Solution 34acd865-8ff8-4a97-a8d5-d74ce32f095a

Points: 2 pts

Lemma 1: If $p\mid n$, then $n$ has two prime factorizations

By definition, $p\mid n$ implies
\[
n=pm
\]
for some integer $m$. We can then prime factorize $m$ as
\[
m=q_1^{b_1}q_2^{b_2}\cdots q_k^{b_k}.
\]
Therefore,
\[
n=p\left(q_1^{b_1}q_2^{b_2}\cdots q_k^{b_k}\right).
\]
Since
\[
p\notin\{p_1,p_2,\ldots,p_r\},
\]
this gives a prime factorization different from the original prime factorization
\[
n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}.
\]
Thus, if $p\mid n$, then $n$ has two different prime factorizations. This contradicts the fundamental theorem of arithmetic, which says that the prime factorization of $n$ is unique.

Final answer:
p\nmid n