If $p$, $q$ and $r$ are prime numbers such that their product is $19$ times their sum, find $p^2$ + $q^2$ + $r^2$.
I came across this question in a Math Olympiad Competition and had no idea how to do it. Can anyone help?
One of the primes must be $19$, so WLOG $r=19$. Then $(p-1)(q-1)=20$. There aren't too many ways to factorise $20$...
Expanding on Ivan Loh's answer:
This expression can be written as: $pqr=19(p+q+r)$
As p, q, r are prime numbers, its product will just have p, q and r as prime divisors. Hence, one of them must be 19. Say $r=19$. Then, we have:
$pq19 = 19(p+q+19)$
$pq = p+q+19$
$pq-p-q = 19$ (*)
On the other hand, we always have the expression: $(p-1)(q-1) = pq-p-q+1$
which is equivalent to:
$(p-1)(q-1)-1 = pq-p-q$
Joining with (*)
$(p-1)(q-1)-1 = 19$
$(p-1)(q-1) = 20$
20 can be factorised in: $20=2*2*5$
in blocks of two:
$20=4 * 5$
$20=2 * 10$
$20=1 * 20$
so we have to options:
$p-1=4 -> p=5$
$q-1=5 -> q=6$
This cannot happen, because $p$, $q$, $r$ have to be prime and $p=6$ is not.
$p-1=1 -> p=2$
$q-1=20 -> q=21$
This cannot happen, because $p$, $q$, $r$ have to be prime and $q=21$ is not.
$p-1=2 -> p=3$
$q-1=10 -> q=11$
Both $p=3$ and $q=11$ are prime numbers, so we can go ahead. In such case,
$p^2 + q^2 + r^2 = 3^2 + 11^2 + 19^2 = 9 + 121 + 361 = 491$