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Question Number 189022 by mr W last updated on 10/Mar/23

if ((√(1+x^2 ))+x)((√(1+y^2 ))+y)=1 with x,y ∈R, find (x+y)^2 =?

$${if}\:\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+{x}\right)\left(\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+{y}\right)=\mathrm{1}\: \\ $$$${with}\:{x},{y}\:\in{R},\:{find}\:\left({x}+{y}\right)^{\mathrm{2}} =? \\ $$

Answered by Frix last updated on 11/Mar/23

Obviously y=−x ⇒ answer is 0 [(√(x^2 +1))+x=(1/( (√(x^2 +1))−x))]

$$\mathrm{Obviously}\:{y}=−{x}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$$$\left[\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}}\right] \\ $$

Commented by mr W last updated on 11/Mar/23

can we be sure that no other solutions exist?

$${can}\:{we}\:{be}\:{sure}\:{that}\:{no}\:{other}\:{solutions} \\ $$$${exist}? \\ $$

Answered by Frix last updated on 11/Mar/23

(√(x^2 +1))+x=(1/( (√(y^2 +1))+y)) (√(x^2 +1))+x=(√(y^2 +1))−y x+y=(√(y^2 +1))−(√(x^2 +1)) x^2 +2xy+y^2 =y^2 +1+x^2 +1−2(√(x^2 +1))(√(y^2 +1)) xy=1−(√(x^2 +1))(√(y^2 +1)) 1−xy=(√(x^2 +1))(√(y^2 +1)) 1−2xy+x^2 y^2 =x^2 y^2 +x^2 +y^2 +1 −2xy=x^2 +y^2 (x+y)^2 =0

$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}=\frac{\mathrm{1}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}+{y}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}=\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}−{y} \\ $$$${x}+{y}=\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} ={y}^{\mathrm{2}} +\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$${xy}=\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{1}−{xy}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{1}−\mathrm{2}{xy}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} ={x}^{\mathrm{2}} {y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1} \\ $$$$−\mathrm{2}{xy}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

Commented by mr W last updated on 11/Mar/23

nice solution, thanks!

$${nice}\:{solution},\:{thanks}! \\ $$

Answered by MJS_new last updated on 11/Mar/23

z=sinh u ∧y=sinh v e^u e^v =1 e^(u+v) =1 u+v=0 v=−u sinh (−u) =−sinh u y=−x

$${z}=\mathrm{sinh}\:{u}\:\wedge{y}=\mathrm{sinh}\:{v} \\ $$$$\mathrm{e}^{{u}} \mathrm{e}^{{v}} =\mathrm{1} \\ $$$$\mathrm{e}^{{u}+{v}} =\mathrm{1} \\ $$$${u}+{v}=\mathrm{0} \\ $$$${v}=−{u} \\ $$$$\mathrm{sinh}\:\left(−{u}\right)\:=−\mathrm{sinh}\:{u} \\ $$$${y}=−{x} \\ $$

Commented by mr W last updated on 11/Mar/23

nice solution, thanks!

$${nice}\:{solution},\:{thanks}! \\ $$

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