I'm having a problem interpreting and starting this problem with primes.

Question

Find all triples $(p, q, r$) of primes such that $$pq + qr + rp + 34 \max(p, q, r) = 2018.$$

1) Do p, q, and r all have to be distinct?
2) I know that all of them must be less than 60, but I don't know how to solve it systematically.

Answer 1

I don't think there are such primes if they cannot repeat.

Namely, suppose at most one of $p,q,r$ is $2$ (which is in particular true if $p,q,r$ are all different). Thus, $pq,qr,pr$ are either all three odd, or one is odd and two are even. In any case, the sum on the left side is odd and, on the right side, we have 2018, which is even.

The only other option is that at least two of $p,q,r$ are equal to $2$. Without loss of generality, let $p=q=2$ and $r\ge 2$. This gives us:

$$4+2r+2r+34r=2018$$

i.e. $38r=2014$, i.e. $r=53$, which is prime. It follows that the only solutions, even allowing repetition, are the permutations of $(2,2,53)$.

Answer 2

Since $34 max(p,q,r)$ is even, it means that $p q + q r + r p$ must be even since $2018$ is even. Now, all three numbers of $p q, q r, r p$ cannot be odd, thus at least one is even, i.e. say $p=2$, then $p q$ and $r p$ are even. Therefore another number has to be even, say $q=2$.
Therefore we have: $$4 + 2 r + 2 r +34 r =2018 $$ Thus $38r=2014$ therefore we have $r=53$. Finally, all permutations of the triple $(2,2,53)$ are solutions.