Generalising a question about divisibility of two numbers raised to a power

Question

Let $x$ and $y$ be integers greater than or equal to two. It is fairly straightforward to prove that if $x^3|y^2$ then $x|y$. I am interested in a more general question: when does $x^n| y^m$ imply that $x|y$ for $n,m\in\mathbb{N}$ with $n\geq m$? If we choose $n=4$ and $m=5$ then the statement is clearly false. But I can't figure out how to prove such a statement.

Answer 1

You might as well just consider the case where $x$ and $y$ are powers of the same prime $p$. Cases where they have more factors can be reduced to this. Let $x=p^j, y=p^k$ then $x^{n}=p^{nj}, y^m=p^{mk}$ and $x|y$ when $j \le k$. $x^m|y^n$ when $mj \le nk$ so any time $m \ge n$ you will have $x^n|y^n \implies x|y$

Answer 2

Simple enough.

Let $x = \prod p_i^{k_i}$ be the prime factorisation of $x$.

Then $x^m|y^n$ means $p_i^{m\cdot k_i}|y^n$ so $p_i|y$ for each prime factor $p_i$ of $x$..

Let $l_i$ be "the power of $p_i$ in $y$". i.e. Let $p_i^{l_i}|y$ so that $p_i^{l_i + 1} \not \mid y$.

Then $p_i^{m\cdot k_i}|p_i^{n\cdot l_i}$ so $m\cdot k_i \le n\cdot l_i$ so $k_i \le \frac nm l_i$.

So $\ldots$ t

If $x^m|y^n$ then the prime factors of $x$ are also prime factors of $y$ and the powers of those prime factors are equal or less than $\frac nm$ the powers of the prime factors in $y$. Or the powers of the prime factors of $y$ are at leas $\frac mn$ of those in $x$.

In particular if $n \le m$ then $x|y$. But if $n > m$ we have limits of the powers of the prime factors.

Example if $54^5|y^7$ then as $54 = 3^3\cdot2$ then $3^{\lceil \frac 57\cdot 3\rceil} =3^3 |y$ and $2^{\lceil \frac 57\cdot 1\rceil} =2 |y$ so $54|y$ but if $54^5|w^8$ then $3^{\lceil \frac 58\cdot 3\rceil} =3^2 |w$ and $2|w$ so $18|w$ and indeed if $w = 18$ we have $18^8 = 3^{16\cdot2^8} = (3^3\cdot 2)^5\cdot(3\cdot2^3)= 54^5\cdot24$.