Let $1≤n <m$ be two fixed integer.
(a) Prove that if $a \mid b$, then $a^n \mid b^m$.
(b) Prove that $ \exists$ positive integers $a,b$ where $a \nmid b$ but $a^n \mid b^m$
I can get the answer for part (a), but I dont have any ideas about part (b). Can someone give me some hints for this question? Thanks!
$(a)$ We have $a \mid b \implies a^n \mid b^n$. Since $n<m$, we also have $b^n \mid b^m$. This gives us $a^n \mid b^m$.
$(b)$ Just set $a=x^m$ and $b=x^n$ for some positive integer $x>1$. We can clearly see that since $m>n$, $$ x^m>x^n \implies x^m \nmid x^n \implies a \nmid b$$ But we have $a^n=b^m=x^{mn}$ which clearly shows that $a^n \mid b^m$ since every number divides itself.