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How to find $3$ prime numbers $a,b$ and $c$ such that: $$a| (bc+b+c)$$ $$b|(ac+a+c)$$ $$c|(ab+a+b)$$

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Do you want distinct primes? Else $a=b=c=\text{prime}$ will do the job. –  user17762 Jun 30 '12 at 20:18
    
3 distinct prime numbers –  Frank Jun 30 '12 at 20:31
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up vote 11 down vote accepted

There are no such primes:

Without loss of generality assume $a<b<c$. Clearly 2 cannot be one of the primes, so $a\geq 3$ , $b\geq 5$ and $c\geq 7$.

Now $abc+ab+bc+ca+a+b+c$ is divisible by $a$, by $b$ and by $c$ so it is also divisible by $abc$,

but this is impossible since $1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}\leq\frac{4}{3}\frac{6}{5}\frac{8}{7}<2$

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