Prove: If a natural number $n$ doesn't has the form of $k^{m}$, where $k$ and $m$ are natural numbers, then the equation $x^{m}=n$ has no rational root.
How do I start to prove with contradiction (or another method if it doesn't work)?
My first thought of this proof is Rational Root Theorem. I know by substituting $n=k^{m}$ into the equation above implies $x=k$, which is also a natural number. However, this is not the proof that I wish to proceed, so anyone have some idea to start my proof?
On
Hint: Suppose there is a rational root - call it $\frac{p}{q}$ where $\gcd(p,q)=1$
Plug that in the equation and see that it results in the equation:
$$p^m=nq^m$$
Now use the fact that $\gcd(p,q)=1$, by (for example) looking at a prime divisor of $q$ (if such exists)
On
What about trying to prove the contrapositive? Suppose that $\exists x\in \mathbb{Q}$ such that $x^m=n$. Since $n>0$, Assume that $x>0$ and $x=\frac{b}{a}$ where $a,b\in \mathbb{Z}$, $a,b>0$, gcd$(a,b)=1$. Then $(\frac{b}{a})^m=n$ or $b^m=a^mn$.$\;$ Now consider prime factorizations of $a,b,n$ and recall the fundamental theorem of arithmetic.
If $\frac{p}{q}$, with $(p,q)=1$, is a rational root of $x^m = n$ we conclude that
$\frac{p^m}{q^m} = n$ or equivalently $$n\cdot q^m = p^m\tag{1}$$
Now let $p_i$ be any prime divisor of $n$, say with multiplicity $k_i$. Then $p_i^{k_i}| n$ and $n | n \cdot q^m = p^m$, so that $p_i | p$. If $l_i$ is the multiplicity of $p_i$ in $p$ then $m*l_i = k_i$ by the unique prime decomposition theorem, as $p_i$ does not divide $q$, so that $p_i$ fits exactly $k_i$ times in the left hand side of (1). This shows that each prime factor of $n$ occurs a multiple of $m$ times, which means that $n$ is an $m$-th power.