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The statement is: Let $q$ be an integer. Suppose that $q \geq 2$, and that for any integers $a$ and $b$, if $q|ab$ then $q|a$ or $q|b$. Prove that $q$ is a prime number. This is my proof:

Suppose $q$ is not prime. Then there exist integers $n$ and $k$ such that $q = nk$, satisfying $1 < n, k < q$. Now $q|q$ holds. Thus $q|nk$ By hypothesis, $q|n$ of $q|k$. In either case, we get a contradiction because $n, k < q$. Therefore $q$ is prime.

My friend gave me the numbers $q = 20$ and $ab = 4000$. It seems to be a counterexample. Any explanation would be appreciated.

Bill Dubuque
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IamRigour
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  • Did your friend say what $a$ and $b$ is? It is not a counterexample. Your statement is about arbitrary $a$ and $b$ with this property, not only for $ab=4000$, to test $q=20$. "For any integer" means for all here. Take $a=2$ and $b=10$ for example. Then your criterion fails, so that $q=20$ is not prime. – Dietrich Burde Feb 02 '25 at 09:57
  • You already sketched the standard linked proof that a composite fails Euclid's Lemma, so simply apply it to the composite $20$ to get some values $,a,b,$ where Euclid fails. – Bill Dubuque Feb 02 '25 at 10:22

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Your proof is correct. $q=20$ and $ab=4000$ is not a counterexample. You proved that if the integer $q\geq 2$, and it satisfies the property $$\forall a,b\in\mathbb{Z},\text{ if }q\mid ab,\text{ then }q\mid a \text{ or } q\mid b$$ then $q$ must be prime.

Thus, a valid counterexample consists of a non-prime $q$ such that the property holds. Even if we fix $ab=4000$ and ignore the $\forall$, the statement does not hold when $q=20$. Take $a=250$ and $b=16$. $20\nmid 250$ and $20\nmid 16$, but $20\mid 4000$.

In fact, $a$ and $b$ can be any integers. For the property to hold, $q\mid ab \Longrightarrow q\mid a \text{ or }q\mid b$ must be true for every pair of integers $a$ and $b$. Taking $a=4$ and $b=5$ shows that $q=20$ fails to satisfy the property. So it is not a counterexample.

Christina 2005
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    This question seems not to meet the standards for the site (it's a FAQ with many duplicate targets). Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments suggesting improvements. Please also read the meta announcement regarding quality standards. – Martin Brandenburg Feb 02 '25 at 11:22