The converse of Euclid lemma : counter example?

Question

Converse of Euclid lemma : if (p|ab implies p|a or p|b), then p is prime, where p,a,b are integers. But counter example that I have, 4|8*2 and 4|8 and 4∤2, then 4 should be a prime number, but it is a composite number. Please let me know, if it is wrong counter example and why?

Answer 1

Euclid's lemma states that if $p$ is prime, then $p \mid ab$ implies $p \mid a$ or $p \mid b$.

The lemma makes absolutely no comment on the case when $p$ is composite. In fact, the statement can be both true or false depending on the composite number.

With $p = 4$, $a = 8$ and $b=2$ (your example), $p \mid ab$ and $p \mid a$ so the statement holds.

But if I took $p=10$, $a=4$ and $b=15$, $p \mid ab$ but $p \nmid a$ and $p \nmid b$, so the statement doesn't hold.

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The converse of Euclid's lemma states that if for all $a,b > 1$, $p \mid ab \implies p \mid a$ or $p \mid b$, then $p$ is prime. The example you give is only for specific values of $a$ and $b$. You need to show the implication holds for all $a$ and $b$.

Answer 2

If $p$ is some number that can divide another number $a$, $b$, or both, that fact has no bearing on whether $p$ is prime. All we can say is that $p$ is a divisor of $a$ or $b$. Observe also that the divisibility of the product $ab$ is irrelevant; if $p$ divides one of $a,b$, then, of course, it must also divide their product since divisibility is transitive.

As an example, $p=2$ does divide one of the numbers $a=4$ and $b=5$ (and then necessarily also divides their product $ab=4\times5$). And here, we have $p$ prime. But then scale everything up by $\times3$, so that we have $p=6$ dividing one of $a=12$ and $b=15$. The system is symmetric but now $p$ is no longer prime.

The relevance of primality to Euclid's lemma is that prime factors are the atomic building blocks of the natural numbers and their products. A composite number dividing a product need not be a factor of any individual in that product since its prime factors could be distributed among different parts. For instance, $6$ divides $12=4\times3$ but does not divide either $4$ or $3$, since its prime factors are split up between them. Euclid's lemma relates to the prime factors themselves, which can only go in one place.