n)} = 1$ (mod n).

Version ii) However, I have seen the following version described as Eulers theorem too:

if $a = b$ (mod ø(n)), then $x^a = x^b$ (mod n) if gcd(x,n)=1.

My question is then, if they are equivalent, how can i show that version i) implies version ii) ? Thanks,

$(i)\Rightarrow(ii)$ is a special case of modular order reduction, and $(ii)\Rightarrow(i)$ follows by specializing $,a=\phi(n),\ b = 0\ \ $ — Bill Dubuque, Oct 03 '20 at 17:22

score 1 · Accepted Answer · answered Oct 03 '20 at 13:48

1

$a \equiv b \pmod{\phi(n)} $ means $$ a = b +k \phi(n) $$ for some integer $k$. Therefore $$x^a = x^b (x^{\phi(n)})^k. $$

answered Oct 03 '20 at 13:48

user1337

Thanks, could you elaborate why $=+()$ need to be true? – ghetto_department Oct 03 '20 at 13:55
That is because the difference $a-b$ is known to be a multiple of $\phi(n)$. – user1337 Oct 03 '20 at 14:00
Ah, of course. Thx a lot. – ghetto_department Oct 03 '20 at 14:00

score 0 · Answer 2 · answered Oct 03 '20 at 14:20

0

By $i)$, you can reduce the exponent $\bmod\varphi(n)$. Thus we get $ii)$.

answered Oct 03 '20 at 14:20

2 Answers2