Questions mathématiques diverses

Question 55:
Justifier rigoureusement l'équivalence suivante :
$\left\{{\begin{aligned}&{x^2-y^2=a}\\&{2xy=b}\end{aligned}}\right.$ $\iff \left\{{\begin{aligned}&{x^2-y^2=a}\\&{x^2+y^2=\sqrt{a^2+b^2}}\\&{(xy)b\geqslant 0}\end{aligned}}\right.$

Réponse 55:
$\left\{{\begin{aligned}&{x^2-y^2=a}\\&{2xy=b}\end{aligned}}\right.$ $\Longrightarrow \left\{{\begin{aligned}&{x^2-y^2=a}\\&{(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2}\\&{2xy=b}\end{aligned}}\right.$ $\Longrightarrow \left\{{\begin{aligned}&{x^2-y^2=a}\\&{(x^2+y^2)^2=a^2+b^2}\\&{(2xy)b=b^2}\end{aligned}}\right.$ $\Longrightarrow\left\{{\begin{aligned}&{x^2-y^2=a}\\&{x^2+y^2=\sqrt{a^2+b^2}}\\&{(xy)b\geqslant 0}\end{aligned}}\right.$
Réciproquement
$\Longrightarrow\left\{{\begin{aligned}&{x^2-y^2=a}\\&{x^2+y^2=\sqrt{a^2+b^2}}\\&{(xy)b\geqslant 0}\end{aligned}}\right.$
$\Longrightarrow \left\{{\begin{aligned}&{x^2-y^2=a}\\&{(2xy)^2=(x^2+y^2)^2-(x^2-y^2)^2=b^2}\\&{(xy)b\geqslant0}\end{aligned}}\right.$ $\Longrightarrow \left\{{\begin{aligned}&{x^2-y^2=a}\\&{2|xy|=|b|}\\&{(xy)b\geqslant0}\end{aligned}}\right.$ $\Longrightarrow \left\{{\begin{aligned}&{x^2-y^2=a}\\&{2xy=b}\end{aligned}}\right.$