Final Round

Answer: 2014. By condition, $r{p}^3=-(p^2+p)+2r{q}^2+(q^2+q)$, and $p^2+p$, $q^2+q$, $2r{q}^2$ are even for all natural $p$ and $q$, then $r{p}^3$ is also even. Therefore, since $r$ and $p$ are prime, we see that either $p=2$ or $r=2$.

If $p=2$, then the initial equality has the form $8r+4+2=2r{q}^2+q^2+q$. But this is impossible because the right-hand side of this expression is greater than its left-hand side. Indeed, since $p=2$ and $q\ne p$ is prime, we have $q>2$, so $2r{q}^2+q^2+q>8r+4+2$.

Thus $r=2$ and the initial equality has the form $2p^3+p^2+p=4q^2+q^2+q$, or $$p(2p^2+p+1)=q(5q+1).(1)$$ Since $p$ and $q$ are distinct prime numbers, we see that $p$ and $q$ are coprime. Then from (1) it follows that $2p^2+p+1$ is divisible by $q$, and $5q+1$ is divisible by $p$. Therefore $2p^2+p+1=mq$ and $5q+1=mp$ for some $m\in \mathbb{N}$. We put $q=(mp-1)/5$ into the former equality, then $$10p^2+(5-m^2)p+(m+5)=0.(2)$$ Consider this equality as a quadratic equation with respect to $p$. Since the coefficients of (2) are integer and its root $p$ is integer, we obtain that the discriminant of (2) is necessarily a perfect square, i.e. $$D=(m^2-5{)}^2-4\cdot 10\cdot (m+5)=m^4-10m^2-40m-175=n^2(3)$$ for some nonnegative integer $n$. Since $D<m^4-175$, we see that for $m=1,2,3$ the discriminant is negative. Moreover, for $m=4$ we have $D=256-160-160-175<0$. Therefore $m\ge 5$.

From (3) it follows that $(m^2-5{)}^2>n^2$. Show that $n^2>(m^2-11{)}^2$. If not, then from (3) it follows that $$(m^2-5{)}^2-40(m+5)\le (m^2-11{)}^2\iff 3m^2-10m-74\le 0.(4)$$ It is easy to see that the latter inequality holds only for $m\le 6$. But $m\ge 5$, so to prove that $n^2>(m^2-11{)}^2$ it remains to show that for $m=5$ and $m=6$ equality (2) does not hold for prime $p$.

If $m=5$, then (2) has the form $10p^2-20p+10=0$ and has exactly one root $p=1$; but $1$ is not prime number. If $m=6$, then the discriminant of (2) is equal to $D=1296-360-240-175=521$; but $521$ is not perfect square.

Therefore $(m^2-11{)}^2<n^2<(m^2-5{)}^2$, i.e., taking into account that $m\ge 7$ and $n\ge 0$, we have $m^2-11<n<m^2-5$. Note that from (3) it follows that $m$ and $n$ have different parity, so $n$ can admit only two values $n=m^2-9$ or $n=m^2-7$.

If $n=m^2-9$, then from (3) we obtain $m^4-10m^2-40m-175=(m^2-9{)}^2$ or $m^2-5m-32=0$. However, it is easy to see that the discriminant of this equation is equal to $153$, so the equation has no integer solutions.

If $n=m^2-7$, then from (3) we get $m^4-10m^2-40m-175=(m^2-7{)}^2$ or $m^2-10m-56=0$. We see that $m=-4$ and $m=14$ are the roots of this equation. Since $m\in \mathbb{N}$, we have $m=14$.

Therefore $m=14$, then $n=m^2-7=189$. From (2) we obtain $p=(m^2-5\pm n)/20=(191\pm 189)/20$. Since $p$ is integer, we have $p=19$ which is prime. Hence $q=(mp-1)/5=(14\cdot 19-1)/5=265/5=53$.

Thus the required value of the product is $pqr=19\cdot 53\cdot 2=2014$.