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I'm working out of the text "Elementary Number Theory" by James K. Strayer and I've run into the following problem:

Find integers $a$, $b$, and $c$ such that $a \mid bc ~$ but $a \nmid b \space $ and $a \nmid c$

When I first saw this problem I thought it was weird because I thought for sure I've seen a theorem somewhere that says that this would never happen. The only thing is that usually if a question isn't valid they say "Prove or Disprove", which makes me think I'm overlooking the obvious. Is there an easy solution to this problem that I'm overlooking? If not, how do you prove its falsity. Thanks

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$\begin{eqnarray}\rm{\bf Hint}\ \ To\ solve\ \ \rm a\mid bc,&&\rm\ a\nmid b,c\\ \rm notice\ that\quad\ \ \ bc\mid bc,&&\rm\! bc\nmid b,c\ \ for\ \ b,c > 1\end{eqnarray}$

Remark $\ $ It shows: reducible elements are not prime; or, contrapositively, primes are irreducible. See here for further discussion.

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  • $\begingroup$ This makes sense, thanks! $\endgroup$ – Amateur Math Guy Feb 10 '13 at 3:32
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Hint: Consider integers $b, c,\;\; b\neq 1, c\neq 1$ such that $a = bc$. Then then clearly $a = bc\nmid b$ and $a = bc\nmid c,\;$ while, of course, $a = bc\mid bc$.

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Hint: A number $a$ is said to be prime if whenever $a \mid bc$, either $a \mid b$ or $a \mid c$

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Hint: $6 \mid 6$.

The two theorems you might have seen are:

T If $a$ is prime and $a \mid bc$ then $a \mid b$ or $a \mid c$.

OR

T If $\gcd(a,b)=1$ and $a \mid bc$ then $a \mid c$.

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  • $\begingroup$ Yeah I was thinking of the first one involving primes $\endgroup$ – Amateur Math Guy Feb 10 '13 at 3:33
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EXist. let a = 6, b=3 and c=4 then b.c=3.4=12. we see that 6 divides 12 but, 6 not divides 3 and 6 not divides 4.

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