Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

up vote 1 down vote accepted

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

share|improve this answer

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 $k>0$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.