$p , q , r$ are prime numbers, where $ {p}\times{q}\times{r} = 19(p+q+r)$ and $ {p}\lt{q}\lt{r} $ Find the value of ${r}-{q}-{p}$

Question

This was a question I got in a Math Competition

More clearly it was :

$p , q , r$ are prime numbers, where $ {p}\times{q}\times{r} = 19(p+q+r)$ and $ {p}\lt{q}\lt{r} $ Find the value of ${r}-{q}-{p}$

It is obvious that either $p,q,r$ is equal to 19 since the given identity is a multiple of 19 and $p,q,r$ are primes. Then brute work (testing out the other 2 numbers one by one) gave me the answer which was $3$ and $11$ fits into the identity $ {3}\times{11}\times{19} = 19(3+11+19)$. Hence the answer to the question is $19-11-3 =5 $

Now what I want to know is how do we find these answers more rigorously or is this question is designed to be done by testing out the 2 other numbers ? My lack of knowledge in number theory has prevented me from trying to solve this mathematically/rigorously. I am also interested in a modified equation of this question

$p , q , r$ are prime numbers, where $ {p}\times{q}\times{r} = p(p+q+r)$ and $ {p}\lt{q}\lt{r} $

What are the values of $q$ and $r$ if given some arbitrary prime $p$ ?

Edit: To the second question or an additional question. What I meant was

$p , q , r$ are prime numbers, where $ {p}\times{q}\times{r} = r(p+q+r)$ and $ {p}\lt{q}\lt{r} $

What are the values of $q$ and $p$ if given some arbitrary prime $r$ ?

Answer 1

Let us rearrange $p,q,r$ without any particular order into $a,b,c$.

Assume that $a=19$. Then we have to solve $$bc-b-c=19 \implies (b-1)(c-1)=20$$ So $(b,c)=(2, 21), (3, 11), (5, 6), (6,5), (11,3), (21, 2)$.

As $b$ and $c$ are both prime, we have $\{a,b,c\}=\{19,3,11\}=\{p,q,r\}$. Since $p

As for $$pqr=p(p+q+r) \implies (q-1)(r-1)=p+1$$ However, $q-1 \ge p, r-1 \ge p$ from $r>p, q>p$. So $$(q-1)(r-1)=p+1 \ge p^2$$ However, no prime $p$ satisfy this inequality. So there are no such $p,q,r$.

Your first question only yielded solutions because the right hand side had contained $\color{red}r$, not $\color{blue}p$. So, the question may have solutions if you are trying to solve $$pqr=r(p+q+r)$$ In which case the problem becomes $$(p-1)(q-1)=r+1$$ Solutions include $(3,5,7), (3,7,11), (3,11,19) , \dots$. There are probably a infinite amount of solutions, but I cannot prove it.

Answer 2

Hint: one of the primes must be $19$.

Now lets say that $p = 19$, then $qr = 19 + q + r \implies (q-1)(r-1)=20$

There are only finitely many choices to factorize $20$ and you can check explicitly which pair satisfies the condition.