Let $e$ and $n$ be positive integers. Suppose $e$ and $φ(n)$ are relatively prime. Prove if $x^e ≡ 1\bmod(n)$, then $x ≡ 1\bmod(n)$

Question

My thoughts so far:

Since $φ(n)$ is always even, $e$ must be odd $\implies e$ can be written as $2d + 1$. $\implies x^{2d+1} - 1 = kn$

However, I don't know how to proceed from here. Any tips?

Please see How to ask a good question: Good Titles. Your question should not refer to information in the title and try to make your title more interesting. — Simply Beautiful Art, Mar 11 '18 at 22:53
Use https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — lab bhattacharjee, Mar 12 '18 at 00:55

score 2 · Accepted Answer · answered Mar 11 '18 at 22:52

2

Well we know $x^{\phi(n)}\equiv 1\mod n$, so we can use that, since now we know for all positive integers $a,b$ that $$(x^{\phi(n)})^a(x^e)^b\equiv1\mod n$$ which can be rewritten to $$x^{a\phi(n)+be}\equiv1\mod n$$

for every $a,b$. Now since $\phi(n)$ and $e$ are coprime, we can find $a,b$ such that $a\phi(n)+be=1$, and so if we pick those, the congruence reads

$$x\equiv 1\mod n$$

exactly what we wanted.

answered Mar 11 '18 at 22:52

2

Isn't this answer wrong since Euler Theory only works if $x$ and $n$ are coprime? That is not necessarily the case – helpneeded Mar 12 '18 at 00:10
1

Well, the fact that there exists an $e$ with $x^e\equiv 1\mod n$ implies $\gcd(x,n)=1$. – Mar 12 '18 at 10:27

Let $e$ and $n$ be positive integers. Suppose $e$ and $φ(n)$ are relatively prime. Prove if $x^e ≡ 1\bmod(n)$, then $x ≡ 1\bmod(n)$

1 Answers1