How to find all integer solutions of $p^2+q^2=((2q+1)^2+q+1)^2+1$ $$p^2+q^2=((2q+1)^2+q+1)^2+1$$
How do I find integer solutions to this equation? I've already found $(p,q)=(11,1)$. How do I go about finding new ones? 
 A: For $q \geq 21,$
$$ (4q^2 + 5 q + 1)^2 < p^2 < (4q^2 + 5q + 2)^2  $$
because
$$ q \geq 21 \Longrightarrow 41 q^2 > 40 q^2 + 20 q + 5  $$
Therefore 
$$ (4q^2 + 5 q + 1) < p < (4q^2 + 5q + 2)  $$
and $p$ cannot be an integer.
Should be something similar for negative $q.$
Yes, 
$$ q \leq -3 \Longrightarrow 0 < 7 q^2 + 20 q + 5,  $$
so both sides of the inequality work. Also, we get the necessary $4 q^2 + 5 q> 0$ when $q \leq -3.$
Here is page 268 in Mordell's book, Diophantine Equations. He gives the few examples where all solutions can be found (but no inequalities apply) in the following pages.

A: $$\begin{align}p^2+q^2&=((2q+1)^2+q+1)^2+1\\\implies p^2+q^2&=16q^4+40q^3+41q^2+20q+5 \\\implies p^2&=16q^4+40q^3+40q^2+20q+5\end{align}$$
So for an integer $q$ if $16q^4+40q^3+40q^2+20q+5$ is a perfect square, then you will get integer solutions for your equation.


*

*$q=1\implies16q^4+40q^3+40q^2+20q+5=121=11^2$

*$q=-1\implies16q^4+40q^3+40q^2+20q+5=1=1^2$


So $(11,1)$ and $(1,-1)$ are solutions, others may exist.
