I know this is about looking at the unique prime factorizations for each variable but I keep getting mixed up when it comes to showing that the factorizations are the same.
EDIT: Sorry I wrote the wrong exponent for y the first go around.
If for all primes $p$ which divide $n$, $p^6$ also divides $n$, then there exists positive integers $x,y$ such that $n=x^3y^4$
0
$\begingroup$
elementary-number-theory
1 Answers
2
I think this is not necessarily true. For example take $n=128$. The only prime $p$ which divide $n$ is $p=2$, and $p^6=64$ also divides $n$. But there do not exist such positive integers $x,y$.
Edit (for the updated problem): Write $n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k}$. You know that $e_i\geq 6$ for all $i$. Now, all integers $m\geq 6$ can be written as $3a+4b$, where $a,b\geq 0$ are integers. (Why?) So suppose that $e_i=3a_i+4b_i$. Then let $x=\prod_i p_i^{a_i}$ and $y=\prod_i p_i^{b_i}$.
-
1As a hint to the OP, try showing that three consecutive numbers (i.e. $6,7,8$) can be written as $3a+4b$. Then for any positive integer greater than $8$, you can take one of these cases and add to $a$ as necessary. – 2017-02-22