How to find all integer solutions of $p^2+q^2=((2q+1)^2+q+1)^2+1$

Question

$$p^2+q^2=((2q+1)^2+q+1)^2+1$$ 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?

Answer 1

$$\begin{align}p^2+q^2&=((2q+1)^2+q+1)^2+1\\\implies p^2+q^2&=16q^4+40q^3+41q^2+20q+5 \\\implies p^2&=16q^4+40q^3+40q^2+20q+5\end{align}$$

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.

Answer 2

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.