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From Euler's theorem I know that $a^{\phi(n)} \equiv 1 \pmod n$ if $\gcd(a,n) = 1$. However I can't find any proof/explaination of the proof in the title.

Bill Dubuque
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meowmeowxw
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2 Answers2

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Hint: Consider that the the only thing that changes from $a^k$ to $a^{k\!\pmod{\phi(n)}}$ is that the exponents differ by a multiple of $\phi(n)$.

Arthur
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  • Since $\Z_n^*$ is a finite set, than we can define a number $y$ such that $a^y \equiv 1 \pmod n$, from euler's theorem we know that $y = \phi(n)$, and this also means that $a^{k \pmod{\phi(n)}} \equiv a^k \pmod n$ for a generic $k$ , since $\phi(n) \mid k$ (?) – meowmeowxw Sep 30 '19 at 20:06
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    Not quite. And a bit more complicated than I was thinking. We do not know that $y=\phi(n)$. We know that setting $y=\phi(n)$ makes $a^y\equiv 1$, but the converse doesn't hold (for instance, if $a=1$, then any $y$ will work). However, note that if we set $y=\phi(n)$, then elementary exponent properties give us that $a^k=(a^y)^j\cdot a^{k!\pmod{\phi(n)}}$ for some integer $j$. Applying modulo $\phi(n)$ to this equality and using Euler's theorem gives you your result. – Arthur Sep 30 '19 at 20:12
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The key idea is to use modular order reduction on exponents as in the Theorem below. We can find small exponents $\,e\,$ such that $\,a^{\large \color{#c00}e}\equiv 1\,$ either by Euler's totient or Fermat's little theorem (or by Carmichael's lambda generalization), along with obvious roots of $\,1\,$ such as $\,(-1)^2\equiv 1.$

Theorem $ \ \ $ Suppose that: $\,\ \color{#c00}{a^{\large e}\equiv\, 1}\,\pmod{\! m}\ $ and $\, e>0,\ n,k\ge 0\,$ are integers. Then

$\qquad\ \ \ \ n\equiv k\pmod{\! \color{#c00}e}\,\Longrightarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}.\: $ The converse holds if $\:\color{#c00}e = {\rm ord}\,a,\,$ i.e.

$\qquad\ \ \ \ n\equiv k\pmod{\! \color{#c00}e}\,\Longleftarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}\ $ and $\, a\,$ has order $\,\color{#c00}e\,$ mod $\,m$

Proof $\ $ Wlog $\,n\ge k\,$ so $\,a^{\large n-k} \color{#0a0}{a^{\large k}}\equiv \color{#0a0}{a^{\large k}}\!\iff a^{\large n-k}\equiv 1\iff n\equiv k\pmod{\!e}\,$ by this Corollary, where we cancelled $\,\color{#0a0}{a^{\large k}}\,$ using $\,a^{\large e}\equiv 1\,\Rightarrow\, a\,$ is invertible so cancellable (cf. below Remark).

Corollary $\ \ \bbox[7px,border:1px solid #c00]{\!\bmod m\!:\,\ \color{#c00}{a^{\large e}\equiv 1}\,\Rightarrow\, a^{\large n}\equiv a^{\large n\bmod \color{#c00}e}}\,\ $ by $\ n\equiv n\bmod e\,\pmod{\!\color{#c00}e}$

Remark $ $ If you are familiar with modular inverses then it is not necessary to restrict to nonnegative powers of $\,a\,$ above since $\,a^{\large e}\equiv 1,\ e> 0\,\Rightarrow\,$ $a$ is invertible by $\,a a^{\large e-1}\equiv 1\,$ so $\,a^{\large -1}\equiv a^{\large e-1}$.

Bill Dubuque
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