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Let $a_1,a_2 \in \mathbb{N}$ = $\{1,2,3...\}$ and let p be a prime. If $a_1$ and $a_2$ are co-primes to p, does the equation:

$r = \frac{a_1a_2}{p}$ have a solution in $\mathbb{Z}$ here? And how do I sufficiently answer that it hasn't?

Mathaniel
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    So $p$ is a prim not being a factor of $a_1$ or $a_2$? Then you can't simplify the ratio $a_1a_2/p$ which means it's a non-integer. – skyking Mar 17 '17 at 12:19
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    $p\mid ab\Leftrightarrow p\mid a$ or $p\mid b$. But you say that $p$ divides neither $a_1$ nor $a_2$. – Mathematician 42 Mar 17 '17 at 12:20
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    Here's one way: (1) assume that such an $r$ exists; (2) write the equation as $rp=a_1a_2$;(3) Then $p$ is a factor of the product $a_1a_2$; (4) Then $p$ must be one of the primes in the unique factorization of $a_1a_2$;(5) But unique factorization implies that $p$ must be one of the prime facotrs of $a_1$ or $a_2$, contrary to the condition that $a_1,a_2$ are coprime to $p$. – quasi Mar 17 '17 at 12:26

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