Prove: If $p$ is prime and $p \mid a_1a_2\dots a_n$, then $p \mid a_i$ for $i\in\{1, \dots, n\}$.

Question

Euclid's lemma: If $p\mid ab$, then $p\mid a$ or $p\mid b$.

Prove: If $p$ is prime and $p \mid a_1a_2\dots a_n$, then $p \mid a_i$ for some $i = 1, 2, \dots, n$.

Proof by Induction:

P(1): From $p\in\mathbb{P}$ and $p\mid a_1$ it follows that: \begin{equation*} p \mid a_i, \end{equation*} for $i = 1$. P(1) is true.

P(2): If $p\in\mathbb{P}$ and $p\mid a_1a_2$ it follows from the lemma that if $p \nmid a_1$, then: \begin{equation*} p \mid a_2, \end{equation*} so \begin{equation*} p \mid a_i \end{equation*} for some $i = 1, 2$.

On the other hand, if $p\mid a_1$, then the same conclusion is reached. P(2) is true.

Now assume P(k) inductively: \begin{equation*} p \mid a_1a_2\dots a_k \rightarrow p \mid a_i \end{equation*} for some $i = 1, 2, \dots, k$.

P(k+1) is also true because from $p \mid (a_1a_2\dots a_k)a_{k+1}$ two cases arise:

Case I: $p\nmid a_1\dots a_k$.

Using Euclid's lemma it follows that if $p\nmid a_1\dots a_k$, then: \begin{equation*} p\mid a_{k+1} \end{equation*} and thus: \begin{equation*} p\mid a_i, \end{equation*} for some $i = 1, 2, \dots, k+1$.

Case II: $p\mid a_1\dots a_k$.

By the inductive hypothesis, the conclusion may be reached: \begin{equation*} p \mid a_i, \end{equation*} for some $i = 1, 2, \dots,k, k+1$.

Answer 1

Here is your argument written more concisely/clearly:

For all arguments for $n=1$ case is trivial.

If $p \mid a_1\cdots a_n \implies p \mid a_{i}$ for some $n$ then suppose $ p \mid a_1\cdots a_na_{n+1}=(a_{1}\cdot\cdot\cdot a_{n})a_{n+1}$

By the $n=2$ case we must have that $p$ divides atleast one of them.

Then if $p \mid (a_1\cdots a_n)$ we are done by the inductive hypothesis, and if $p \nmid (a_1\cdots a_n)$, $p \mid a_{n+1}$, and the $i$ that satisfies the implication is $n+1$.

Your proof does not use the inductive hypothesis because you are using using an additional $\gcd$ result based on properties of primes. This not necessary as you may use the $n=2$ case of the theorem known as Euclid's lemma. Although, it is a similar result.

The issue is, you are having to note basic results and insights when you could have just proven the $n=2$ case and done a standard induction.

Below are my initial proofs:

One proof is to use strong induction on $n$ and once again Euclid's lemma for the $n=2$ case.

Then $p\mid a_1\cdots a_n \implies p\mid(a_1\cdots a_{n-1})a_n$. If $p \nmid a_n, $then $p\mid(a_1\cdots a_{n-1})$.

You may fill in the details, but by repeating this process of moving the paranthesis inductively, one finds that $p\mid a_i$ for some $i$.

Also, consider the proof where we start with $a_{1}\cdot\cdot\cdot a_{n}=pk$ for some integer $k$.

Then by the Fundamental Theorem of Arithmetic, $p$ must be in the prime expansion of the $LHS$, and hence in the prime expansion of at least one of the $a_{i}$'s.