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How can I use the method of infinite descent to prove that if $n$ is a positive integer and $n$ is not a perfect cube, then $\sqrt[3]{n}$ is an irrational number. This question originates from a problem in an elementary number theory book. The problem is: show that if $n$ is a positive integer and $n$ is not a perfect square, then $\sqrt{n}$ is an irrational number. The proof in the book is given like this:
Proof. We present the proof by contradiction. Suppose $\sqrt{n}=\frac{p}{q}$, where $p$ and $q$ are positive integers. Thus we get that $p^2=nq^2$. Due to $n$ is not a perfect square, there exists a positive integer $m$, such that $m<\frac{p}{q}<m+1$, namely $0<p-mq<q$. We next subtract $mpq$ from both sides of the equation $p^2=nq^2$ to obtain $p^2-mpq=nq^2-mpq$. The equation is equivalent to the following equation: $\frac{p}{q}=\frac{nq-mp}{p-mq}$. Let $p_1=nq-mp$, $q_1=p-mq$. Hence $q_1$ is a positive integer and $q_1<q$. So, we get that $p_1$ is also a positive integer and $p_1<p$. As a result, we get that $\frac{p}{q}=\frac{p_1}{p_1}$ with $p_1<p$ and $q_1<q$. By the well-ordering property, we know that among above positive fractions whose numerators and denominators are positive integers, there is a fraction with the smallest value of the numerator. However, we have shown that from this fraction we can find another fraction with a smaller value of the numerator, leading to a contradiction. This completes the proof by the method of infinite descent. I want to prove the question I raised in this way. Suppose $\sqrt[3]{n}=\frac{p}{q}$, where $p$ and $q$ are positive integers. Thus we get that $p^3=nq^3$. Due to $n$ is not a perfect cube, there exists a positive integer $m$, such that $m<\frac{p}{q}<m+1$, namely $0<p-mq<q$. But in next step, I can't figure out what polynomial should I subtract. How could I continue…

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  • $\begingroup$ Generalize it with prime factors. If $n$ is not a perfect cube there is a prime factor $p$ whose highest power dividing $n$ is not a multiple of three. So if $p^{3k+j}|n$ then $p^j|\frac n{p^{3k})$ so .... $\endgroup$
    – fleablood
    Jun 4 '20 at 3:44
  • $\begingroup$ I think the direct generalization of this method might not be appliable for irratioality of $\sqrt[3]{n}$. $\endgroup$
    – dust05
    Jun 4 '20 at 4:45
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    $\begingroup$ Do you need to use infinite descent in your proof? If you believe the Fundamental Theorem of Arithmetic, then a proof can be given in fewer than a dozen lines. $\endgroup$
    – Lubin
    Jun 4 '20 at 6:04
  • $\begingroup$ Yes, I know the Fundamental Theorem of Arithmetic will make the proof much simpler, but I need to use infinite descent in my proof. $\endgroup$
    – Bob
    Jun 4 '20 at 6:46
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It seems that you've asked this question twice, though the previous version was closed as you didn't show your attempts. In any event, I'm reposting what I wrote in response to your previous question:

This is basically the same way that you would prove $\sqrt{2}$ is rational. Let us assume $n^{1/3} = a/b$ for some integers $a$ and $b$, and further assume that $a$ and $b$ are as small as possible. Then $$n = a^3 / b^3 \iff n b^3 = a^3$$ Suppose the unique prime factorization of $n$ is $p_1^{k_1} \cdot p_2^{k_2} \cdots p_m^{k_m}$. Then it must be the case that $p_1^{q_1} \cdot p_2^{q_2} \cdots p_m^{q_m}$ divides $a$, where $q_i = \lceil k_i / 3\rceil$; otherwise $a^3$ could not be a multiple of $n$ as suggested. Since by assumption $n$ is not a perfect cube, at least one of the $k_i$ must not be a multiple of $3$. WLOG, let $k_1$ be this number that is not a multiple of $3$. Let us count factors of $p_1$ on both sides of the equation $nb^3 = a^3$. On the left-hand side, we have $k_1$ factors from $n$, and some factors from $b^3$ (we'll come back to this later). On the right-hand side, we have $3 q_1 = 3 \lceil k_1 / 3\rceil$, which is either $3 k_1 + 1$ or $3 k_1 + 2$ factors, depending upon whether $k_1$ was $1$ or $2$ modulo $3$. Of course, for the equation $nb^3 = a^3$ to hold, the number of factors of $p_1$ must be the same on each side. Therefore, it must be the case that $p_1$ divides $b^3$ (otherwise we simply do not have enough factors of $p_1$ on the left-hand side to make up for the number of $p_1$'s we have on the right-hand side). However, if $p_1$ divides $b^3$, then $p_1$ must divide $b$ since $p_1$ is a prime. But $p_1$ divides $q$ too, and as $p_i$ is a prime (and hence at least two), it follows that $$n = \frac{(a / p_1)^3}{(b / p_1)^3} = \frac{(a')^3}{(b')^3} \implies n^{1/3} = \frac{a'}{b'}$$ with $a'$ and $b'$ integers. Note also that $a' < a$ and $b' < b$, so we have constructed a fraction equal to $n^{1/3}$ with a strictly smaller numerator and denominator. Hence, $n^{1/3}$ must be irrational. $\square$

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