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$x^n=N$, where n and N are positive integers.

Hence, $x=N^{(1/n)}$

Let $x=N^{(1/n)}$=$\prod_{i=1}^{j} a_{i}^{(x_{i}/n)}$, where $a_{i}$ is a distinct prime factor of $N$ and $x_{i}$ is $a_{i}'s$ multiplicity.

Hence, ${x_{i}/n}>0$, as the prime factors multiplicity is positive, and n is positive.

Considor $x=a/b$, where a and b are positive integers.

In this scenario, x is only an integer if all primes in b are also in a, and the resulting multiplicity of all prime factors is positive.

Hence, if x is not an integer, there must be some primes in x that have negative multiplicities.

But, we have seen before that for $x=N^{(1/n)}$, x must contain only prime factors with positive multiplicities, hence any number of the form $a/b$, not being an integer, hence a rational number, cant be a solution for x.

Hence, only irrational or integers can be a solution.

Thanks in advance.

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  • $\begingroup$ It's pretty unclear as it stands - everything after "consider x=a/b" is unjustified, since it implicitly assumes that the factorization of a and b is related to the given one for x - which is true, but basically circular. $\endgroup$ Sep 29, 2017 at 14:02

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I think it is incorrect, or at least incomplete. The problem is, we could not guarantee that $$ x = \prod_{i=1}^{j} a_{i}^{x_{i}/n} $$ is an unique factorization (of which some generalized form, i.e., $x_i/n$ could be rational), unless you prove it explicitly. And also we should note that in this stage of the proof, we can not assert that $x_i/n$ is an integer.


An efficient remedy is changing $$a/b = x = \prod_{i=1}^{j} a_{i}^{x_{i}/n}$$ to $$(a/b)^n = x^n = \prod_{i=1}^{j} a_{i}^{x_{i}}.$$ Then we can easily apply the unique factorization theorem to both LHS and RHS now.

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  • $\begingroup$ Thanks, that does improve it $\endgroup$
    – Sam Gregg
    Sep 29, 2017 at 14:39

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