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If $\gcd(a,n)=1$, Show that $$a^b\equiv a^{b\bmod{ \phi(n)}}\pmod n $$ Where $\phi$ is Euler’s totien function.

I saw this on my number theory book but the author didn’t provid any proof of it. We probably would use the fact that $$a^{\phi(n)} \equiv 1\pmod n$$ But I don’t know how to use it properly.

Arthur
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PNT
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    Hint: $b=(b\mod\varphi(n))+k\varphi(n)$ for some $k$. – abiessu Jan 21 '22 at 17:43
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    First perform integer division of $b$ divided by $\phi(n)$, and then let $b=q\cdot \phi(n) + (b\bmod \phi(n))$. – peterwhy Jan 21 '22 at 17:43
  • Euler's theorem is in fact the key. Hint : Write $b$ as $p\cdot \phi(n)+q$ with $0\le q<\phi(n)$ and show $a^b\equiv a^q\mod n$ – Peter Jan 21 '22 at 17:43
  • Special case $,e=\phi(m),$ of the Corollary in the linked dupe. – Bill Dubuque Jan 21 '22 at 18:28
  • I’m not against marking the post as duplicated, but I have one question, if someone in the future wanted to ask the same question he won’t find my question in “Similar questions” while asking it. So someone will have to mark their question as duplicated. And this loop won’t end. @BillDubuque – PNT Jan 21 '22 at 18:30

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Proof:

Let $ b= k\phi(n) + r$, with both $k$ and $r$ integers, and $0 \le r < b$. Then: $$a^b = a^{k\phi(n) +r} = a^ {k\phi(n)} a^r \equiv a^r \equiv a^{b \mod \phi(n)} \pmod n$$

Test
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Jan 21 '22 at 17:53
  • What do you mean by $r\in \mathbb R$? – PNT Jan 21 '22 at 17:57
  • $\mathbb{R}$ is the set of real numbers, $\in$ is the symbol for denoting inclusion in a set. – Test Jan 21 '22 at 18:07