2
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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 $q \in \Bbb{Z}$.

$\endgroup$
0
$\begingroup$

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 $\, a\mid b.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.