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Let $n \in \mathbb{N}_{\geq 2}$. Prove that the following statements are equivalent

(1) For every $x \in \mathbb{Z}_n,$ if $x^2 \equiv 0,$ then $x \equiv 0.$

(2) There is no prime $p$ such that $p^2 | n.$

To prove (1) $\implies$ (2), I attempted $\lnot (2) \implies \lnot (1).$

So, I assumed that there does exist some prime $p$ such that $p^2|n.$ Since the statement should be true for all $n \geq 2,$ it seemed ok to me to specify a particular case: $n = 5.$

Here, $5^2|25.$ Also, $5^2 \equiv 0$ but $5 \not\equiv 0.$

Does such a counter example work to prove this? I have a lot of doubt.

Thank you

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Rafael Vergnaud
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    I don't believe you can specify $n=5$. The statement is of the form "For all $n\in\mathbb{N}_{\geq2}$, such-and-such" and you have only proved "For $n=5$, such-and-such". – 79037662 Nov 08 '19 at 17:41
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    You are right be doubtful: first, you need to prove that $\neg(2)\implies\neg(1)$ for all $n$. Second, even what you tried to do is wrong, because if $n=5$, then you are checking congruences modulo $5$. – Arturo Magidin Nov 08 '19 at 17:41
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    Specifically, you are trying to prove that "for all $n$, (1)$\implies$(2)". The proof by contrapositive is "for all $n$, $\neg$(2)$\implies\neg$(1)". But you seem to have tried to prove "there exists $n$ for which $\neg$(2) and $\neg$(1)", which as you can see is very much not what you need to prove. – Arturo Magidin Nov 08 '19 at 17:43
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    This equivalence is one of many characterizations of a squarefree integer. – Bill Dubuque Nov 08 '19 at 18:21
  • It works for any square factor $d^2$ not only squares of primes - see my answer. – Bill Dubuque Nov 08 '19 at 20:37

2 Answers2

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A correct way to word your idea would be as follows.

Suppose that $n$ is divisible by the square of the prime $p$. Then we can write $n=p^aq$ where $a\ge 2$ and $q$ is coprime to $p$.

Now consider $x=p^{a-1}q$.

$x^2$ is divisible by $n$ but $x$ is not.

  • The argument has nothing to do with primes - it works for any square factor $> 1$, see my answer. – Bill Dubuque Nov 08 '19 at 20:37
  • It has everything to do with primes because the OP wishes to prove that there is no prime $p$ such that $p^2$ divides $n$. In any case, it is obvious that if there is no such prime $p$ then there is no such $d>1$. –  Nov 08 '19 at 22:58
  • The point is $(1)$ implies $n$ is squarefree i.e. it has no nontrivial square factors $d^2$ so it has no $p^2$ factors. In fact the proof in your answer is a special case of the general proof in my answer. – Bill Dubuque Nov 09 '19 at 00:24
  • Having a nontrivial square factor is completely equivalent to having a prime square factor. Neither statement is a special case of the other. –  Nov 09 '19 at 01:05
  • They are equivalent in $\Bbb Z$ and UFDs but not in more general domains (where prime factors needn't exist). There we must use the more general definition of squarefree (which has nothing to do with primes). Your proof is the special case $,d = p,$ of mine, since your $, x = n/p = n/d = {\rm my}\ x$. – Bill Dubuque Nov 09 '19 at 01:26
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More generally $\,d^2\mid n\,\Rightarrow\, d^2n\mid n^2\Rightarrow n\mid (n/d)^2\,\overset{(1)}\Rightarrow\,n\mid n/d\,\Rightarrow\,d \mid 1.\,$ OP is case $\,d = p\,$ prime.

Bill Dubuque
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