# 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?

• You can show that if $n$ is a perfect cube, then so is $\sqrt n$.
– MJD
Commented Jul 20, 2015 at 0:04
• Commented Jul 20, 2015 at 7:41
• It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself? Commented Jul 20, 2015 at 8:56
• More generally see here Commented Oct 29, 2023 at 22:06

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.

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. $\mathbb{C}[t^2, t^3]$.

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.

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 $$(b^m)^n=b^{mn}$$.

$$(n^3)^2=(n^2)^3=n^6$$