If $n = \frac{ab}{c}$ with $n, a, b, c$ positive integers, $a \gt c$ and $b \gt c$. Is it true that $n$ is not prime?

Question

Is the following statement true? And if so, how can one prove it?

Let $n$ be an integer, $n \gt 1$.
Assume that there exist three postive integers $a, b, c$ with $ a \gt c$ and $b \gt c$  such that $$n = \frac{ab}{c}$$ Then n is not a prime number.

I am pretty sure that one possible solution is to resort to the prime factorization of numbers (of either $a, b$ and $c$) but I would also appreciate a more elegant solution that does not rely on this idea.

Answer 1

You can aim for a contradiction. Write $$cn=ab$$

If $n$ were prime it divides either $a$ or $b$. Without loss of generality let $n\mid a$, then $$c=\left(\frac{a}{n}\right)b$$ This is a contradiction as $b>c$.

Answer 2

As $c<\min(a,b)$, then $n\le\max(a,b)$ implies that $cn<\min(a,b)\max(a,b)=ab$.

Therefore we have that $n>\max(a,b)$

As $n$ prime implies that it is only divisible by $1$ and itself, neither $a$ or $b$ would divide $n$, therefore from

$$\frac{cn}{a}=b$$

we would have $a$ divides $c$ but $a>c$ ad so this is impossible.

Therefore $n$ is not prime.

Answer 3

The proof of Bézout's identity does not require prime factorization. A consequence of this theorem is Gauss' lemma, which is the following generalization of Euclid's lemma:

If $a,b,c$ are three integers such that $a | bc$ and $a$ is prime to $b$, then $a|c$.

Indeed, given the situation of the lemma, by Bézout's identity there exists some integers $u,v$ such that $au + bv = 1$. Multiply this identity by $c$ to deduce that $c = auc + bcv$. Now, $a$ divides $auc$ and $bcv$, so that $a$ divides their sum, that is $c$.

So far, nowhere did we use prime factorization.

Now, the proof just goes as written in the first answer. Assume towards a contradiction that $n$ is prime. If $n$ is prime to $a$ then $n$ divides $b$ by Gauss' lemma. The identity $$c = \left(\frac{b}{n}\right)a$$ then contradicts the fact that $c < a$.

Otherwise, $n$ is not prime to $a$. It means that there exists a common divisor $d>1$ to both $n$ and $a$. Since $n$ is prime, we must have $d=n$. Thus, $n$ divides $a$. The identity $$c = \left(\frac{a}{n}\right)b$$ then contradicts the fact that $c < b$.