Prove $ \ \sqrt[5]{17} \notin \mathbb{Q}$

Question

Question:

Prove that: $$ \ \sqrt[5]{17} \notin \mathbb{Q}.$$

My attempt:

Assume $ \ \sqrt[5]{17} \in \mathbb{Q}$. Then $ \sqrt[5]{17} = \dfrac{m}{n}$ where $ \ m,n \in \mathbb{Z}, n\neq 0$ and $ \ m,n$ have no common factors.

$ \ \sqrt[5]{17} = \dfrac{m}{n} \implies 17 = \dfrac{m^5}{n^5} \implies 17n^5 = m^5 \implies 17 | m^5 \implies 17|m \implies \exists a \in \mathbb{Z}$ s.t $ \ m = 17a$

Then,

$ \ 17n^5 = m^5 \implies 17n^5 = 17^5. a^5 \implies n^5 = 17^4.a^5$

I am stuck here. $17^4$ is not prime. I need to show that $17 |n^5 \implies 17|n$

Answer 1

In your proof if $n^5$ divided by $17^4$ then $n^5$ divided by $17$ and from here $n$ divided by $17$, which gives the contradiction.

I like the following way.

If $\sqrt[5]{17}=\frac{m}{n}$, where $m$ and $n$ are natural numbers then $17n^5=m^5$, which says $m$ divided by $17$.

Thus, $m=17m_1$, where $m_1$ is a natural number and from here we obtain: $$n^5=17^4m_1^5,$$ which says that $n^5$ divided by $17$

and from here we obtain that $n$ divided by $17$ and $n=17n_1$, where $n_1$ is a natural number.

Thus, $17n_1^5=m_1^5$, where $m>m_1$.

Since we can repeat now this thing more and more,

we'll get an infinite series of natural numbers $\{m_i\}$ for which $$m>m_1>m_2>...,$$ which is impossible.

We got a contradiction.

Thus, $\sqrt[5]{17}$ is an irrational number.

Answer 2

If $p|x^n$ where $p$ is prime and $x,n$ are integers, then the prime factorization of $x^n$ has a $p^j$ term in it. The $n$th root of $x^n$ is an integer, so the $n$th root of $p^j$ must also be an integer (can you verify this?). Therefore $n|j$ and $j/n \geq 1$ since $p$ does not divide $1$.