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As the title suggests:

Prove that for any natural number, if it is a perfect square and a perfect cube, it is also a perfect sixth power.

For some reason I just have hit a road block on this question. Any help would be appreciated.

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    $\begingroup$ What have you tried? Can you convert the sentences into mathematical statements? $\endgroup$ – Calvin Lin Oct 25 '13 at 0:58
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Suppose that $n$ is an integer such that $n = a^2 = b^3$ for some integers $a, b$.

Let $p$ be any prime.

How many times does $p$ divide $n$?

Can you show that it must be a multiple of $6$?

Hence conclude that $n$ must be of the form $c^6$, i.e. a perfect sixth power.

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Without los of generality we can assume that $a,b >0$.

Let $x= \sqrt[6]{n}$. Then $x^2=b$ and $x^3=a$. Thus

$$x=\frac{x^3}{x^2}=\frac{a}{b}\,.$$

As $x$ is rational, we can write it in the reduced form $x=\frac{c}{d}$ with $\gcd(c,d)=1$. Then $x^6=\frac{c^6}{d^6}=n$ implies that $d^6|c^6$.

Now prove that $\gcd(c^6,d^6)=1$ to conclude that $d^6=1$, which shows that $x$ is an integer.

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Consider the prime factorization of your number.

If it a perfect square, what do you know about the exponent on each of the prime factors?

If it a perfect cube, what do you know about the exponent on each of the prime factors?

So, now what combined simple fact do you know about the exponent on each prime factor?

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