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Suppose I have $a, b \in \mathbb Z^+$ such that $a^3 = b^2$. How does this imply that $a$ is a perfect square, i.e. $\exists e \in \mathbb Z^+$ such that $a = e^2$ ? I know I have to use prime factorization somewhere in my argument, but I'm stuck.

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  • $\begingroup$ I'm not sure that you have to use prime factorization. A basic property of exponents might suffice. Let's say $a^3 = c^6$. Then $c^3 = b$ and $b^2 = (c^3)^2 = a^3$. For example, $4^3 = 8^2$, and indeed $(2^3)^2 = 64$. $\endgroup$
    – Bob Happ
    Oct 12, 2015 at 21:01

3 Answers 3

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Hint. So, start writing down $a$'s and $b$'s prime decomposition, say $$ a = \prod_p p^{\nu_p(a)}, \quad b = \prod_p p^{\nu_p(b)} $$ Now $a^3 = b^2$ reads $$ \prod_p p^{3\nu_p(a)} = \prod_p p^{2\nu_p(b)} $$ Uniqueness gives you $$ \forall p: 3\nu_p(a) = 2\nu_p(b) $$ What does this tell you about the parity of $\nu_p(a)$?

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  • $\begingroup$ I don't quite understand the notations used. I have just learnt about the construction of integers and just started discussing on prime numbers. Intuitively, i see this as saying that the prime factors of $a$ are even, but how do i know that every prime factor $p$ of $a$ is of the form $p^{2k}$? $\endgroup$
    – lemonCylon
    Oct 12, 2015 at 14:22
  • $\begingroup$ I mean to say there for all prime factors $p_i$ of $a$ such that $a$=$p_1p_2...p_n$, we have $n$ to be even $\endgroup$
    – lemonCylon
    Oct 12, 2015 at 14:36
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Divide both sides by $a^2$, you get $$a=\frac{b^2}{a^2}=\left (\frac ba \right)^2$$ you say it's integer ,it's rational , it can only be integer if $a$ divides $b$ . But if $a$ divides $b$ then so do all prime powers dividing $a$ . Further, it means $b=ac$ . Rewriting the equation gives:$$a^3=a^2c^2$$ or $$a=c^2$$

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The prime factorization of $a=p_1^{n_1}p_2^{n_2}......p_n^{n_n}$

$\implies a^3=p_1^{3n_1}p_2^{3n_2}......p_n^{3n_n}$

Since, $a^3=b^2 \implies a$ and $b$ will have same primes

$\implies b =p_1^{m_1}p_2^{m_2}......p_n^{m_n}$

$\implies b^2 =p_1^{2m_1}p_2^{2m_2}......p_n^{2m_n}$

As, $a^3 =b^2 \implies p_i^{3n_i}= p_i^{2m_i} $, for $i \in \{1,2,....,n\}$

$\implies 3n_i=2m_i \implies n_i=2t_i$

$\implies a=p_1^{2t_1}p_2^{2t_2}......p_n^{2t_m}$

So, $a$ is a perfect square

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  • $\begingroup$ Shouldn't the argument be extended to show that for all prime numbers such that $p|a$, one has $p^{2k}$ $\endgroup$
    – lemonCylon
    Oct 12, 2015 at 14:07
  • $\begingroup$ Also, from $p^{3n}|b^2 \implies p^{3n/2}|b$, how do we ensure that $p^{3n/2}$ is an integer such that it divides $b$? $\endgroup$
    – lemonCylon
    Oct 12, 2015 at 14:14
  • $\begingroup$ Using $a=9$ and $b=27$, i have $9^3=27^2$ and $3$ is a prime such that $3|9 \implies 3^3|9^3 \implies 3^3|27^2 \implies 3^{3/2}|27 $, but $3^{3/2}$ is not an integer am i doing something wrong? $\endgroup$
    – lemonCylon
    Oct 12, 2015 at 14:28

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