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Hey, I'm a little stumped on this basic number theory question. The solution is probably staring me in the face, but for some reason it's eluding me...

If [tex]a^n | b^n[/tex], prove that [tex]a|b[/tex].

So, say that [tex]a^n \cdot x = b^n[/tex] for some integer [tex]x[/tex], there's not a lot I can go to from there. We do get that [tex]a | b^n[/tex] and that [tex]a^k | b^n[/tex] for all [tex] k \leq n[/tex], but I can't find a way in which that's useful.

I also tried using induction on [tex]n[/tex]. The base case is trivial. For the inductive case, assume that [tex]a^n | b^n[/tex] implies that [tex]a | b[/tex]. Then we must prove that [tex] a^{n+1} | b^{n+1} [/tex] implies that [tex] a|b[/tex]. So assume that [tex] a^{n+1} | b^{n+1}[/tex]. Then we get [tex]a^n | b^{n+1}[/tex], but I actually would need to show that [tex]a^n | b^{n} [/tex] and I can't figure out how to get there.

Am I totally missing something? I think I'm overthinking this. Nothing is coming. I'd appreciate any ideas on what to try next. Thanks.

If [tex]a^n | b^n[/tex], prove that [tex]a|b[/tex].

So, say that [tex]a^n \cdot x = b^n[/tex] for some integer [tex]x[/tex], there's not a lot I can go to from there. We do get that [tex]a | b^n[/tex] and that [tex]a^k | b^n[/tex] for all [tex] k \leq n[/tex], but I can't find a way in which that's useful.

I also tried using induction on [tex]n[/tex]. The base case is trivial. For the inductive case, assume that [tex]a^n | b^n[/tex] implies that [tex]a | b[/tex]. Then we must prove that [tex] a^{n+1} | b^{n+1} [/tex] implies that [tex] a|b[/tex]. So assume that [tex] a^{n+1} | b^{n+1}[/tex]. Then we get [tex]a^n | b^{n+1}[/tex], but I actually would need to show that [tex]a^n | b^{n} [/tex] and I can't figure out how to get there.

Am I totally missing something? I think I'm overthinking this. Nothing is coming. I'd appreciate any ideas on what to try next. Thanks.

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