If a solution exists, how can I construct another solution with coprime integers?

Question

Suppose , $n$ is an odd positive integer and positive integers $a,b$ not being coprime satisfy the equation $$a^2-2b^2=n$$

Can I always find a solution of $$c^2-2d^2=n$$ with coprime integers $c,d$ ?

I know that I can construct infinite many solutions from one solution of this pell-like equation, but the problem is that the usual procedure gives integers not being coprime , if I start with a solution with integers not being coprime. What can I do ?

Answer 1

No. Consider $n=9$.

There are non-coprime solutions gotten by taking solutions to $x^2-2y^2=1$ and multiplying them by $3$.

Suppose $c^2-2d^2 = 9$. Mod 3, $d^2 = 0$ or $1$. Therefore $2d^2 = 0$ or $2$, so $c^2 = 0$ or $2$. This can only hold if $c^2 = 0$, so $c = 0$ mod 3. Therefore $d = 0$ mod 3 so $c$ and $d$ are not relatively prime.

Answer 2

Knowing the first solution $(x_0 ; y_0)$ for the equation. $$x^2-qy^2=n$$

It is necessary to use any solution of the Pell equation. $$p^2-qs^2=1$$

Then knowing what some solution $(x_0 ; y_0)$ we can find the following for equation $x^2-qy^2=n$

$$x=px_0+qsy_0$$

$$y=sx_0+py_0$$