I want all the solutions to this Diophantine equation. p and q are integers.
$$13pq \pm 1 = p^2 + 21 q^2$$
Solutions always come in pairs. $p=2$ and $q=1$ is one solution and so necessarily $p=-2$ and $q=-1$ is also a solution. Note that p and q must either both be positive or both be negative.
$13 \cdot 2\cdot 1 = 26$
$2^2 + 21\times 1 ^2 = 4 + 21 =25$
$26 - 1 = 25$
I want to find four other pairs of solutions to this equation. If no other solutions are possible, I want a proof of the impossibility of any other solutions.
Can anyone help ?!
1- equation $13pq-1=p^2+21q^2$
We rewrite it as:
$p^2-13pq+21q^2+1=0$
We solve this equation for p:
$\Delta=(13q)^2-4(21q+1)=85q^2-4$
Let $85q^2-4 =k^2$, this is a Pell like equation which may have some solutions, for example for $q=1$ we have:
$p^2-13p+22=0 \rightarrow k=\Delta=81\rightarrow p=11, p=2$
You may find more solutions by searching or using brute force to find $\Delta$ as a perfect square.
2- Equation $13pq+1=p^2+21q^2$
We rewrite it as:
$p^2-13pq+21q^2-1=0$
$\Delta=(13q)^2-4(21q-1)=85q^2+4$
Now you have Pell like equation $85q^2+4=k^2$ to be solved. For example one solution is :
$q=9\rightarrow k=\Delta=83^2$ which gives $p=100$ and $p=17$.
You may find more solutions by solving the Pell like equation.