# If $x$ is a perfect square and a perfect cube, is $x$ a perfect power of $6$?

Given $a,b$ are natural numbers s.t.

$a^2 = y$ and $b^3 = y$,

is it true there must exist a natural number $c$ such that

$c^6 = y$?

• Hint: consider the unique factorization of $y$, $y=\prod p_i^{a_i}$ . Show that each $a_i$ is divisible by both $2,3$. – lulu Sep 11 '17 at 0:03
• Hello and welcome to math.stackexchange! That's a nice question. Please tell the reader what the prigin of th question is and also what your work on this problem is so far. – Hans Engler Sep 11 '17 at 0:20
• The title should ask not "is $x$ a perfect power of 6" but rather "is $x$ a perfect 6th power." – John Blythe Dobson Sep 11 '17 at 21:14

## 2 Answers

The answer is yes.

Set $c = \frac{a}{b}$. Then $$c^6 = \frac{a^6}{b^6} = \frac{y^3}{y^2} = y \, .$$ It remains to show that $c$ is an integer. If that were not the case, then there would exist a prime number $p$ that divides one of the numbers $a, b$ but not the other one, e.g. $p \mid a$ and $p\nmid b$. But then $p \mid a^2 = y$ and $p \nmid b^3 = y$ which is silly. Therefore $c$ is an integer.

Yes. $$a^2 =b^3=y \implies a=\sqrt{b^3}=\sqrt{y}$$

Since $a$ is a natural number, $\sqrt{b^3}$ is as well.

I guess the next step is to show that if $b^3$ is a square, then $b$ is a square.

Then the number $z$ such that $z^2=b$, is the number that $z^6=y$.