# Does $a^3 = b^2$ imply $a$ is a perfect square? [duplicate]

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.

• 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 $(2^3)^2 = 64$. Oct 12, 2015 at 21:01

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 $\nu_p(a)$?

• 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 $p^{2k}$? Oct 12, 2015 at 14:22
• 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 Oct 12, 2015 at 14:36

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 $$a=c^2$$

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

• Shouldn't the argument be extended to show that for all prime numbers such that $p|a$, one has $p^{2k}$ Oct 12, 2015 at 14:07
• 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 $b$? Oct 12, 2015 at 14:14
• 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? Oct 12, 2015 at 14:28