Finding a set of 3 positive prime numbers that satisfy a polynomial equation What is the set of positive primes p,q,r that satisfy the equation: $p^{4}+2p+q^{4}+q^{2}=r^{2}+4q^{3}+1$. It can be easily shown (using parity concepts) that either p or r must be 2, but it is not clear to me how to advance any further since after proper substitution for p or r did not find an appropriate factorization.   
 A: There are no solutions with $r=2$ because we get $p^4+2p+q^4-4q^3+q^2=5$ and the left side is always too large.  I find $p=2,q=5,r=13$ is a solution.  There are no more with $p=2$ and $q,r$ naturals, ignoring the requirement that $q,r$ be prime.  With $p=2$ we get $r^2=q^4-4q^3+q^2+19$  Once $q\gt 5$ , we have $(q-1)^4 \lt r^2 \lt ((q-1)^2+1)^2=q^4-4q^3+8q^2-8q+4$
A: We have
$$ r^2=p^4+2p+q^4-4q^3+q^2-1$$


*

*If $q\ge 5$, then $q^4-4q^3+q^2-1\ge q^3+q^2-1>0$ and hence $r^2>p^4\ge 16$, so $r\ge 5$.

*If $q=3$, then $r^2=p^4+2p-19$, so $r=2$ leads to $(p^3+2)p=23$, contradiction

*If $q=2$, then $r^2=p^4+2p-13$, so $r=2$ leads to $(p^3+2)p=17$, contradiction.


We conclude that $r\ne 2$, hence by your parity argument
$$ p=2.$$
The equation becomes
$$r^2=q^4-4q^3+q^2+19 $$
Again we check small values of $q$ separately: $q=2$ leads to $r^2=7$; $q=3$ leads to $r^2=1$,so in both cases no prime $r$.
We conclude $$q\ge 5.$$
Now
$$ r^2=q^4-4q^3+q^2+19<q^4-4q^3+2q^2+4q+1=(q^2-2q-1)^2.$$
On the other hand,
$$ (q^2-2q-2)^2=q^4-4q^3+8q+4=r^2-q^2+8q-15=r^2-(q-4)^2+1\le r^2$$
with equality only for $q=5$, which makes $r=13$. Thus the only valid solution is
$$p=2,\qquad q=5,\qquad r=13. $$
