Find all pairs of prime numbers $p, q$ such that $p+q = 18(p−q)$. Find all pairs of prime numbers $p, q$ such that $p+q = 18(p−q)$.
It is clear that $p-q$ must be an even number since if we consider $q$ as $2$, we won't get any solution. So any pair of odd prime does the work. I got $p=19$ and $q=17$ as one pair, considering the fact that $p+q$ must be a multiple of $18$. So considering numbers $18\cdot 2$, $18\cdot4$ , $18\cdot6$ ... as $p+q$ and then testing whether such pair of odd prime exist or not. 
Is this the right approach since I am stuck up?

$p+q = 18p - 18q$
$17p = 19q$
Therefore $p = 19n$ and $q=17n$. 
Then $p$ and $q$ are prime only when $n=1$.
Hence the only solution in primes is $(p,q) = (19,17)$
