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If $a \gt 1$ and if $a$ has the property that whenever $a|bc$ then $a|b$ or $a|c$, show that $a$ must be a prime.

Not sure how tot prove this, so I started out by checking a few values..

$2|7*4$, where $2 \nmid 7$ but $2|4$ so this $a=2=$prime checks out.

$3|7*6$, where $3 \nmid 7$ but $3|6$ so this $a=3=$prime checks out.

$5|7*15$, where $5 \nmid 7$ but $5|15$ so this $a=5=$prime checks out.

What about not a prime?

$6|4*3$, where $6 \nmid 4$ and $6 \nmid 3$ so $a$ must be a prime

Before, everyone freaks out that "proof by example, isn't a proof"... I know. This is all I could manage. I find writing proofs to be difficult.

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You just wrote what is probably the most common definition of prime element without resourcing to ideals and stuff, so: what's your definition of "prime"? – DonAntonio Dec 7 '12 at 4:07
A prime is only divisible by $\pm$ itself and $\pm 1$ – Dmitri.Mendeleev Dec 7 '12 at 4:09
up vote 4 down vote accepted

If $a$ is composite, then $a=pq$ where $1<p<a$ and $1<q<a$. But although $a$ divides $pq$, it does not divide $p$ or $q$.

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Hint $\rm\,\ bc\mid bc\:$ but $\rm\:bc\nmid b,c\:$ if $\rm\:b,c>1$.

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