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I already try to solve this with divisibility rule $p = ka, a^n$ = $ka*ka$ as much as $n$ times $a^n≡a (mod n)$, then $p^n=ka*ka$ therefore $p^n= a^n$ be $^n√p^n|^n√a^n$ which $p|a$ but im not sure how right am i or wrong

Bill Dubuque
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Rahayu
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Nov 01 '23 at 11:32
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    If you can use the fact that a prime $p$ divides $x y$ iff it divides either $x$ or $y$, (Euclid's lemma) this is just induction. – George C Nov 01 '23 at 11:50

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There are several ways. You could use the contraposition for example. But I think the best way to understand the relation is the direct way.

Let $p$ be a prime with $p\mid a^n$ for some natural $a$ and $n$. If $a = p$, we are done. Now look at the unique prime decomposition of a: $$ a = p_1\cdot p_2 \cdot \dots \cdot p_k $$ where every $p_i$ is a prime number. These do not have to be distinct. Then $$ a^n = (p_1\cdot p_2 \cdot \dots \cdot p_k)^n = p_1^n \cdot p_2^n \cdot \dots \cdot p_k^n. $$ Since $p$ is a divisor of $a^n$, there is a prime number $p_i$ in this decomposition, such that $$ p\mid p_i^n. $$ But since $p$ and $p_i$ are prime, it follows that $p = p_i$ (because multiplying two numbers does not lead to new divisors). So $$ p\mid p_i\mid a \Rightarrow p\mid a. $$

Lereu
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