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Given a natural number say $n \in \mathbb{N}$ with a prime factorization $p_1^{m_1} \cdot p_2^{m_2} \dots p_k^{m_k}$. If you take product of the prime factors $p_1 \cdot p_2 \dots p_k$ then the following holds:

$$ \exists i,j \in \mathbb{N} ~~ n^i = (p_1 \cdot p_2 \dots p_k) \cdot j $$

I think that is correct. Is it? If have no idea how to proof it.

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Huh? Consider $12=2^2\cdot 3$; $12^i=2^{2i}\cdot 3^i\neq 6^j$ for any $j$... – Yongyi Chen Apr 16 '12 at 17:40
Sorry I made a mistake. It must be a multiple not a power! – joachim Apr 16 '12 at 17:52
up vote 1 down vote accepted

To your edited question: It is obviously true; set $i=1$ and $j=p_1^{m_1-1}\cdots p_k^{m_k-1}$.

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Let $n=20$. Then $n^i =2^{2i} 5^i$ for all $i$. Suppose that this is equal to $(2\cdot 5)^j = 2^j 5^j$. Then you immediately see that $j=i$ and $j=2i$. Contradiction. So it's incorrect.

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Consider $12=2^2\cdot 3$. For any $k$ we have $6^k=2^k3^k$; if this were equal to $12^n$ for some $n$, we’d have $2^{2n}3^n=2^k3^k$. Unique factorization implies that $n=k$, since there must be the same number of factors of $3$ on each side, but it also implies that $2k=n$. This is impossible if $k>0$.

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