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For $a,b,n,m\in \Bbb N$, if $a$ and $b$ are coprime,

than for $m^n=a\cdot b$

$\exists a',b'\in \Bbb N: a=a'^n \land b=b'^n$

I thought it would suffices to show that $m=a'\cdot b'$, but I'm not sure that's why I also thought of :

As $a,b$ are coprime than we can write $$\begin{align}a&=p_1^{u_1}\cdots p_i^{u_i}\\b&=q_1^{v_1}\cdots q_j^{v_j}\end{align}$$

where $p_g\neq q_g \forall g$ so$$\quad m^n=a\cdot b=p_1^{u_1}\cdots p_i^{u_i}\cdot q_1^{v_1}\cdots q_j^{v_j}$$ As the $u_i,v_i$ doesn't necessary need to be distinct, we can rewrite to $$m^n=(p_1q_1)^{n_1}\cdots(p_iq_j)^{n_{ij}}$$

we define $m:=\prod p_hq_h$ and $n:=\prod n_{ij}$

Not sure how to continue or either it's correct or not :(

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kratos88
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    There is some confusion here (by associating $p_i$ and $q_j$ and so on). It may be simpler to start with the unique factorization of $m$ and distribute the prime factors of $m^n$ in $a\cdot b$. – Raymond Manzoni Jul 06 '17 at 20:00
  • Something which adds to the confusion, you are trying to prove this for any $m,,n,$ which means you can't define them in your proof. – mdave16 Jul 06 '17 at 20:02

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