I couldn't find anything to use apart from the fundamental theorem of arithmetic.

Here is my proof :

Let $a,b \in N$

Suppose $a^5 | b^5$

Let $S = \{ \text{ n is prime } , n | a \lor n | b \} $

$ P_i \in S, \alpha_i\ge0, a=P_1^{\alpha_1}.P_2^{\alpha_2}...P_k^{\alpha_k} \rightarrow a^5=P_1^{5\alpha_1}.P_2^{5\alpha_2}...P_k^{5\alpha_k} $

$P_i \in S,\beta_i\ge0, b=P_1^{\beta_1}.P_2^{\beta_2}...P_k^{\beta_k} \rightarrow b^5=P_1^{5\beta_1}.P_2^{5\beta_2}...P_k^{5\beta_k} $

$\frac{b^5}{a^5} = P_1^{5\beta_1-5\alpha_1}.P_2^{5\beta_2-5\alpha_2}...P_k^{5\beta_k-5\alpha_k} $

$a^5 | b^5 \rightarrow \frac{b^5}{a^5} \in N \rightarrow 5\beta_1-5\alpha_1 \ge 0 \rightarrow 5\beta_1 \ge 5\alpha_1 \rightarrow \beta_1 \ge \alpha_1 $

$ \frac{b}{a} = P_1^{\beta_1-\alpha_1}.P_2^{\beta_2-\alpha_2}...P_k^{\beta_k-\alpha_k} $

$ \beta_1 \ge \alpha_1 \rightarrow \frac ab \in N \rightarrow a|b $

Is the proof correct ?


Yes, your proof is correct, but it uses the unique prime factorization theorem which is pretty strong. You could use the following result:

If $q \in \Bbb{Q}$ with $q^5 \in \Bbb{Z}$ then $q \in \Bbb{Z}$. The proof is close to the classical proof of the irrationality of $\sqrt{2}$.

Suppose that $q=a/b$ with $a,b$ coprimes. If $q^5 = \pm 1$ then you are done. If not, then $a^5 = kb^5$ with $k \in \Bbb{Z}, k \neq 1$ and $k$ not a fifth power of an integer. (This is too complicated: Then $k$ has a prime factor $p$, which does not have an exponent which is a multiple of $5$. This prime number will divide $a$ and then it will divide $b$, contradicting the fact that $a,b$ are coprimes. )

Or simply if $b \neq 1$ then $b$ has a prime factor which will divide $a$ directly, so $a,b$ are not coprime.

So in your case $q = (a/b) \in \Bbb{Q}$ verifies $q^5 \in \Bbb{Z}$. Therefore $q \in \Bbb{Z}$.


Yes, you can prove it using unique prime factorizations. Simpler: clear if $\,a = 0.\,$ Else for $\, q = b/a\,$ we have $\, q^5 = n\in\Bbb Z\,$ so $\,q\,$ is a root of $\,x^5-n.\,$ By the Rational Root test, $\, q = b/a \in\Bbb Z,\,$ so $\, a\mid b.$


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