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Let $n \geq 2$ be an integer. $[x]$ represents the equivalence class mod $n$. Prove the following are equivalent:

$1)$ $n=p_1\cdots p_k$ where $p_1,...,p_k$ are distinct primes, that is, in the prime factorization of $n$, all exponents are $1$.

$2)$ If $[x]\in\mathbb{Z}_n$, $[x]\neq[0]$, then $[x]^2\neq [0]$.

Proving: $1)\implies 2)$ confuses me since the statements barely seem related. To begin my proof, I assume $1)$ and then additionally assume $[x]\in\mathbb{Z}_n$, $[x]\neq[0]$. Is it sufficient to prove that $[x]$ is not a zero divisor? Even if it were, I'm not sure how I'd go about since the properties of which I am aware deal with unique aspects of $\mathbb{Z}_p$ where $p$ is prime.

user26857
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1 Answers1

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Try contrapositive and assume that $[x]^2 = 0$. Then $x^2$ would be divisible by $p_1p_2\cdots p_k$. Further, each prime factor in a square must appear twice.

J126
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  • First, thank you for your answer. To prove by contrapositive, would I assume If $[x]\in\mathbb{Z_n}$, $[x]=0$, then $[x]^2=0$? Moreover, what tells you then that $[x]$ is divisible by the distinct primes? –  Sep 22 '16 at 00:30
  • @EthanZell: The contrapositive of “$[x] \neq [0] \implies [x]^2 \neq [0]$” is ”$[x]^2 = [0] \implies [x] = [0]$.” Equivalently, $n | x^2 \implies n | x$. – Matthew Leingang Sep 22 '16 at 00:42
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    For contrapositive you assume $[x]^2 = 0$ and derive $[x] = 0$. Well, if $[x]^2 = ap_1p_2 \cdots p_k$, and all primes must appear twice, what would happen if you took a square root? – J126 Sep 22 '16 at 00:43
  • @JoeJohnson126 Would it be that $[x]=n$? –  Sep 22 '16 at 00:45
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    You've mixed notation. You can't have $[x] = n$, only $[x] = [n]$, which is what would happen. – J126 Sep 22 '16 at 00:53
  • @JoeJohnson126 I think I have it now. We have $[x]^2=0 \implies x^2 \equiv 0 \text{mod} n \implies p_1...p_k|x^2$. By Euclid's Lemma, $p_1...p_k|x$. Then, $[x]=[n]$ and $n$ must be the product of some nondistinct primes since $n=ap_1...p_k$. –  Sep 22 '16 at 01:21
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    @EthanZell Yes. – J126 Sep 22 '16 at 12:25