How to find $3$ prime numbers $a,b$ and $c$ such that: $a| (bc+b+c)$ $b|(ac+a+c)$ $c|(ab+a+b)$
Finding $3$ prime numbers $a| (bc+b+c)$ $b|(ac+a+c)$ $c|(ab+a+b)$
4
$\begingroup$
number-theory
-
03 distinct prime numbers – 2012-06-30
1 Answers
12
There are no such primes:
Without loss of generality assume $a. 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 $1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}\leq\frac{4}{3}\frac{6}{5}\frac{8}{7}<2$