Given that each of p and q is a prime number:
Determine all possible pairs (p,q) that satisfy this equation:
p(p4+p2+10q) = q(q2+3)
(In reply to
Analytic solution with computer verification by Larry)
"Therefore q is odd which means RHS is even."
"Which means p is even."
I'm not seeing the logical leap between these two statements.
The LHS can be even regardless if p is odd or even:
p^4+p^2 = p^2*(p^2+1), these are consecutive integers so this will be even regardless of the parity of p. And 10q is even trivially, so then the sum p^4+p^2+10q is always even.
re: Analytic solution with computer verification
