Question Number 50747 by behi83417@gmail.com last updated on 19/Dec/18
$$\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{a}},{a}\neq\mathrm{0} \\ $$$$\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{z}}^{\mathrm{2}} =\boldsymbol{{b}},{b}\neq\mathrm{0} \\ $$$$\boldsymbol{{z}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{c}},{c}\neq\mathrm{0} \\ $$$${solve}\:{for}\::{x},{y},{z}.\:\: \\ $$
Commented by behi83417@gmail.com last updated on 19/Dec/18
$${yes}.{sir}.{fixed}. \\ $$
Commented by ajfour last updated on 19/Dec/18
$${a}+{b}+{c}\:=\mathrm{0}\:\:\:\left({obviously}\right). \\ $$
Answered by mr W last updated on 20/Dec/18
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\left({b}+{c}\right) \\ $$$${if}\:\:{a}\neq−\left({b}+{c}\right)\:\Rightarrow{no}\:{solution} \\ $$$${if}\:\:{a}=−\left({b}+{c}\right)\:\Rightarrow{infinite}\:{solutions}: \\ $$$${x}={t} \\ $$$${y}=\pm\sqrt{{t}^{\mathrm{2}} −{a}} \\ $$$${z}=\pm\sqrt{{t}^{\mathrm{2}} +{c}} \\ $$