Prove that if $b^m$ divides $a^n$ and $m>n$, then $b$ divides $a$.

Question

Help! I've tried over and over to solve this and can't quite get it...

Let $n>0$. Prove that if $b^m$ divides $a^n$ and $m>n$, then $b$ divides $a$.

Answer 1

HINT: Use the prime decompositions of $a$ and $b$. If $p_1,\ldots,p_r$ is a list of all of the prime factors appearing in either $a$ or $b$, we can write $b=p_1^{k_1}\ldots p_r^{k_r}$ and $a=p_1^{\ell_1}\ldots p_r^{\ell_r}$ for some non-negative integers $k_1,\ldots,k_r,\ell_1,\ldots,\ell_r$. (We have to allow for the possibility that some $k_i$ or $\ell_i$ are $0$; why?)

• For $i=1,\ldots,r$, what is the exponent of $p_i$ in $b^m$? In $a^n$?
• What does the hypothesis that $b^m\mid a^n$ tell you about the relative sizes of the exponents that you calculated for the previous point?
• Now use the hypothesis that $m>n$ to show that $b\mid a$.

Answer 2

In fact this question provides an answer already:
If $m\gt n,$ and $b^m\mid a^n,$ then also $b^n\mid a^n,$ so, by the linked question, $b$ divides $a.$

Hope this helps.