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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.

Bill Dubuque
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Julien
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    What is "not convincing enough" with the answer already posted? It is correct.... – Mike Nov 17 '22 at 16:05
  • Or perhaps more to the point, I imagine it is hard to show a number is prime without relying on prime factorization... – Mike Nov 17 '22 at 16:21
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    "I am pretty sure that one possible solution is to resort to the prime factorization of numbers but I would also appreciate a more elegant solution that does not rely on this idea". So .... you want to prove a statement about prime factorization that does not rely upon the concept of prime factorization. – fleablood Nov 17 '22 at 16:31
  • Okay, reading the comments on the answer it seems the OP was, and still is, unfamiliar with enter link description hereEuclid's Lemma – fleablood Nov 17 '22 at 16:36
  • Ok, I forgot Euclid's lemma (or never heard of it) – Julien Nov 17 '22 at 16:39
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    Euclid's Lemma is the statement: if $p$ is prime and $p$ divides $ab$ then $p$ must divide either $a$ or $b$. It's very fundamental. – fleablood Nov 17 '22 at 16:43
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    So if $n =\frac {ab}c$ then $c =\frac {ab}n$ and $n|ab$. If $n$ is not composite than $n$ is prime. ANd if $n$ is prime then $n\mid a$ or $n\mid b$. But if $n$ divides either of those the result is a positive integer. So either $c = \frac an b; \frac ac=k$ is a positive integer $\ge 1$ or $c =a\frac bn; \frac bn=j$ is a positive integer $\ge 1$. Those contradict $c>a; c>b$. That is all in the very convincing answer. – fleablood Nov 17 '22 at 16:48
  • I suppose you also need to know that if $k$ is a positive integer then $k \ge 1$ and that if $a > 0$ then $ak \ge a$. ($m < n$ and $c>0\implies mc < nc$ is a given axiom. So if $a >0$ and $k \ge a$ then $ak \ge a\cdot 1 = a$.) – fleablood Nov 17 '22 at 16:52

3 Answers3

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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 $b>c$.

cansomeonehelpmeout
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  • I think you just write $p$ instead of $n$ Is this right ? – Julien Aug 08 '21 at 10:16
  • @Julien Yes, thank you! Fixed. – cansomeonehelpmeout Aug 08 '21 at 10:18
  • Can you justify the statement : If n were prime it divides either a or b? – Julien Aug 17 '22 at 11:55
  • @Julien Suppose it is not true for a prime $p$. Then you may write $a=a'x, b=b'y$ where $xy=p$ and $x,y<p$. This is a contradiction. – cansomeonehelpmeout Aug 17 '22 at 15:37
  • I still don't get it. I guess that one has to use the prime factorization of numbers in some form – Julien Aug 17 '22 at 16:07
  • "Can you justify the statement : If n were prime it divides either a or b? " That's Euclid's Lemma. It should be the first or second thing taught in any number theory class. – fleablood Nov 17 '22 at 16:33
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    "I guess that one has to use the prime factorization of numbers in some form" Considering "composite" means "not prime" means "not not factorable" means "factorable" it seems perverse to think you could avoid prime factorization. – fleablood Nov 17 '22 at 16:40
  • I will not accept this answer since it relies on Euclid's lemma and it did not cite this lemma explicitly – Julien Nov 17 '22 at 16:44
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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.

JMP
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  • It's not clear, can you use implications instead of "otherwise" – Julien Nov 19 '22 at 15:23
  • Is this any clearer? – JMP Nov 19 '22 at 21:36
  • How $n>\max(a,b)$ implies that $n$ is not prime – Julien Nov 20 '22 at 07:10
  • That depends on how you have defined 'prime'. $ab$ can't contain the larger $n=p$ as a factor for example. – JMP Nov 20 '22 at 07:22
  • @JMP And this last argument requires Euclid's lemma, which the OP doesn't accept as an answer ... ! (for reasons obscure to me) – Suzet Nov 21 '22 at 08:11
  • I didn't say that I don't accept Euclid's lemma as an answer. I prefered a solution that didn't require a direct use of prime factorisation. I don't know if the proof of Euclid's lemma requires prime factorisation – Julien Nov 21 '22 at 09:28
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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 $c < b$.

Suzet
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