a natural number that is both a perfect square and a perfect cube is a perfect sixth power? I really can't get a grasp on how to prove this, because if $x$ = $\sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help? 
 A: Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
A: This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write 
 $$n = p_1^{e_1}p_2^{e_2} \ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} \ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $\mathbb{C}[t^2, t^3]$.
A: The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$, 
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is 
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
A: The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, \cdots, n^6
$$
for $n\in\mathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
