Proof that if $p$ divides $bc$, then $p$ divides $c$ or $p$ divides $b$

Question

I came up with the following as an answer to a question in a book. I think there is a problem with this answer but I can't figure out what. The solutions have a very different proof. Can you help me identify my mistake?

I note that, in previous questions I proved that ever positive integer has at least 1 prime divisor and every positive integer can be written as the product of prime divisors and the number 1.

Here is my proposed proof:

Let $p$ be a prime that divides $bc$. Note that $bc = pm$ for some positive integer $m$. If $b$ and $c$ are prime and the theorem doesn't hold, $p$ does not divide $bc$, which is a contradiction.

Let $S_b$ be the set of primes that divide $b$ and let $S_c$ be the set of primes that divide $c$. If $p$ does not divide $a$ or $b$, then $p \notin S_a$ and $p \notin S_b$. Now, let $P_1\cdot...\cdot P_n$ and $Q_1\cdot...\cdot Q_k$ be prime factorizations of $b$ and $c$ respectively. Then $(P_1\cdot...\cdot P_n)(Q_1\cdot...\cdot Q_k) = c = pm$. So we have $m = \frac{(P_1\cdot...\cdot P_n)(Q_1\cdot...\cdot Q_k)}{p}$. The LHS is an integer. However, since $p$ does not divide any of $P_1,...,P_n,Q_1,..., Q_k$, the RHS is not an integer. This is a contradiction

Answer 1

You say ,“However, since $p$ does not divide any of $(P_1,...,P_n),(Q_1\cdot...\cdot Q_k)$, the RHS is not an integer. This is a contradiction”
How do you prove that $p$ does not divide all them multiplied together? You can use unique factorisation but then how do you prove unique factorisation? Most proofs of unique factorisation use this result in the proof so it would be circular …

Answer 2

By contradiction. Suppose $b_0$ is the least $b\in\Bbb N$ for which there exists $c\in \Bbb N$ such that $p|bc$ but $p\not |b$ and $p\not |c.$ So take $c_0\in \Bbb N$ such that $p|b_0c_0$ but $p\not |b_0$ and $p\not |c_0.$

Step 1. We prove by contradiction that $b_0<p.$

Suppose $b_0\ge p.$ Then there exists a unique $m\in \Bbb N$ such that $pm\le b_0<p(m+1)$ and we must have $pm<b_0$ (otherwise $p| pm=b_0$). But then $b_0> b'=^{def} b_0-pm\in\Bbb N$ with $p|(b_0c_0-pmc_0)=b'c_0,$ so the minimality of $b_0$ (in the definition of $b_0$) requires $(p|b'\lor p|c_0).$ And we have $p\not |c_0.$ So $p|b'=b_0-pm,$ implying $p|b_0,$ contrary to the initial hypotheses.

Step 2. Since $b_0<p$ here exists a unique $n\in \Bbb N$ such that $b_0n\le p<b_0(n+1).$ More strongly, $$0<b_0n<p<b_0(n+1).$$ Because if $b_0n=p$ then, as $p$ is prime, either $n=1$(implying $p|p=b_0n=b_0$) or $b_0=1$ (implying $p|b_0c_0=c_0$). So $b_0>b_1=^{def} p-b_0n\in\Bbb N.$

Now $b_1c_0=(p-b_0n)c_0=p\cdot c_0 -n(b_0c_0)$ is divisible by $p,$ but $0<b_1<b_0,$ so the minimality of $b_0$ (in the definition of $b_0$) requires $(p|b_1\lor p|c_0).$ And we have $p\not |c_0.$ So $p|b_1.$

But then $p|(p-b_1)=b_0n$ with $0<b_0n<p,$ which is absurd. QED.

Answer 3

I assume unique factorisation in a product of primes is already established.

We can write $a=pq_1 \dots q_k$ for primes $q_1, \dots, q_k$. We can also write $b= s_1\dots s_m$ and $c= s_{m+1} \dots s_{m+n}$ where all the $s_i$'s are prime numbers. Then we see that $$pq_1\dots q_k = s_1 \dots s_{m+n}$$ and since prime factorisations are unique, there is $i \in \{1, \dots, m+n\}$ such that $p = s_{i}$. But if $i \le m$, then $p$ divides $b$. Otherwise, if $i > m$, then $p $ divides $c$.