Let $D>1$ be a non-square-integer and assume that $$x^2-Dy^2=k$$ with integer $k$ has a solution with integers $x,y$.
How can I find the complete set of fundamental solutions , if I know one solution ?
I tried to use that , assuming that $a^2-Db^2=1$, with the solution $(u/v)$ , the pair $$(au-bvD,av-bu)$$ is a solution as well. If a solution $(x/y)$ is known, can I calculate the fundamental solutions with this approach, or do I need more ?
When $k$ is one of $1,-1,p,-p$ for $p$ a prime number, your idea is enough. For prime, you just need to apply your idea to both $(x,y)$ and $(x,-y).$
As the number of prime factors of $k$ increases, more is needed. The reliable method is Conway's Topograph. For the special case of Pell type, we can predict inequalities. All solutions of $x^2 - 2 y^2 = 84847$ with both $x,y > 0$ can be constructed from the first sixteen solutions below by repetitions of the mapping $$ (x,y) \mapsto (3x+4y, 2x+3y) \; .$$ That is to say, the first 16 solutions below all have either $3x-4y \leq 0$ or $-2x+3y \leq 0.$ For these, since $84847 > 0,$ it is always the second inequality, which can be written $y \leq \frac{2}{3} x,$ or $v \leq \frac{2}{3} w$ using the letters in the output. When both numbers are large, $w - v \sqrt 2 = \frac{84847}{w + v \sqrt 2}$ tells us that $v \approx \frac{w}{\sqrt 2} \approx 0.7071 \; w \; ,$ therefore $v$ becomes larger than $\frac{2}{3} w \; $ as both numbers increase.
Here is a picture, I put a much lower target number, $x^2 - 2 y^2 = 17$ and $y \leq \frac{2}{3} x$ Shows where the "seed" solutions lie with target 17.
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Why not, here is what happens when I negate the value of $k$
Another picture, this time $x^2 - 2 y^2 = -17$ and $x \leq \frac{4}{3}y$