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Let $p$ be prime, $n \in \mathbb{Z}$, $\alpha \in \mathbb{Z}^{+}$. Suppose that $n$ shares a common factor with $p$. I wish to conclude that $p|n$.

The proof follows: If $p$ is prime, then the factors of $p^{\alpha}$ are exactly given by $1, p, p^2, ..., p^{\alpha}$ (should I give any further justification for this?). In supposing that the common factor is given by $p^{j}$, then since $p|p^{j}$ and $p^{j}|n$, we have that $p|n$.

Is there a better proof for this?

Bill Dubuque
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V. Elizabeth
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    Looks good at first glance to me. I'm guessing your in a Discrete Math course, not an Abstract Algebra course. Whether you need to justify the factors will depend on context, but probably don't need it. – nickalh Sep 27 '23 at 22:46
  • You can prove it that way using unique factorization. But it's more general to use Euclid's Lemma as in the dupe, i.e. if $p$ were coprime to $n$ then so too would be $p^n$ (being a product of coprimes). You can also invoke Euclid's lemma directly: prime $,q\mid (n,p^n)\Rightarrow q\mid p^n\Rightarrow q\mid p\Rightarrow q=p.\ \ $ – Bill Dubuque Sep 27 '23 at 23:01
  • Explicitly: OP contrapositive is $,(p,n)=1\Rightarrow (p^k,n)=1,,$ which by Bezout is equivalent to $!\bmod n!:\ p^{-1}$ exists $\Rightarrow (p^k)^{-1},$ exists, true by $,(p^k)^{-1}\equiv (p^{-1})^k.,$ That's the arithmetical view behind the proof presented here in the dupe. $\ \ $ – Bill Dubuque Sep 27 '23 at 23:28
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    If the gcd is not $1$ , there must be a prime both dividing $n$ and $p^\alpha$. The only possible such prime is $p$ since $p^\alpha$ has only one prime factor. – Peter Sep 28 '23 at 06:09
  • @Peter That's the same as the "direct" proof in my first comment. – Bill Dubuque Sep 28 '23 at 17:12

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