Triple Primes Equation

Relevant wiki: General Diophantine Equations - Problem Solving

We begin by considering $p,q,r \neq 2$ which means $p,q,r \equiv 1,3,5,7 \pmod{8}$ .

It is easy to verify that $r \equiv 1,3,5,7 \pmod{8} \implies r^3 \equiv 1,3,5,7 \pmod{8}$ and hence $p+r^3$ will leave an even residue $\pmod{8}$ as $\text{odd}+\text{odd}=\text{even}$ .

We have $q \equiv 1,3,5,7 \pmod{8} \implies q^2 \equiv 1 \pmod{8}$ . This means $p+q^2+r^3$ leaves an odd residue $\pmod{8}$ but this contradicts $2016 \equiv 0 \pmod{8}$ so there are no solutions in this case.

We now consider the cases where one or more of $p,q,r$ is equal to $2$ :

Case 1: $r=2 \implies p+q^2=2008$ .

Setting $q=3$ gives the solution $p=1999,q=3$ .

$q \neq 3 \implies q \equiv 1,2 \pmod{3} \implies q^2 \equiv 1 \pmod{3}$ but as $2008 \equiv 1 \pmod{3}$ this would mean $p \equiv 0 \pmod{3} \implies p=3$ but this yields no solution.

Case 2: $q=2 \implies p+r^3=2004$

This restricts $2 \leq r \leq 12$ giving us $5$ cases to check yielding $p=1669,r=7$ .

Case 3: $p=2 \implies q^2+r^3=2006$ .

Firstly we can check $q,r \neq 7$ . Now we have $q^2 \equiv 1,2,4 \pmod{7}$ and $r^3 \equiv 1,6 \pmod{7}$ . As we have $2006 \equiv 4 \pmod{7}$ it is quick to check none of these pairs of residues will yield a solution.

The only solutions are $(p,q,r)=(1999,3,2),(1669,2,7)$ so in total there are $\boxed{\boxed{2 \text{ solutions}}}$ .

We have $r^3\le 2016$ , yeilds $r\le12$ . So we consider 5 following cases:

Case 1 : $r=11$ , so $p+q^2=685$ . Hence $p$ or $q$ must be equal to $2$ .

If $p=2$ then $q^2=683$ , which is impossible.

If $q=2$ then $p=681=3\cdot227$ , which is not a prime.

Case 2 : $r=7$ , so $p+q^2=1673$ . Hence $p$ or $q$ must be equal to $2$ .

If $p=2$ then $q^2=1671$ , which is impossible.

If $q=2$ then $p=1669$ , which is a prime. We get $\boxed{(p,q,r)=(1669,2,7)}$ is a solution.

Case 3 : $r=5$ , so $p+q^2=1891$ . Hence $p$ or $q$ must be equal to $2$ .

If $p=2$ then $q^2=1889$ , which is impossible.

If $q=2$ then $p=1887=3\cdot629$ , which is not a prime.

Case 4 : $r=3$ , so $p+q^2=1989$ . Hence $p$ or $q$ must be equal to $2$ .

If $p=2$ then $q^2=1987$ , which is impossible.

If $q=2$ then $p=1985=5\cdot397$ , which is not a prime.

Case 5 : $r=2$ , so $p+q^2=2008$ .

If $q\ne 3$ then $q^2\equiv 1\ (\bmod{3})$ , yeilds $p\equiv 0\ (\bmod{3})$ . Hence, $p=3$ then $q^2=2005$ , which is impossible.

If $q=3$ then $p=1999$ , which is a prime. We get $\boxed{(p,q,r)=(1999,3,2)}$ is a solution.