Find three prime numbers

Question

I have to find three prime numbers where, the product of them is equal to seven times of sum of them.

So I wrote the equation:

$x$-1st of them;

$y$-2nd;

$z$-3rd;

$xyz=7(x+y+z)$

${xyz\over 7}=x+y+z$

Where $x,y,z$ are prime numbers, so one of them must be $7$.

For example $x$

$yz=(7+y+z)$

Now I have to find these two numbers

I have found $5$ and $3$

$3*5*7=7(15)$

Which is correct.

My question is; Are there any other prime numbers, which meet the equation?

Answer 1

You have concluded that we may assume that $pq=7+p+q$ and $p\le q$ for primes $p,q$. For $p\ge 7$ we see that $pq>p+q+7$. So it is enough to check the cases $p=2,3,5$. We see that the only solution is $p=3$ and $q=5$, up to permutation of $p$ and $q$.