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I have encountered a statement which says: for natural numbers $h$ and $k$ such that $(h,k)=1$,we will have $h^{\phi(k)}\equiv 1(\textrm{mod}\ k)$, where $\phi(k)$ is the Euler's totient function that shows the number of positive integers not exceeding $k$ relatively prime to $k$. I wonder is there any formal theorems that illustrate this statement? And how will this theorem be proved?

I think this statement is related closely with cyclic group in group theory, but I can't think of any useful facts to prove this statement. Thank you very much in advance for any replies!

Chang
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    https://en.wikipedia.org/wiki/Fermat%27s_little_theorem – one potato two potato Sep 03 '21 at 03:38
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    It's called Euler's Theorem (one of many theorems meriting that name), and you'll find a proof in every intro Number Theory text ever written, also on one out of every three webpages, and probably in many posts on this website. – Gerry Myerson Sep 03 '21 at 03:39
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    https://en.wikipedia.org/wiki/Euler%27s_theorem also duplicate of https://math.stackexchange.com/questions/450260/showing-that-a-phin-equiv-1-pmod-n-when-a-and-n-are-relatively-prime?rq=1 – Brauer Suzuki Sep 03 '21 at 03:57
  • Gcd of powers is related to gcd of base,and modulus since it's always 1 in this case the remainders must be coprime to the modulus, how many unique remainders are there ? – Roddy MacPhee Sep 03 '21 at 12:48
  • @Roddy, "gcd of powers", what are you talking about? – Gerry Myerson Sep 03 '21 at 12:55
  • $\gcd(a^n,b)\mid \gcd(a,b)^n$ – Roddy MacPhee Sep 03 '21 at 14:01
  • That plus there being $\phi(n)$ possible coprime remainders, means by pigeonhole principle that ${a^0, \ldots , a^{\phi(n)}}$ contains two powers giving same remainder ... – Roddy MacPhee Sep 03 '21 at 19:43
  • @Roddy, I would not have guessed that that was what you meant. Also, two powers giving same remainder implies $a^r\equiv1$ for some $r$, $1\le r\le\phi(n)$, but doesn't show $a^{\phi(n)}\equiv1$. – Gerry Myerson Sep 04 '21 at 01:52
  • Yeah, I'm a bad communicator, and verbose. I like this method more, just because it basically generalizes to that point for all gcd's between modulus and base of power. – Roddy MacPhee Sep 04 '21 at 02:06

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