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Show that $\text{rad}(n)=n$ when $n$ is a positive integer if and only if $n$ is square-free.

I came so far:

$n$ is square free $\Rightarrow$ no $x^2$ divides $n \Rightarrow$ $n=p_1* ...*p_n\Rightarrow\text{rad}(n)=n=p_1* ...*p_n$.

Teddy38
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WinstonCherf
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    If $n = \prod p_i^{r_i}$, $p_i$ primes, $r_i\in\mathbb{Z}$, $r_i>0$, then $\text{rad}(n) = \prod p_i$. Isn't the result clear? – rogerl Nov 15 '17 at 14:52

1 Answers1

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You have the right idea, but I would write it down a bit more formally. Your core lemma is the following:

An integer is square-free if and only if every exponent in the prime factorization is $1$.

Prove this both ways, one way by contradiction by dividing through $x^2$, and the other way by explicit counterexample if $p^k, k> 1$ is in the prime factorization of $n$.

Then you can use $\text{rad}(n) = p_1 \cdots p_k = n$.

orlp
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