Résoudre p=(q-n)(q+n)/n^2-2

$$p = \frac { ( q - n ) ( q + n ) } { n ^ { 2 } - 2 }$$

Calculer n (solution complexe)

$$\left\{\begin{matrix}n=-i\left(p+1\right)^{-\frac{1}{2}}\sqrt{-2p-q^{2}}\text{; }n=i\left(p+1\right)^{-\frac{1}{2}}\sqrt{-2p-q^{2}}\text{, }&\left(|arg(-i\sqrt{p+1})-arg(-\sqrt{-2p-2})|\geq \pi \text{ and }|arg(i\sqrt{p+1})-arg(-\sqrt{-2p-2})|\geq \pi \right)\text{ or }\left(p\neq -1\text{ and }q\neq \sqrt{2}\text{ and }q\neq -\sqrt{2}\right)\\n\in \mathrm{C}\setminus -\sqrt{2},\sqrt{2}\text{, }&\left(p=-1\text{ and }q=\sqrt{2}\right)\text{ or }\left(p=-1\text{ and }q=-\sqrt{2}\right)\end{matrix}\right,$$

Calculer p (solution complexe)

$$p=-\frac{n^{2}-q^{2}}{n^{2}-2}$$

$$n\neq -\sqrt{2}\text{ and }n\neq \sqrt{2}$$

Calculer n

$$\left\{\begin{matrix}n=\sqrt{\frac{2p+q^{2}}{p+1}}\text{; }n=-\sqrt{\frac{2p+q^{2}}{p+1}}\text{, }&|q|\neq \sqrt{2}\text{ and }\left(p<-1\text{ or }p\geq -\frac{q^{2}}{2}\right)\text{ and }\left(p>-1\text{ or }p\leq -\frac{q^{2}}{2}\right)\text{ and }p\neq -1\\n\in \mathrm{R}\setminus \sqrt{2},-\sqrt{2}\text{, }&p=-1\text{ and }|q|=\sqrt{2}\end{matrix}\right,$$

Calculer p

$$p=-\frac{n^{2}-q^{2}}{n^{2}-2}$$

$$|n|\neq \sqrt{2}$$